Sep 08, 21
On WoV, GunBj wonders: Can anyone please explain to me what happens "inside" a slot machine when an operator decides to change the RTP? Does the paytable change? Do the probabilities of the combinations change?
If I'm not wrong I have seen par sheets with "set" RTPs where it's either 96% or 94% or 85% (random numbers) depending on the particular probability/paytable combination, but what if one wanted to have some custom 92.76% RTP? Is this possible or are slots only settable to these preset RTPs?
When some websites state the RTP range of a certain slot (for example 92%-96%) is it just gathering sources from online casinos or does it mean that it's impossible to find anywhere in the world that same slot with an RTP lower than 92%?
What about Cleopatra here? Does this analysis only pertain to this specific version found at the specific site or is this for all Cleopatras in the world?
It depends on the specific machine, but is often the probabilities that change. Other things that can change are the following:
- Free Games Reels (which would be the probabilities, essentially, only during Free Games which would cause them to return better or worse)
- Progressive Meters-if applicable. (This is the amount of each bet that increases the progressive(s).)
In theory, a slot could be designed to be capable of more or less specificity, but generally speaking, a given slot title is only going to have a handful of RTP settings.
On some occasions, slot games might show an RTP range if the amount that the player is betting is a factor. A player can sometimes be eligible for certain jackpots or features on greater amounts bet than lower ones, or maximum lines played as opposed to fewer than maximum. Otherwise, it could just be that the probabilities of certain events are different with higher bets.
It's the same thing with Video Keno. You might provide a range for all of the possible bet amounts and spots picked for a particular title---but the specific return of the game would depend on how many spots the player is playing and/or what bet amount the player is playing. Denomination can also be a factor with different types of games.
Please note that this post consists of generalizations and is hardly comprehensive.
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Sep 08, 21
Choirboy asks: My local card room runs a 'High Hand' promo hourly. You must have aces full or better to qualify. In Holdem, both hole cards must play. Omaha 08 is also eligible for this promotion but must qualify using only the flop.
My question is: Which game has the better percentage of hitting a qualifying hand based on the above information (not considering number of hands played per hour or any other factors)?
Edit: All the games are all fixed limit or 2-100 spread limit. I'm looking for someone that can is skilled enough to give me an actual percentage (like 51.5% omaha vs. 48.5% holdem), we've discussed the topic to death at the tables and believe its close but would like know the numbers.
He was unsatisfied with the short and simple explanation of a certain concept, so he got a longer and slightly annoyed one:
You wouldn't. Why do you even care about analyzing this promotion if you are not willing to read the answers until you understand them?
THE YOUR HAND: 1, 2
COMMUNITY CARDS: 3, 4, 5, 6, 7
OMAHA 8 YOUR HAND: 1, 2, 3, 4
COMMUNITY FLOP: 5, 6, 7
Hands: 1, 2, 3, 4, 5, 6, 7 OR 1, 2, 3, 4, 5, 6, 7
A priori, these are seven random cards. WHERE THEY ARE DOESN'T MATTER INITIALLY!!! They are seven random cards. It is a random seven-card hand.
Now, we will look at a Full House. Trips and a pair. The probability of a dealt full house in ANY SEVEN CARDS is roughly 2.6% in frequency. That is for all Full Houses, not just the aces, but it's the same concept.
In one instance, I have two (underlined) cards that must be in the hand. In the other, I have three bolded cards that must be part of the Full House (or other).
That means, in both instances, two cards are NOT part of the hand.
THEREFORE, five cards out of seven are part of the completed hand, and two are not.
nCr(3,3)*nCr(4,2)/nCr(7,5) = 0.2857142857142857
nCr(2,2)*nCr(5,3)/nCr(7,5) = 0.4761904761904762
Therefore, ASSUMING a completed Full House of any kind, the probability in Texas Hold 'Em that two of the cards in your pocket cards, all else being equal, will be part of that Full House is 47.62%, rounded. In the meantime, the probability that three specific cards will be part of the Full House, all else being equal, is 28.57%, rounded.***
THUS, the question then becomes one of opportunities to see more such hands. What are opportunities? Opportunities are hands that you do not fold.
THAT YOU HAVE FOUR CARDS IN YOUR POCKET COMPARED TO TWO MEANS NOTHING!!!! IT'S STILL SEVEN RANDOM CARDS!!!
So, how many flops do you see? Do you see so many more flops in Omaha8 that it offsets the fact that you need three cards to be part of the hand rather than two?
HERE IS THE MATH: Three is a bigger number than two. That's the math.
***This isn't exactly correct because there could be a Full House on the board AND a player could be holding a pocket pair in the same hand. Also, board such as AA699 with an Ace and Six in the players hand is a worse Full House than AAA99, unless it still plays for the purposes of this promotion.
Check out the thread if you want, we came up with 1.667x more likely in the Texas Holkd ‘Em Game.
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Sep 08, 21
WINO84, on WoV, asks: Dear Members,
I’m new at No Limit Hold’em and my question is, if someone is playing say a top pair of some kind and the opponent is on a draw either flush or straight, theory says that for the person on the draw, they need “adequate” pot-odds before it’s worthwhile to proceed on the turn or flop or what have you. My questions are, isn’t putting money into that pot as the person on a draw -EV at that point and so isn’t more money bet just putting more money under influence of -EV? Then, when there’s “enough” money in the pot, why would it be a “good” risk to be drawing after a certain point? When the person with the strong pair puts out a large bet to dissuade the person drawing, does it help the person with the pair? does it help the person on the draw or both? And why? My conceptual understanding needs some help. Thanks for all of your replies.
DISCLAIMER: I am not a great Texas Hold 'Em poker player, therefore, my explanation on this should be taken with a grain of salt. If anyone comes along who plays THE more regularly than I do, please feel free to make any corrections to my explanation, offer a better one or simply phrase mine in a better way. /DISCLAIMER
In the simplest terms, "Pot Odds," refer to what you stand to gain, or lose, relative to the draw that you are making, amount you have to bet and the amount in the pot. This is an expected value analysis that has to be done on the fly that many people won't be able to do in their heads.
Let's imagine a 2-5 NLHE game that would have you protecting a big blind with a hand such as 9c7c, so here's the action:
We're going to say that the player before the small blind makes it $15 to see the flop, the small blind calls and you also call. Thus, there is now $45 in the pot.
FLOP: 6c Kc 2h
Okay, so the Small Blind checks post-flop, you throw out a $50 bet to see if you can get folds, the pre-flop raiser shoves all-in for $260 (total), the Small Blind folds...you have enough to cover the pre-flop raiser, so should you call?
For that, you want to look at your expectation.
PreFlop: Pot---$45 Yours: $15
Postflop: Pot---$355 (you bet $50 and the pre-flop raiser made it $260 plus the $45 from before) Yours: $65
If you fold, then you definitely lose $65, so you want your Expected Value for this hand to be better than losing $65 if you were to call.
