Maximum Aggregate Payouts (Gambling Math)
PREFACE---For those of you who might not gamble in land-based casinos too often, the concept of a Maximum Aggregate Payout is that the casino can set a limit as to the Maximum Amount of Winnings that it will pay on a single result.
The short version of the Maximum Aggregate Payout rules is that the casino must pay in full on Table Minimum bets, but beyond that, can set the Maximum Aggregate Payout to whatever they want to. Effectively, this increases the House Edge (reduces the Return to Player) because it results in the casino having the ability to pay out at less than the advertised odds.
On Wizard of Vegas, the following question was asked by Member, SeaBarrister, which I am going to shorten slightly:
SeaBarrister: Caesars seems to have placed a payout cap on its Mississippi Stud Tables of $50,000.
At a $5 or $10 ante this is a non-issue as even if you go against optimum strategy and raise to 3x on all rounds, you've put $100 on the table and the 500:1 Royal payout doesn't cross the $50K threshold.
However, at a $15 ante (and I have seen this as the table minimum) any Royal Flush should be played at Ante-1X-3X-3X or $120 at risk. The $50K cap would then cost the player $10K in payout.
Two questions: 1) (How) Does this affect the optimum strategy? and 2) how much worse does this make the house edge?
Does anyone care to critique my math or method? Also, it seems that this would stay consistent until the straight flush hits the cap which would be a $62.50 ante at optimum strategy (50,000/100 (payout of the SF) /8 (antes at 1x-3x-3x)).
As for the effect on house edge, and assuming that my analysis above is correct and that the cap has no effect on optimum strategy, the return for the royal drops from .006156 to 5/6 of that at $15 (100/120) or .005130 and increases the house edge from 4.9149% to 5.0175% (and this change worsens as the ante increases above $15).
Anyone care to take a stab and the impact of trying to modify the 2nd street bet to avoid the cap instead?
Mission146 ANSWERS: Excellent post!
This is a concept that is typically referred to as, "Maximum Aggregate Payout," and is often applied to the base payouts (as opposed to side bets with Progressives) for many Table Games.
Question 1: How Does this impact Optimal Strategy?
Answer: Don't let it. Simply don't bet such that you would go over the Maximum Aggregate Payout. If the Table Minimum is such that the winnings would otherwise be over the Maximum Aggregate Payout, don't play. If the casinos don't want to pay the paytable, then they should not accept the bet.
Question 2: How much worse does this make the House Edge?
Answer: The question is complicated. By necessity, the more you are betting, the worse the House Edge is because that's a greater proportion of the applicable payouts that you are NOT being paid, and thereby, are essentially playing against a reduced paytable. Secondly, if you were absolutely determined to bet such that the MAP could come into play, then there might perhaps be deviations from (otherwise Optimal) strategy which could themselves differ based on how much is being bet.
The easiest way to figure it out without changing strategy is just to look at the applicable wins according to the probabilities on Wizard of Odds, then adjust accordingly for the modified payouts. You'd only have to do this for one bet level as it would also tell you how much you are losing per each dollar bet over what the maximum would otherwise be.
As non Royal Flush payouts get added, then you would have to account for those as well.
Either way, the best thing to do is for everyone to simply NOT bet more than what the casino is willing to pay the paytable on. If they can't afford to pay the paytable, then they shouldn't allow you to bet that much.
MIDDLE: SeaBarrister wanted to see a mathematical analysis as to the effects of the Maximum Aggregate Payout, at this point, and I was only too happy to comply:
MISSION146 ANSWERS: Okay, here we go with doing it for the Royal Flush:
The first thing that we stipulate is that only one Royal Flush is normally possible and happens in the 1-3-3 betting configuration.
We will use a more precise probability than is available on the site because it becomes relevant when we are talking about the Maximum aggregate payout.
0.00000153907
Obviously, .000002 is not an acceptable roundup for the purposes of this specific question.
Okay, so this would have us betting a total of eight units. The Royal Flush pays 500-to-1, and we do not want payouts to exceed $50,000, in total, on the Royal Flush. With that, we have to look at 50000/8 = $6,250, which is the amount that we do not want to exceed in payout relative to unit bet.