The nine flush outs are easy and you know that the king on the board is a club. You are pretty sure he has top pair, but at least you don't have to worry about his king eventually giving him a better flush with running clubs. We're going to go ahead and say that we assume getting any flush will cause us to win the hand.*
*(By the way, I've also created a hand with runner-runner straight draw possibilities in it involving at least one eight. I did that so I could conveniently ignore that the player might draw to a better flush if his kicker is a higher club and the board comes running clubs.)
Okay, so we have two cards and there are three on the board. That leaves 47 cards that are unknown, except we are also very confident that the other player has the pair of Kings, especially since he shoved our bet that opened post-flop action...he could have a high inside pair, but probably would have bet more preflop had it been Aces or Kings, so with anything else and us betting, the king on the board should have scared a lower inside pair. The point is, we are acting as if we know he has a king and we know that King cannot be a club...therefore, there are effectively 46 cards that we don't know.
(37/46) * (36/45) = .64347826087 or 64.347826% that neither the turn nor the river will be a club. That means there is a .35652173913 probability that you will catch a club for a flush that will win the hand.
(By the way, one shortcut to approximating outs is to multiply your total number of outs by four on the flop and two on the turn)
Okay, so let's just use that for now. You have to call another $210 into a pot of $355 to make a total pot of $565 of which $275 is your money.
(.35652173913 * 290) - (.64347826087 * 275) = -73.5652173916
Okay, so with that your expectations are probably slightly worse than just folding and losing the $65 guaranteed. Of course, we did take your straights out of account and haven't been accounting for runner-runner hands with your hole cards that are generally going to be winning.
For example, I decided to use the WoO Texas Hold 'Em Calculator:
And I gave our opponent KhJs, and KJ offsuit seems like a reasonable hand for this behavior. It gives you a 38.89% probability of winning if that is the case:
(.3889 * 290) - (.6111 * 275) = -55.2715
Okay, so now as you can see you ARE expected to lose either way, but factoring in all of the ways that you could win, this is better than folding given the money that you have in there.
Granted, you would almost definitely fold to a shove here if you didn't have money that you put in after the flop, but that's okay, because the raise was a good play. The Small Blind player checked and, even though we knew the preflop raiser didn't have total garbage, it's more likely than not that he doesn't have a king in his hand and our bet would even give an inside pair pause if he didn't catch something on the board (66 in the hole, for example).
So, you made the right play*** but got stuck in a jam where you have to make one negative expectation decision or the other. In this case, with the hands you do win, you seem to do better by calling his shove.
***Maybe not the absolute, 'Right play,' but you made a defensible play. You should probably bet into this in this situation at least sometimes.
Of course, this is only one example.
K9 and K7 are slightly worse for you, but it's not as bad as you might think. The vast majority of your wins are flushes, anyway. Even though runner-runner clubs becomes less likely with a club in the other player's hand, an over club in his hand is a disaster for your prospects...but it's unlikely and there's no reason in particular to believe that he has a high club...his King can't be a club, for one thing.
Okay, so let's take this example as far as we can with your questions:
I’m new at No Limit Hold’em and my question is, if someone is playing say a top pair of some kind and the opponent is on a draw either flush or straight, theory says that for the person on the draw, they need “adequate” pot-odds before it’s worthwhile to proceed on the turn or flop or what have you. My questions are, isn’t putting money into that pot as the person on a draw -EV at that point and so isn’t more money bet just putting more money under influence of -EV?
Yes, but then what matters is whether or not your expected loss is more or less than it would be if you were to fold. In the example above, it is -EV to call, but it loses less than folding in the long run. This is assuming that we are talking about a cash game and I could list about 100 other qualifiers.
Sometimes the best decision is simply the, "Least worst."
(.35652173913 * 305) - (.64347826087 * 275) = -68.2173913046
So, with another pre-flop caller, now it's almost the same thing considering only the flush. Most of these pot odds things will be estimates unless you're a human calculator.
Also, in the long-term poker sense (assuming you'll be playing with these people again) it's good to showdown some draws. You don't want your competition to think that they can bluff you out of every pot just by repping high pair post-flop every time, or you're going to get pushed around a lot.
Then, when there’s “enough” money in the pot, why would it be a “good” risk to be drawing after a certain point?
Because you are expected to lose less one way than you are the other way, when it comes to average result. If you fold with $65 in the pot, then you always lose $65.
Let's pretend for a second that the Small Blind checks, but is actually slow-playing K6 (Two Pair) and hoping to get action, you raise the $50 (as before), the preflop raiser shoves (as before) the Small Blind calls.
Small Blind: Kd6d6cKc2h
Big Blind (You): 9c7c6cKc2h
PreFlop Raiser: Kh10c6cKc2h (Notice I am giving him an over club here)
You're still 30.9% to win this hand and now:
(.309 * 550) - (.691 * 275) = -20.075
Okay, so you're still losing money on this situation in the long run, but it's now MUCH better than folding for a guaranteed loss of $65. Let's say that neither of the other two players has an over club, so now the calculator says (I changed PFR from 10c to 10s) you are 36.77% to win:
(.3677 * 550) - (.6322 * 275) = 28.38
Hell, look at that! You take the over club out of the preflop raiser's hand and now this is profitable! You're expected to win money in the long run in that situation.
What's the probability that one or the other has an over club? I don't know. If they do, it's probably either a 10 or a Jack because Queens/Aces should bet more with a King. Jacks should probably also bet more. I don't think the small blind is likely to have an over club.
Also, for that player with the pair of Kings and 10c, it probably barely occurs to him that he can catch a running flush.
When the person with the strong pair puts out a large bet to dissuade the person drawing, does it help the person with the pair?
It does help him. The thing about this situation is, other than the fact that you don't know if you're going to get outdrawn, it's win-win for top pair.
1.) You fold and top pair takes the pot down without a fight.
2.) You call and top pair (at least, presumes he) is at an advantage.
Top pair will either take the guaranteed win or take the value on this one, but generally, top pair not only WANTS you to call, but he's hoping you're calling with some sort of draw rather than a made hand. Imagine he was playing K10 and you were playing 66 on the board in our example, he's DEFINITELY not happy to run into trips here!
If it sounds mutually beneficial that you call the flush draw (in most situations), it's not. The value calculations that we did above are the same as HIS value calculations for that example, except remove the (-) from the right side of (=) as appropriate. In other words, his expected value is better in our example hand if you fold. Your expected value is less bad if you call.
For me, if I'm playing you for a draw in this situation, then I'm happy to see you fold. If you bet into me and want to see your draw, then it's going to cost you as much as I can bet in order for you to get to do that.
Another thing that you have to understand about the draw, from the perspective of the player with the high pair, what's going to happen if you miss all draws? Suppose that I have the high pair and I DON'T bet post-flop...that's silly.
Why is that silly?
It's silly because I am not maximizing the value of my top pair whatsoever. When you look at that board, here are the things that I know:
1.) You absolutely can not have a flush yet.
2.) You absolutely can not have a straight yet.
Therefore, here are the only hands where you're beating me:
1.) Pair of Kings/Aces in the hole. (Almost definitely not based on you not re-raising preflop and I also have a king.)
2.) Pair of sixes or deuces in the hole. (I doubt it. It's unlikely, but I have to take my lumps when this happens.)
3.) Any two pair. (This is mostly unlikely. It's almost definitely not with a king, because I have a king. Two pair with 6-2? I seriously doubt that. You should know better than to call any preflop raise (even if you only have to put 2x more than what you have in there $15 total with $5 already in there) with 6-2 and only one person calling my raise).
4.) King with a better kicker. (This is probably the most likely way you're beating me...but supposing I raised with KJ or K10 off...I would think that you would re-raise me preflop with AK, KQ and maybe even KJ and definitely KJ suited. For that reason, while it's the most likely hand you have where I have the worst of it, I don't think it's terribly likely.)
So, I'm pretty confident that any money I push in there I am getting the best of it. That leads us to if I don't bet (assumes you checked to me):
If you don't bet, then one of two things can happen if I am somehow putting you on a draw and I check:
1.) You can hit the draw and I am screwed.
2.) You can fail to hit the draw and now I am much less likely to extract any money out of you.
In short, I'm betting my pair here and am betting at least enough to put a lot of pressure on you. That's if I don't shove outright, which I am probably doing.
does it help the person on the draw or both? And why?
The pair wants you to either fold or he wants you to bet with the worst of it. In the case of you betting $50 and him shoving in the example, he's mathematically better off (EV) if you fold, but has the best of it either way.
In the case of the example, you generally call your flush draw because it has a lower expected loss in the long run. Keep in mind you bet $50 post flop. If you checked and the other player shoved, then you would want to fold because the guaranteed loss of $15 becomes less bad based on him probably having at least top pair. (Although, if SB checks, you check and he shoves with SB calling the shove...you're probably back to calling unless you are supremely confident that one of those two shoved with a better flush draw than you have...that would be a hell of a read if you could put one of them on that. I sure as hell couldn't.)
I want to add some stuff about the pre-flop decision, since it's such a borderline case in the example hand.
With two callers, and you can verify using the WoO calculator, you're still getting pot odds only having to call for another $10, rather than folding, against most hands. You're getting pot odds against most hands for the other two players that don't have a seven or a nine in them and the situation where you're not getting them (and no other player has a seven or a nine with an overcard) most frequently is another player having suited clubs with something better than a nine.
In fact, with the Raise to only $15, you're even getting pot odds in some instances where one of the other players has your nine covered!
Wired pairs more over nine for one of the other players would frequently result in you not having pot odds, but such a player would probably bet more than the initial raiser did and the small blind probably would not have just called. One of the aspects of the suited 97 situation out of the BB that you're looking at is that it's very likely that the other two players are betting with either one high card, both high cards or a pair lower than nines.
For example, if you have suited 97, another player has QJ (suited or not) and the third player has 66 in the hole, you're actually more likely to take the pot down than the guy holding 66, unless one of his sixes is a club (remember, we haven't seen the flop yet) but even then you still have pot odds that suggest calling.
Obviously, that's for this specific situation. If the first player to bet Raises to $30, then you should always be folding in this situation...especially if the small blind calls. It can be pretty close, though, so if you're an exceptional post-flop player you could maybe mix some calls in here.
Obviously, if either of them shove pre-flop, you're always folding this.
Some people might consider that postflop raise that you did in the example questionable, so I am going to defend that.
1.) The Small Blind has already checked and can be assumed not to have a king. This will sometimes be wrong, but it happens.
2.) If both players missed the King on the flop, from their perspective, you're repping EITHER a king or a flush draw. The nice thing about being the Big Blind in this particular hand pre-flop is that the pre-flop raise was low enough that they can't really put you on a hand when you call. Would you ALWAYS re-raise a King here? Doubtful. But, with a hand like K8, or whatever, you'd definitely at least take a flop at that price.
So, now you've kind of flipped the script on them and they are deciding whether or not they want to call against a king you might have.
3.) If those hands missed the King and don't have any pairs or flush draws, they should usually be folding to your bet. Do you know that they could even have a pair of sixes or deuces with an overcard (to your nine) and as long as they don't have a better flush draw in the process you're still the odds on favorite to win the hand? You'd win it with a seven or a nine if they didn't hit anything else...not to mention your flush draw.
4.) You KNOW you don't have a king or a pair, so if you check, you're putting the original preflop raiser in a position to rep a king (that he might not have) and are in the same situation anyway. Obviously, if he bets or shoves, you're going to assume he has a king. Still, that's why you play these pot odds hands when we get back to the first post...not only do they have value compared to folding, but you don't want other players thinking they can just rep a king here and bluff you off of all of your four flush draws.
5.) If they don't have a king, pair or flush draw, then they are almost certainly not calling. You'll take the guaranteed money from them folding here because, why do you want to risk not hitting your flush and another high card shows up? You know that the other two players either have pairs or high cards, so any high cards that aren't clubs that come on the turn scare you...not to mention the fact that your hand sucks and the probability of you hitting a flush has dropped pretty dramatically.
If the preflop raiser has a pair or overcards, then he should definitely call or raise. Don't worry, you have the best of it against any pair that can be made using a card on the board (except kings), so you're only concerned about Kings, 2P, Trips or better flush draws if you get a call here...all of those hands should raise you anyway, except kings with a poor kicker might just call...but he shouldn't have a very poor kicker with a King since he raised preflop.
Also, trips aren't particularly likely because they would almost have to be sixes. I don't think someone is doing that sort of preflop raise with deuces.
So, with the small blind checking I like betting your four-flush in this situation.***
***Now, there are STILL going to be exceptions to this! I would say that the biggest exception would be if you know your pre-flop raiser won't make a terribly aggressive bet if you do check, or is very unlikely to be very aggressive (maybe likely just to check himself), then maybe you would check if you think you can see the Turn (and maybe catch your flush) cheap or no extra cost. You'd have to know that was a tendency of the other player, of course.
***What a tough game!!! That only covers the general concept of pot odds and looks at only a few example hands. YIKES! My advice is to be well-studied (moreso than I am) if you want to get your money involved with Texas Hold ‘Em, especially online, where your opponents can be using solvers, optimal decision charts, etc…
Believe it or not, that’s not even all that sophisticated an answer, by Texas Hold ‘Em standards!
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Sep 08, 21
Came from BCMarshall, who wanted to know: I know the conventional wisdom. Always split 8's. Yet instinctively and intuitively, I just don't see how splitting and getting two 18's vs. a dealer's 19 or 20 is a good deal.