6250/500 = $12.50, which tells us that $12.50, if the casino would even allow pinks, is the highest base amount we would want to bet.
For the purposes of a test case, we will use the base bet amounts of $15, $20 and $25 for proof of concept. Even at $25, we don't run into any Straight Flush problems, so this is fine.
If we take the probability of each event for $15, $20 and $25, then we have to determine what the new return would be and can compare it to the old return to see how much we are losing. 8 * 500 will always stay at $4,000, so let's see how much the bet pays normally and multiply that by the probability:
(4000 * 0.00000153907) = 0.00615628 (Agrees with Wizard)
Now, what we have to do is index this to the actual return multiplier for the various bet amounts. The way that we are going to do that is to take $50,000 (max cash return) and divide it by the amount being bet, so:
50000/(8*12.50) = 500
50000/(8*15) = 416.66667
50000/(8*20) = 312.5
50000/(8*25) = 250
This represents the actual amount that you are getting paid for each of these bets on a per unit basis.
Okay, so the next thing that we have to do is essentially see how this per unit return impacts the overall return:
(416.6667*8) * 0.00000153907 = 0.00513023374 ($15 bet)
(312.5*8) * 0.00000153907 = 0.003847675 ($20 bet)
(250*8) * 0.00000153907 = 0.00307814 ($25 bet)
Contrary to what I thought, it doesn't go down a fixed amount per additional dollar bet. I think the reason for that is that it is more important the percentage between the amount that you are betting and the maximum amount that you could bet that insures the full payout.
Change in return percentage per bet amount
0.00615628 - 0.00513023374 = 0.00102604626
0.00615628 - 0.003847675 = 0.002308605
0.00615628 - 0.00307814 = 0.00307814
At that point, you can express the change in return as a negative and subtract it from what would normally be the return of the game. We will do this for the $15 bet:
-0.049149 - 0.00102604626 = -0.05017504626
As we can see, the House Edge increases to a little more than 5%.
The second question is: At what point would you only bet 7 credits, instead of eight, on the Royal Flush? This would be by making the 1x bet the second bet around...not the third. Remember, you're only doing the 3x bet on 2nd Street BECAUSE OF THE ROYAL DRAW; you would not do this for just a three-flush, nor would you for almost half of converted 3SF draws.
The first thing that we want to determine is what the return would be if we were only betting seven credits in the first place, that's easy:
(3500 * 0.00000153907) = 0.005386745
Now, let's look at the difference in return:
0.00615628 - 0.005386745 = 0.000769535
Right, so we can see that we are already losing less than if we bet $15 (x8) with the $50,000 total payout cap. We know $12.50 is where it is exactly the maximum return, as far as bet amount goes, if they would let us bet that. With that, we will try $13:
50000/(8*13) = 480.769230769
(480.769230769*8) * 0.00000153907 = 0.0059195
Okay, so we see that this is more of a return than we would get from betting seven credits. Not included is the fact that we would also lose a few 1-3-3 straight flushes and flushes otherwise. Of course, we were never doing this for the flushes, anyway, though we sometimes do for the straight flush. In other words, Three-Flush without a 3SF or 3RF would be negative value anyway on its own and even is with some 3SF.
50000/(8*14) = 446.428571429 ($14 per unit)
(446.428571429*8) * 0.00000153907 = 0.00549667857
At $14 per unit, we appear to be slightly better off to still make the 3x raise. In fact, I'm going to say that you're probably going to be better off because there are also going to be some single and double gap straight flush draws that would normally be satisfied by these conditions. The only hand that would make a straight flush impossible is any hand that has an Ace in it. Even then, if 100x is good with conditions that would make a royal flush impossible, then it must be good if a Royal Flush is possible.
For that reason, some other raises are probably still going to be 3x raises, even if you're losing some value on the Royal.
Essentially, you can't look at the Royal in isolation when deciding the new 3x raise point, but everything that can possibly come from the 1-3-3 betting. We know that flushes don't matter, because you don't bet just three-to-a-flush, so the flush being there does not actually add any value.
We know the potential for High Cards, Trips and Two Pair do not add any value on their own because you would not raise something like J-Q-K, unless suited, which actually has the most potential for an unsuited hand of royal cards.