I understand that splits/doubles present themselves as a result of initial split, yet I just have a very hard time wrapping my head around the advantages of risking so much more money, at least twice, and with doubles could be 3, 4, or even 5 x original bet, vs. just treating it as a 16 and taking either one win or one loss.
The videos posted, especially the Excel video, seems to imply that the calculations could be done rather efficiently by someone of Michael's ability.
Can someone please offer a numeric comparison between the odds of a simple hit vs. the split?
Thank you in advance.
Honestly, you HOPE that you end up betting more than 2x your initial bet, because that is a very good sign for you. Part of what makes the EV in favor of splitting (rather than hitting) is the fact that it results in sometimes winning more on the hands that you do win.
The reason that this happens is because the splitting now sometimes leads to situations that are +EV. You should want to get into a situation where you have to double. You split 8's against a nine or ten and one of your eights draws a three, so now you can double against a dealer nine, which is a great situation for the player.
However, the calculator even says it's a split with no DAS (double after split), eight decks and only being allowed to split once. It doesn't really get any worse than that for 8-8 and is still better than surrendering and much better than just taking a card. The only way hitting becomes better is those same rules against a ten and the dealer does not check for Blackjack, so you can't eliminate the dealer from having an ace, and in fact, the other rules don't matter and you would still only take one card if the dealer doesn't check (unless you could surrender).
So, how can it be better than hitting with no DAS? Let's break this one down with minimal math:
Hit 16: You have eight ranks that bust you and the hand ends immediately. You have five ranks (A-5) where you make a standing hand.
- For each eight, you have six ranks where you make a standing hand (9, 10, J, Q, K, A), so your prospects for at least making a hand that can win or push a playing dealer hand are immediately better.
- Each eight can draw a two or three, which even without DAS, gives the opportunity for a hit on those hands that literally can not bust and can lead to making hands that can win or push the dealer.
- With exception to another eight, all the other ranks (4-7) lead to you having the opportunity to hit to a different hand total (12-15) that is less likely to bust than your sixteen was.
One other thing you can use the calculator to do is look at a player Hard Eight v. Dealer Ten for a close enough comparison. If you double the negative EV on the Hard Eight hand's best decision (hitting, obviously) it still comes out to a lower Expected Loss than does hitting sixteen against a dealer ten. Actually, doubling the -EV on hitting the Hard Eight has a lower Expected Loss (slightly) than does surrendering a Hard 16, but again, that's because you cannot immediately bust hitting a Hard Eight and also that Hard Eight can turn into a good hand to draw to, such as eleven.
Why this comparison? Well, I think the ability to DAS makes the decision obvious, but even if you couldn't...two hands that are each Hard Eight have a lower combined expected loss than one hand that would see you either hit or stand on 8-8. If you can't DAS, then you're basically splitting 8's to make two separate hands that are both Hard Eight.
So, I hope this explanation helps without getting too far into the weeds. I also don't like Blackjack, so Blackjack discussion isn't really my strong point.
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Sep 08, 21
On WoV, BillRyan wants to know: Am I doing this wrong?
The next card has a 4/13ths chance of being a Ten. As does the next one.
4/13 +4/13= 8/13.
At a zero count, aren't the chances much higher than 50-50?
Eight Decks: 416 Cards
Player 8-8: (Two Non-Ten Cards)
We will say the dealer's upcard hasn't come out yet. There are (16 * 8 = 128) ten value cards remaining of 414 total remaining cards. 414-128 = 286
nCr(286,3)*nCr(128,0)/nCr(414,3) = 32.8611% (Rounded)---Zero Dealer Tens
nCr(286,2)*nCr(128,1)/nCr(414,3) = 44.4319% (Rounded)---One Dealer Ten
nCr(286,1)*nCr(128,2)/nCr(414,3) = 19.7995% (Rounded)---Two Dealer Tens
nCr(286,0)*nCr(128,3)/nCr(414,3) = 2.90761% (Rounded)---Three Dealer Tens, even though that can't happen.
However, by the time this information actually matters (when it comes time for the player to make a decision) the dealer's upcard either is already a ten or is not a ten. For that reason, you either remove the ten (or not a ten), as well as any other known player cards, and do the math accordingly.
For one example, let's say the dealer's card is a nine.
Eight Decks: 416 Cards
Player 8-8: (Two Non-Ten Cards)
Dealer Upcard: 9
nCr(285,2)*nCr(128,0)/nCr(413,2) = 0.4756811396600766 or 47.5681% that the next two cards are not tens. It will be slightly more likely neither are tens if a ten is the dealer's upcard.
Anyway, none of this is really pertinent to anything. If the dealer's upcard is a nine, then 8-A all end dealer actions and we do not even see the dealer draw any cards.
Let's go with the dealer showing nine and look at more precise numbers:
1.) The undercard is already a ten:
128/413 = 30.9927%
2.) The undercard is 8, 9, or A and the hand ends anyway (Two eights and a nine are gone already):
93/413 = 0.22518159806 or 22.5182%
3.) The hole card is 2, 3, 4, 5, 6 or 7 and the dealer takes (at least) one other card, which is a ten:
192/413 = 46.4891% (Overall)
Ten: (192/413 * 128/412) = 0.14443216812 or 14.4432%
Not Ten: (192/413 * 284/412) = 0.32045887303 or 32.0459%
Therefore, in an eight-deck game and with the dealer showing a nine, the probability of the dealer either having a ten undercard OR having an undercard that would require the dealer to draw AND drawing a ten on the next card is: .309927 + .144432 = .454359 or 45.4359% (Based on rounded numbers)
A dealer could also end up with a ten in a sequence such as 9 + 3 (Undercard) + 2 (Drawn Card) + 10...or any other number of ways that I am not going to go through.
I hope that helps. Either way, it's less than 50/50 mainly because the dealer could have (not a ten) and still not have to draw any cards.
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Sep 08, 21
On WoV, MikeSTL wonders: I full agree there is no system to overcome the house edge with a great enough sample. However, I can’t figure out what I’m missing in the following system for roulette. Any ideas?
Method: pick any two competing dozens and place equal bets. So let’s say $5 on first and second 12. If you win, repeat. If you lose martingale. But unlike some martingales you don’t automatically go back to the base bet. You go down one level since the martingale isn’t a win, rather a previous loss eraser (because of the hedge with the second dozen). So, example, lose at 5, bet 10, lose and bet 20, win and go back to 10, win and now you’re back to 5.
So the idea is you cancel out losses (as long as you can avoid losing it all with the martingale) and every win at the base level is one bet. So on American roulette with 100 average spins, you would win 63 and lose 37. Let’s say 37 of those wins have to be dedicated to canceling out the 37 losses at break even, so you are left with 26 units profit. Or say it’s a less lucky night and it’s 55-45. 45 of the wins give no profit due to cancelling so you have 10 units profit.
I know I must be missing some simple issue here, but I can’t figure out what it is. It seems like it would be the case you’d often have small to medium winning sessions and occasional total loss sessions when you lose at the top end of the martingale.
Bet two dozens at $5 each and lose: -$10
Bet two dozens at $10 each and lose: -$20 (-$30 total)
Bet two dozens at $20 each and win: +$20 (-$10 total)
So, it is what you think it is. What is it that you think you are missing? If the same dozen hits a bunch of times in a row, or hits an inordinate number of times within a relatively small number of spins, you're screwed.
"Small and medium winning sessions," is a very relative term. You would have various winning, "Sessions," that add up to less than your inevitable devastating losing session(s) would.
It doesn't seem like you're missing anything. Suppose your total bet could go:
10-20-40-80-160-320-640 (half of each bet on each dozen) and then the table limit would come into play.
(14/38)^7 = 0.00092132121
1/0.00092132121 = 1 in 1085.39778434
So, for every 1,085 (roughly) initial spins, you'll expect to have a loss of $1,270 without even winning any. Unfortunately, as you will see below, that's not your only problem.
From Random.org: (37's and 38's represent 0 and 00)
Let's look at this. We're going to pretend that we were betting the first two dozens. Here's how it went:
Okay, so this particular session would have had $145 in profits. That's on 65 winners and 35 losers (assuming I counted correctly), so pretty good session.
Let's do one more:
W-W $10 (2)
L-L-W-W-W $5 (5)
L-W-W $5 (7)
L-L-L-W-L-W-W-L-W-L-W-W-W $5 (14)
Six Wins $30 (20)
L-L-W-W-W $5 (23)
L-L-L-W-W-L-W-W-W $5 (28)
***L-W-L-L-W-W-L-W-L-L-L-L-L-W-L-L-L-W-W-L-W-W-W-W-W-W-L-W-W $5 (43)
W $5 (44)
L-L-W-W-W $5 (47)
W $5 (48)
L-W-L-W-W $5 (51)
W $5 (52)
L-W-L-W-L-W-W $5 (56)
After this set of 100 spins, you have one that is still unfinished and only made $100 in profits. That's only if we conveniently ignore the fact that you hit the Table Max that I originally proposed (see below). You could not have made another bet as you were seven levels behind, at one point.
The one we didn't finish went L-L-W-W-L-W-L-L-W-W-L-W-W.
I guess the, "Part you're missing," is that the results are not going to distribute perfectly because result distribution does not care to go out of its way to make betting systems work.
Every attempt that takes more than one spin starts with a loss, because it has to. At that point, whatever you're prepared to do bankroll-wise (and the house is prepared to let you do by table limits) is to get to the point where you win one more than you lose. The second session above was completely randomly selected and I did it this way so that I could illustrate to you why this system does not work.
The second session wasn't great in that you only had 59/100 winning spins, but is that really that unusual? 14/38 = 0.36842105263
The answer is: Not really. 41 (or more) losses with 14/38 ways to lose will happen more than 20% of the time in 100 spins.
***HERE is the other problem. You're screwed in this session. The losses don't have to be in a row because your Martingale does not cancel out all losses. In this particular series, you've effectively lost the seventh one in a row and the series would have ended.
Therefore, I conclude if you are missing anything, they are these two things:
- It doesn't take seven losses in a row for this system to fail. This system is not a Martingale, even, it's kind of like a Martingale/Labouchere hybrid.
- Even your winning sessions aren't going to be distributed such that you consistently get 26 units in profits.
But, you know the important part: That Roulette has a big house edge working against you and that you cannot expect any system to beat the game. I am very happy that your question was more, "Why does this system not work?", as opposed to actually thinking it might work.
I hope my post has helped.
And, my post did help! Yay! I love it when people accept math!
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Sep 08, 21
On WoV, Local 871 Asks: BetUS has a promotion on Thursdays where a Blackjack pays 9/5, instead of 3/2 or the dreaded 6/5. I've looked and can't find any information on what this does to the house edge. I know that 6/5 increases the house edge on average from 0.45% to 1.82%.
As 9/5 is better than 3/2, does anyone know how to compute the improved edge?
This is actually a simpler problem than you would think. The first thing that we want to do is look at this.
I'm just leaving the rules that are already in there the same. 6L5 has a House Edge of 1.78895% and 3:2 is 0.43096%, so:
1.78895% - 0.43096% = 1.35799% is the difference.
With that, we're going to take a trip back in time and concern ourselves with the Least Common Denominator, that way, we will index 6/5 and 3/2 to being the same thing. Let's go ahead and do that:
We start with the fact that our Least Common Denominator is 10, so that makes the adjusted fractions thus:
15/10 (3:2) and 12/10 (6/5)
The next thing we do is ask ourselves, "What would this adjusted fraction be for 9:5 Blackjack?" That's easy: It is 18/10.
We see that 1.35799% is the difference between 15/10 (3:2) and 12/10 (6:5) Blackjack, and relative to the Least Common Denominator of 10, the difference in our numerator (top number) is three.
The next thing we posit is that the probability of a winning player natural does not change, so any changes in the pays are unaffected by strategy.
Our 18:10 Blackjack goes three up in the numerator compared to the base game instead of three down, as a result, 1.35799% is subtracted from the House Edge.
0.43096 - 1.35799 = -0.92703 or 0.92703% Player Advantage
As we can see, it's the same thing as making it 6:5 Blackjack, just the other way.
- This method would NOT work for most strategy-dependent casino games. The only reason it works here is because the probability of a winning player natural cannot be changed by strategy. This would work for adjusting pays on side bets, for instance, provided there are no strategy changes that would make a particular result more or less likely. For most side bets, there are no such changes as many of them are just based on whatever it is you are dealt.
- But, I do hope that this was a useful example of how to use what you already know to get to what you want to figure out.
- You also know now that, having indexed it to ten, every 1/10th unit pay up or down changes it 1.35799/3 = 0.45266333333...at least for the rules linked above. (Only number of decks will change this)
Someone later would confirm this with some kind of software, which I don’t think is as fun.
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Sep 08, 21
On WoV, BigJ proposes a dice game with the following rules:
Thoughts or suggestions appreciated
The game is played on a blackjack table using a chuck a luck roller
Player needs to first qualify to play by rolling a 6 7 8
Player makes an ante wager and the dice are rolled
If player qualifies the dice are rolled again and player wins on 6 7 8
If player missies they can surrender or make another wager equal to the first.
If player missies again they can surrender both wagers or make a third bet
If player missies the 3rd try they loose half the original wager and the game starts over
House edge is .56 on roll once qualified
I went over and above and actually came up with several variations, which appear below, mostly because I didn’t quite follow what he wanted the first time:
Is the first part out of order? I assume the player makes an ante wager and then rolls to try to qualify, otherwise the player is just pointlessly rolling dice. Does this look like what you're going for?:
- Player makes an Ante wager.