At a minimum, we know that we do not make the 3x bet if the Royal would represent a lower return than it would on a seven-unit return that pays 500 credits per unit. After that, it becomes about the return that you are getting from the straight flush potential, which as we know, often asks for a 3x bet on its own even with a Royal being impossible.
Thus, we would not make the 3x on 2nd if the Royal draw has an ace and the Royal returns less on its own than with seven credits bet. After that, it would depend on the bet amount and the particular 3RF draw in question.
WRAP-UP: In short, determining a new playing strategy would be extremely difficult and could only be done based on each possible individual bet amount. However, determining the maximum amount that could be bet is a trivial affair and you could do it for any game using the method above as long as you know the probabilities of the top pays.
Furthermore, you can also use the method above to determine how much in return percentage you are losing on bets that would cause the top pays to exceed the Maximum Aggregate Payouts. Have fun with it! Feel free to shoot me a Private Message if you have a different game you are interested in for this and need help!
Bingo game question (bingo game/math)
PREFACE---Bingo is not usually my strong suit, but fortunately, this question on WoV by Member Family Bingo had to do with setting up a Bingo game that could operate within certain time parameters.
FamilyBingo ASKS: Hi, I'm trying to set up a game of bingo for 21 players that will result in just one winner. Sounds simple right? Well, I want the play time to be 40 minutes to one hour. I'd also like to have some excitement in it (this is for a friendly family competition - competing for a very tiny but very coveted cup so I want the chance to generate some friendly competitive banter haha). I only know the very basics of Bingo but I'm pretty sure someone would get a single row in much less than 40 minutes. So I started thinking maybe the first person to win 3 rounds, but I have no idea how long that would take. Anyone have any suggestions on how I should set this up?? Please help, I'm open to all ideas.
MISSION146 ANSWERS:
- Balls are drawn one every 50 seconds. Until #5
- Each player gets five cards.
- The game is a coverall.
- When any player is only one away from a Bingo, they must verbally declare, "One Away!" This is true even if someone else has declared it or THAT player has declared it on a different card already.
- After, "One Away," is declared, balls are drawn one every 60 seconds.
- If multiple players get a confirmed Bingo, then the game will come down to their remaining four cards. Balls will be drawn every five seconds until one player or the other has a second Bingo, then a third if both get two, etc...There will be no more, "One Away," rule for this part.
MIDDLE: Family Bingo would inquire whether or not I was confident in my time estimate:
FamilyBingo: Thanks Mission 146, you think that will be about 40-60 minutes to get a winner? I'm really glad I asked because I would have guessed it would take much longer for a coverall. Having said that, we might end up faster, my family will be impatient for the next number so I'm not sure I'll be able to wait 50 or 60 seconds before calling another lol.
MISSION146 ANSWERS: I think that would be about the expectation. If everyone had identical cards, or one identical number that would cause all cards to simultaneously win AND it was the last called number (this will not happen) then 75 numbers would be drawn. At 50 seconds each number, this would take 3,750 seconds, which is 62.5 minutes, so we know we're almost certainly not going the full hour.
There are 2400 seconds in forty minutes, so if we divide that by 50 seconds (for each ball drawn), then that would reflect a winner in exactly 48 draws. The probability of a Bingo in that many cards, or fewer, based on 100 cards out (and you're only going to have 105):
Would be well under 1%.
Therefore, it is highly probable that the game will take between 40 minutes and one hour to conclude.
WRAP-UP: Given the time between draws required, my suggestion was to ask trivia questions in between the numbers drawn for small prizes. If you ever want to set up a Bingo game with certain time confines, then you can figure out the best way to do that using the same method I did!
Lottery question (lottery math)
PREFACE---On WoV, Member Seven asks an interesting lottery question in this thread which results in the following exchange:
Seven ASKS: hi @all
let's say a lottery participant has 10 tickets which costs him all in all $1 in a lottery game with 2100 tickets. There will be a draw with 2 winners who get $450 each.
What is the player's EV?