- The player qualifies by rolling a 6, 7 or 8, otherwise the player loses immediately.
- Having qualified, the player must roll another 6, 7 or 8.
- If the player rolls an additional 6, 7 or 8 on the first try, the player wins and is paid the amount of original bet.
- If the player rolls anything but a 6, 7, or 8 after qualifying, then the player may surrender his original bet or make a wager equal to the first.
- If the player makes a second wager and rolls a 6, 7, or 8, then the player wins and is paid on both(?).
- If the player fails to roll a 6, 7, or 8 again, then the player may either surrender both wagers or make a third wager equal to the first bet.
- If the player makes a second additional wager and wins, then the player is paid on all wagers.
- If the player rolls anything but a 6, 7, or 8, then the player loses half of the original wager?
Okay, just to keep this simple, I'm going to make the original wager two units.
The player bets two units and can lose immediately by rolling anything other than a 6, 7 or 8:
(20/36) * (-2) = -1.11111111111
(16/36) * (16/36) * 2 = 0.39506172839 (Player wins two units on the roll after qualifying)
(16/36) * (20/36)---Game Continues. The player would never want to surrender.
(16/36) * (20/36) * (16/36) * 4 = 0.43895747599 (Qualify-Miss-Hit---Player wins four units)
(16/36) * (20/36) * (20/36)----Game Continues. The player would never want to surrender.
(16/36) * (20/36) * (20/36) * (16/36) * 6 = 0.36579789666 (Qualify-Miss-Miss-Hit---Player wins six units)
(16/36) * (20/36) * (20/36) * (20/36) * (-1) = -0.07620789513 (Player loses half of original two unit wager)
Sum of Probabilities: 0.55555555555 + 0.19753086419 + 0.10973936899 + 0.06096631611 + 0.07620789513 = 1
Expected Return: 0.39506172839 + 0.43895747599 + 0.36579789666 - 0.07620789513 - 1.11111111111 = 0.0124980948 (Two Units Bet)
PLAYER Advantage: 0.0124980948/2 = 0.0062490474
I'm getting a Player Advantage of .625%, so that makes me think I am misunderstanding something about the rules. This player advantage (the way I'm interpreting the rules) could be resolved by the player just losing the original bet on series: Qualify-Miss-Miss-Miss and the player, obviously, would still never surrender.
The Expected Return would then look like:
Expected Return: 0.39506172839 + 0.43895747599 + 0.36579789666 - (0.07620789513*2) - 1.11111111111 = -0.06370980033
Overall House Edge: 0.06370980033/2 = 0.03185490016 or 3.1855%
In order to determine the House Edge per roll, we must come up with an average number of rolls that assumes the player will play optimally by never surrendering.
Game Over One Roll: 0.55555555555 * 1
Game Over Two Rolls: 0.19753086419 * 2
Game Over Three Rolls: 0.10973936899 * 3
Game Over Four Rolls: (0.06096631611 + 0.07620789513) * 4
Average Number of Rolls: (0.55555555555 * 1) + (0.19753086419 * 2) + (0.10973936899 * 3) + ((0.06096631611 + 0.07620789513) * 4) = 1.82853223586
House Edge Per Roll: 0.03185490016/1.82853223586 = 0.01742102191 or 1.742%
The Element of Risk (EoR) for this would be based on average amount bet, which would be:
AVERAGE TOTAL BET:
(0.55555555555 + 0.19753086419) * 2 = 1.50617283948
(0.10973936899 * 4) = 0.43895747596
(0.06096631611 + 0.07620789513) * 6 = 0.82304526744
0.82304526744 + 0.43895747596 + 1.50617283948 = 2.76817558288
Element of Risk: 0.06370980033/2.76817558288 = 0.02301508644 or 2.30151%
I assume that I am misunderstanding something about the rules. I'm trying to figure out exactly what that might be, but I think it is one of two things:
A.) The player does not lose immediately if the player fails to qualify.
---If this is true, why bother even having the player need to qualify?
B.) The player loses HALF of the original wager AND any additional wagers that were made if the player misses three consecutive.
Anyway, we are coming up with something different for House Edge, so please clarify my understanding of the rules, if you would be kind enough.
THOUGHTS ON THE GAME
Whatever the House Edge, I must respectfully submit that I do not think this game would work for a number of reasons:
- In general, there appears to be no need for a dice game that might take the place of Craps. Other efforts in this vein have not gone very far.
- The Blackjack Table doesn't leave a heck of a lot of room for side bets, which you are certainly going to want to have for this game to have any chance.
- Unless there is a separate area for the rolling of the dice, one concern is that the dice might knock over chip stacks. Granted, that can also happen at the Craps Table.
- If the player does not immediately lose having failed to qualify, then I don't understand what the point of the qualifying process is.
I think I might have figured this out. The player loses additional wagers AND half of the original wager if Qualify + Miss-Miss-Miss. This assumes the player loses immediately if he fails to qualify.
(16/36) * (20/36) * (20/36) * (20/36) * (-5) = -0.38103947569 (Player Loses Extra Four Units Bet AND half of original wager)
0.39506172839 + 0.43895747599 + 0.36579789666 - 0.38103947569 - 1.11111111111 = -0.29233348576 (Original Bet of Two Units)
AVERAGE TOTAL BET:
(0.55555555555 + 0.19753086419) * 2 = 1.50617283948
(0.10973936899 * 4) = 0.43895747596
(0.06096631611 + 0.07620789513) * 6 = 0.82304526744
0.82304526744 + 0.43895747596 + 1.50617283948 = 2.76817558288
House Edge (Initial Wager): -0.29233348576/2 = -0.14616674288 or 14.61667%
Element of Risk (Based on Average Wager): -0.29233348576/2.76817558288 = -0.10560510957 or 10.5605%
Average Number of Rolls: 1.82853223586
Element of Risk (Per Roll) 0.10560510957/1.82853223586 = 0.05775403216 or 5.7754%
I'm now coming up with a House Edge way higher than yours. Can you please explain the rules of the game step-by-step and preferably with line breaks?
Fortunately, BigJ clarified some things for me:
A hard 6 or 8 pays 3 to 2 A 6 7 8 pays 6 to 5
Player needs a 6 7 8 to qualify Player has 3 tries to qualify before losing
Player makes an ante wager
1st try 6 7 8 hits the player qualifies and the dice are rolled again. Player wins on 6 7 8 losses all other numbers
1st try 6 7 8 missies player may surrender or add a second wager equal to the first to try again
2nd try missies player may surrender both wagers or make a third bet equal to the first and try again
3rd try missies player losses half of the original wager. The other two wagers are a push
Dice are contained in a cage
The qualifying round is a way to build.
Player makes a $5 (simplicity) ante wager.
At this point, the player either rolls a 6, 7, or 8 and has the opportunity to do it again, or the player fails to do that.
A.) The probability that the player rolls one of these numbers is .444444 and the probability that the player does not is .555556.
B.) If the player fails to roll a 6, 7 or 8, then the player may surrender the initial wager, but the player would never want to do that.
C.) If the player successfully rolls a 6, 7 or 8, then the player can do it again and get paid 6:5 for a soft 6, 8 or Any Seven and get paid 3:2 for a Hard Six or Eight.
D.) The player loses the $5 initial bet in the event that the player qualifies, but then fails to roll, a 6, 7 or 8:
(16/36) * (2/36) * 7.5 = 0.18518518518-----Probability: 0.02469135802
(16/36) * (14/36) * 6 = 1.03703703704-----Probability: 0.17283950617
(16/36) * (20/36) * (-5) = -1.23456790123----Probability: 0.24691358024
In order to be on this step, the player now has a total of $10 in bets on the table. The player has failed to qualify (20/36) and has decided to play again. The rules are otherwise the same as Step 1.
(20/36) * (16/36) * (2/36) * 15 = 0.20576131687-----Probability: 0.01371742112
(20/36) * (16/36) * (14/36) * 12 = 1.15226337449-----Probability: 0.09602194787
(20/36) * (16/36) * (20/36) * (-10) = -1.37174211248-----Probability: 0.13717421124
Another possibility is that the player fails to qualify for a second time, which would result in having to do it again.
In order to be on this step, the player now has a total of $15 in bets on the table. If the player fails to qualify for a third consecutive time, then the additional bets will be pushed back to the player AND the player will lose only half of the original $5 bet. The rules are otherwise the same as Step 1:
(20/36) * (20/36) * (20/36) * (-2.5) = -0.42866941015-----Probability: 0.17146776406
(20/36) * (20/36) * (16/36) * (2/36) * 22.5 = 0.17146776406-----Probability: 0.00762078951
(20/36) * (20/36) * (16/36) * (14/36) * 18 = 0.96021947873----Probability: 0.05334552659
(20/36) * (20/36) * (16/36) * (20/36) * (-15) = -1.14311842707-----Probability: 0.07620789513
Sum of Probabilities
0.02469135802+0.17283950617+0.24691358024+0.01371742112+0.09602194787+0.13717421124+0.17146776406+0.00762078951+0.05334552659+0.07620789513 = 1
Sum of Expected Values ($5 Base Bet)
(0.18518518518 + 1.03703703704 + 0.20576131687 + 1.15226337449 + 0.17146776406 + 0.96021947873) - (1.14311842707+0.42866941015+1.37174211248+1.23456790123) = -0.46616369456 ($5 Base Bet)
House Edge Based on Initial Bet
-0.46616369456/5 = -0.09323273891 or 9.3233%
Expected Number of Rolls
Two Rolls (Resolves on Step 1): (0.02469135802 + 0.17283950617 + 0.24691358024) * 2 = 0.88888888886
Three Rolls (Resolves on Step 2): (0.01371742112+0.09602194787+0.13717421124) * 3 = 0.74074074069
Three Rolls (Resolves on Step 3): 0.17146776406 * 3 = 0.51440329218
Four Rolls (Resolves on Step 3): (0.00762078951 + 0.05334552659 + 0.07620789513) * 4 = 0.54869684492
Expected Number of Rolls: 0.88888888886+0.74074074069+0.51440329218+0.54869684492 = 2.69272976665
House Edge Per Roll ($5 Base Bet)
0.09323273891/2.69272976665 = 0.03462387502 or 3.4623875%
Average Total Amount Bet
Resolves on Step 1: (0.02469135802 + 0.17283950617 + 0.24691358024) * 5 = 2.22222222215
Resolves on Step 2: (0.01371742112+0.09602194787+0.13717421124) * 10 = 2.4691358023
Resolves on Step 3: 0.17146776406 * 15 = 2.5720164609
Resolves on Step 3: (0.00762078951 + 0.05334552659 + 0.07620789513) * 15 = 2.05761316845
Average Total Amount Bet: 2.05761316845+2.5720164609+2.4691358023+2.22222222215 = 9.3209876538
Element of Risk (Expected Loss/Average Total Amount Bet)
-0.46616369456/9.3209876538 = -0.05001226392 or 5.0012264%
Element of Risk Per Roll (Element of Risk/Expected Number of Rolls)
0.05001226392/2.69272976665 = 0.01857307203 or 1.85731%
I'm still coming up with a higher house edge than you are, unless I am missing something else. Does a Hard Six or a Hard Eight on the first roll cause the player to win the 3:2 immediately without qualifying?
- I like this version better than what I thought you were saying before, but I still don't think there's any need to replace Craps with anything. I think Scossa might have had the most success in that regard, and that was something like two tables in one or two casinos.
- I have been made to understand that this would not be patentable without there being some sort of electronic component to it.
- The layout not having much room for side bets, as before.
- The thing about the dice knocking over chips, unless there is a walled-off rolling area.
ADDENDUM TO REDUCE THE HOUSE EDGE (INCREASE RTP):
By the way, since I was looking at it anyway, here is something that occurred to me:
The game would still have:
House Edge: 0.00749885688 or 0.75%
Element of Risk: 0.00402256561 or 0.4023%
If instead of losing half of initial bet on failing to qualify three times in a row, the rules instead made failing to qualify three times in a row a total push.
With the modified rules, that would probably make for a really fun home game that can be played without a House Edge (assumes no cheating) as long as players acted as, "The House," on a rotation and got to do so an equal number of times...that also assumes that all players bet equal dollar amounts.
I think the modified rules would also work as a fast-paced online game. The House Edge and EoR are both quite low with the modified rule that three non-qualifying rolls in a row is a total push, but the game would play pretty quickly, so that would make up for the extremely low House Edge.
As far as a side bet, the very rules of the bet prohibit it from being a one roll bet. It cannot resolve in less than two rolls. Technically, it can, but that would require the player to roll a non 6, 7 or 8 (20/36) and then surrender, which would be a terrible decision. Craps also already has a lot going on, so it would be tough to incorporate this on a physical Craps game.
I still think this is a great effort, overall, in coming up with a game. I think it's a theoretically viable online game and would make a great home game.
BONUS SIDE BET!!!:
(Quote above prior to OP edit)
Oh, you meant a single roll by which the player gets:
(14/36) * 6 = 2.33333333333
(2/36) * 7.5 = 0.41666666666
(20/36) * (-5) = -2.77777777778
2.33333333333 + 0.41666666666 - 2.77777777778 = -0.02777777779/5 = -0.00555555555 or 0.5555555556% House Edge
Now I see where that number is coming from! I couldn't figure out why my House Edge was nowhere near yours no matter what I did, but you were talking about two different bets---one single roll and one multi-roll.