I am confused about the 2 winners of the draw
thank you as always in advance for your help
Cheers
(NOTE: Seven later clarifies that the same person cannot win twice)
MISSION146 ANSWERS: Okay, there are three possible combinations of events that can happen in this one. What we're going to do here is express them in the simplest terms:
- The first ticket drawn results in the guy winning: 10/2100 = 0.00476190476
- He does not win on the first ticket, but then he does win on the second ticket drawn: (2090/2100) * (10/2099) = 0.00474148687
- He does not win on either ticket drawn: (2090/2100) * (2089/2099) = 0.99049660836
Okay, so we know that he either wins $450 (which is really $440 profits because he paid for the tickets) or that he loses $10. With that, we have:
((0.00476190476+0.00474148687)*440) - (10 * 0.99049660836) = -5.7234737664
This is only slightly worse than what ThatDonGuy ended up with and the reason why is simply that the guy cannot win twice. It also assumes that the original $10 for the tickets is sunk (pays on a For One basis) even if he wins.
WRAP-UP: The thread goes on to discuss some other scenarios, which I won’t reproduce here, but you’re welcome to click the link above if you want to learn how to do it for other scenarios. I try to use step-by-step approaches to math that anyone can do for themselves.
What are the odds? (craps math)
On WoV, FatGeezus asks:
FatGeezus: I sat down to play the Bubble Craps game at the Tropicana AC casino. I noticed that the Hard 4 had not been thrown in quite a while. The Bubble Craps display board keeps track of the last time a Hardway 4, 6, 8, 10 was thrown. I kept watching to see how long the streak would go.
FINALLY. The last time a Hard 4 was thrown was 212 rolls ago. On the 213th roll a Hard 4 was finally thrown.
What are the odds of not throwing a Hard 4 - 212 times?
MISSION146 ANSWERS: (35/36)^212 = 0.00254856221
1/0.00254856221 = 1 in about 392.38
It's pretty unlikely, but unfortunately not quite as remarkable as one might think it should be. What would be remarkable is if the dice were to avoid every single hard way, Snake Eyes and Midnight for 212 consecutive rolls:
(30/36)^212 = 1.6352187e-17 or 0.000000000000000016352187
1/0.000000000000000016352187 = 1 in about 61,153,899,000,000,000
WRAP-UP: That one was a softball!
Regression Towards The Mean: (Gambling Math)
PREFACE: Wellbush (since banned) made a post on WoV here that I managed to parlay into an interesting commentary on the concept of, “Regression to the Mean,” which I will discuss a bit more after the quotes:
Wellbush Says: The Gambler's Falacy states that every fair play in a game of chance, is independent, and has no better or worse chance based on past play.
How pathetic that the math community believe this! So, the math community unequivocally state the percentages that will eventually occur in each game, for the house, and the gambler.
But how can this be true, due to the Gambler's Falacy???? Another paradox!!!!
How can math say on one hand, you cannot give predictions re chances, but on the other hand predict future chances down to percentage points? Is the math community moronic???
MISSION146 ANSWERS: I guess someone should give this an answer---just in case someone is actually interested in this question.
Okay, so let's give a quick example of Regression Towards the Mean.
Imagine if a Craps Player is betting the Pass Line and it's the first time in his entire life he has ever done so---that person wins (49.29% probability), so now that person has won 100% of his Pass Line bets.
Probability says that event has already happened and that the next attempt is independent and comes with probabilities 49.29% and 50.71%, winning and losing, respectively.
Imagine that our fortunate first-time player wins his next five bets in a row, so he now has six consecutive Pass Line winners: (.4929)^6 = 0.01434012407
Thus, he has experienced an event with 1.434% probability, but it is 100% since it has already happened.
For the next trial, his probability of winning remains 49.29%, which nobody disputes. However, let's suppose that he loses just that one attempt, so he goes from: Win 6, Lose 0---Win 100%, Lose 0%
To:
Win 6, Lose 1---Win 85.7143% (Rounded) Lose 14.2857% (Rounded)
Critically, his winning percentage can never be 100% again, which is a short-run example of how Regression Towards the Mean works in the long run. In other words, the Gambler's Fallacy is relatively focused on the notion that a short-run phenomenon (usually many consecutive losses) will not happen based on the fallacious position that, after a particular series of losing results---the next result(s) to be losing becomes less probable.