Anyway, yes, I think your single-roll idea might work as a Craps side bet. It's kind of like a reverse Field Bet, in a way. I could maybe see where some players would play this and the Field simultaneously where:
2, 6, 7, 8 or 12 = Profit
3, 4, 9, 10, or 11 = Push (One wins, one loses)
5 = Loss
So, I think that's an excellent idea for a one roll side bet.
Yes, doing the same exact thing in my long post, but simply changing the $2.50 loss on three consecutive non-qualifying rolls to a Push would reduce the House Edge on the initial $5 bet to 0.75%. The Element-of-Risk is even lower because it is based on the average total amount bet.
ADDING A FOURTH ROUND!!!:
I'm not sure, but you can add additional rounds by doing exactly what I did in the long post.
I'll do the fourth round for you.
Instead of being a loss:
(20/36) * (20/36) * (20/36) = 0.17146776406
Becomes the probability that a fourth qualifying roll is needed, so if you wanted to stop at four, you would get:
(20/36)^ 4 = 0.09525986892 (Push after four rolls, or probability of Step 5, if you wanted to keep going)
(20/36) * (20/36) * (20/36) * (16/36) * (2/36) * 30 = 0.12701315856----Probability: 0.00423377195
(20/36) * (20/36) * (20/36) * (16/36) * (14/36) * 24 = 0.71127368795-----Probability: 0.02963640366
(20/36) * (20/36) * (20/36) * (16/36) * (20/36) * (-20) = -0.84675439042----Probability: 0.04233771952
Added Expected Loss
(0.12701315856+0.71127368795) - 0.84675439042 = -0.00846754391
Original Expected Loss (Except Three Non-Qualifiers No Longer Loses)
(0.18518518518 + 1.03703703704 + 0.20576131687 + 1.15226337449 + 0.17146776406 + 0.96021947873) - (1.14311842707+1.37174211248+1.23456790123) = -0.03749428441
New Expected Loss on Four Round Proposition
-0.03749428441 - 0.00846754391 = -0.04596182832
New House Edge ($5 Initial Bet)
-0.04596182832/5= -0.00919236566 or 0.91923657%
- You'll need at least one more round to get to over 1% House Edge.
- This would require an additional spot on the layout for a potential fourth bet, as would all other bets.
- As you continue, you will find that less and less gets added to the House Edge, because all added results become increasingly less likely.
***There should be plenty of good stuff on dice math for you readers out there that you can do a lot with to understand other games, or even create games of your own. I hope that becomes a game after all of that work I put into it! PHEW!
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Sep 08, 21
On WoV, Wizard says: I happen to have that book and read that part. My respect for Scarne went down a couple pegs after reading it. Had this casino called me for advice, I would have said roll out the red carpet for said player and keep him playing as long as you can. You'll eventually win it all back and more as long as he keeps playing.
There is no such thing as a "sophisticated" betting system. They are all equally worthless.
CONTEXT: The context was John Scarne advised a casino manager to tighten up the difference between minimum and maximum bets as a player was winning at Roulette using a Negative Progression system. The question posed to the forum was whether or not Scarne believed in the system, which I doubted, by saying:
I think many people tend to be a bit tough on Scarne. Basically, this is an extremely fundamental principle as we look back upon it with today's knowledge, but let's remember that the casino itself is not really there to gamble...they're there to let the House Edge play out. Scarne analyzed the guy's system and encouraged the house to increase the minimum bet (rather than reduce the maximum bet) thereby taking a step or two of the system away before it gets to max.
I don't think Scarne was saying that there's anything magical about Black not losing ten times in a row as opposed to losing eight or nine times in a row...I think he's just tightening up the guy's betting range to make the probability of winning any given system attempt a bit less so that the casino can get itself back to the good more quickly, by expectation.
If asked directly whether or not to let the player keep playing if the player refused the higher minimum, I like to think that Scarne would have said, "Yes, you're always expected to win eventually."
FULL DISCLOSURE: I’m a big Scarne guy and note that he didn’t specialize in any particular gambling study, but rather, looked at gambling and games, in general. He was also a card sharp and figured out a bunch of card game and carnival game gaffes, so he really did a bit of everything. As far as a guy who mostly had paper, pencils and basic calculators at his disposal (whereas we have computers and simulations now, as well as hi-tech calculators), I think Scarne is one of the best to have ever done it and definitely so for his time.
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Sep 08, 21
Wellbush asks, on WoV: So far, from what I've read, the only reason a negative progression strategy doesn't work in a card game, is because the player doesn't have a large enough bankroll to counter long losing streaks. Is this correct?
Before I post the conversation that came about as a result of this, let me give you guys some general thoughts on infinite bankroll theory that you will need to accept to even have this discussion---which is nonsensical in the first place:
1.) If we assume infinite bankroll and infinite players, then that means that EVERY possible mathematical result MUST happen, sooner or later, which means that one player loses infinitely. No matter how much the other players win (one unit at a time) the player that loses infinitely will cover those gains, and then some, for a net loss for all players that will align EXACTLY with the house edge.
Why exactly? The reason is that Variance is taken out of the equation, and Variance is the only reason that a player can win at -EV games (games with an RTP of less than 100%) even in the short-term...basically, it’s what makes gambling gambling.
If you had a 95% RTP game with ZERO Variance, all that would mean is that, whatever amount you bet, you get 95% of it back EVERY SINGLE PLAY.
In an infinite model, every possible sequence of results happens infinitely. While that may result in variance for any given individual player, it results in ZERO Variance for the collective, ultimately, since all possible results are happening at once and constantly. As a result, the net value of all players will always reflect the House Edge.
Of course, introducing a metaphysical concept, such as infinity, is ridiculous because:
2.) Why would you even gamble?
Think about it: If you had an, “Infinite bankroll,” to begin with, then that means that you have all possible money that does exist or can ever exist. Why? Because if you didn’t, then your amount would be quantifiable and would no longer be infinity.
Of course, that would make money without value, because you would be the only person who has any, so people would barter or trade something else for goods and services. Either that, or just agree to create a different currency.
But, most importantly, you can’t add to or subtract from infinity, so the entire concept applied to anything finite (and, therefore, physical) is a nonsensical abstraction designed for no purpose other than in a futile effort to invalidate math.
And, even if money still had value, which it wouldn’t, why would you even need to win more?
Now, here’s my answer:
Brandon James: If he says infinite bankroll, then I say infinite players. If infinite players, then every possibility occurs. One possibility is a player who infinitely loses. Most of our infinite players lose and win, one infinitely wins, but our infinite loser negates our infinite winner (and then some because our infinite winner only wins one unit at a time) and our other infinite players who mostly have an infinite amount of, "Normal," ups and downs will have lost an amount infinitely that equates to the house edge multiplied by the total amount bet.
Actually, the rest of the conversation was total nonsense that doesn’t bear repeating. It got worse from there. Feel free to follow the link and read it for yourself, but don’t say I didn’t warn you.
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