Regression Towards the Mean, on the other hand, describes what happens in the long run. Simulations bear this out, but more importantly, relatively long-term real world results will also bear this out. You can win 80% of your total bets on the Pass Line at Craps, for example, but not for particularly long.
Regression Towards the Mean also does not mean that there will be a series of six losing results in a row to offset the six winning results that happened before. However, in a long-term series of results...long winning AND losing streaks will not be uncommon.
Let's imagine that after our first-time Craps players' first loss the results go W-L for the next ten results, which makes him:
Win 11, Lose 6---Win 64.7059% (Rounded) Lose 35.2941% (Rounded)
Despite the fact that the player opened up with six consecutive wins and has never lost more than once consecutively, we see that his winning percentage is now down from 100% to 64.7059%.
I'm obviously cherry-picking short-term results here to try to do a short-term illustration of a long-term concept. The point is that every loss (when the actual winning percentage is more than 50%) will have a greater impact on the overall percentage than every win. Let's take our 11-6 and see what happens to the winning percentage if the next is a win and if the next is a loss:
Win (12-6): 66.6667% (Rounded)
Difference: 66.667-64.7059 = 1.9611%
Loss: (11-7) 61.1111% (Rounded)
Difference: 64.7059 - 61.1111 = 3.5948%
As you can see, Regression Towards the Mean doesn't require that a long run of winning results is offset by a long run of losing results---though you would expect to see long runs of both in a large sample anyway. Regression Towards the Mean relies on two components:
1.) In a large sample size, any long runs of results are mostly going to be offset by running relatively as expected for a very long time.
And:
2.) If a result is coming up either more, or less, than expected---then the result that is not yet running, "As Expected," will be moved more towards the mean for every time it DOES happen than it is moved away from the mean if it fails to happen again.
Another example is that of a Video Poker player who has never had a Royal Flush in his first 60,000 hands, which means he has run more than one, "Royal cycle," without getting it. He currently has 0.000000% Royal Flushes, but as soon as he gets one, then he is above 0.000000% and can never be at 0% again just by virtue of that one Royal Flush.
The Royal Flush will also pull his overall return percentage up on the game as significantly as any hand could, or more, with only a few exceptions such as Four
Deuces with a Joker on DJW (somewhat rare game) which actually pays more than a natural Royal.
Specifically, it will pull it up by 800x the bet, assuming full coins, so let's imagine that he has played 60,000 hands at an overall return of 96.9% before that Royal:
Credits Played: 300,000 Credits Won: 300,000 * .969 = 290,700
Okay, so let's imagine that he hits the Royal Flush and also what would have happened if the hand had paid nothing instead:
Credits Played: 300,005, Credits Won: 294,700---Actual Return Percentage: 98.232% (Rounded)
No royal---losing hand
Credits Played: 300,005, Credits Won: 290,700---Actual Return Percentage: 96.8984% (Rounded)
What you will notice here is that, as anyone would expect, the Royal has a much more profound impact on the player's overall return percentage than does having a losing hand, which barely moves the percentage. Another thing that you will notice (pretend it's JoB or Bonus Poker) is that a 10 credit return on Two Pair:
Credits Played: 300,005, Credits Won: 290,705---Actual Return Percentage: 96.9017% (Rounded)
Has a slightly more pronounced impact on the overall return percentage (simply because 0% is closer to 96.9% than 200% is) which we also expect to work itself out in the long-run.
Video Poker has more variance than Pass Line bets on Craps, so you might need a more significant sample size of hands to see Regression Towards the Mean fully play itself out. A player could be running a couple Royals, "To the good," or, "To the bad," and resulting,
could run well below (relatively speaking) or well-above, the Expected Return of the game for a very long time.
So, no, the math community is not moronic. The simple answer to this one is that the Gambler's Fallacy (particularly as relates negative expectation betting) is only concerned with short-term results (at least, in a given trial---perhaps some system players have trouble thinking long-term?) and the, "Math Community," primarily as it relates to concepts such as, "Regression Towards the Mean," is more concerned with the long-term.
Anyway, this has been a much longer answer than your post deserves, but there are many posters and readers here that do deserve an explanation of the difference, so there it is.
AND---It's for that reason that I defended something, such as the 18 YO's claim, as being theoretically possible. Do I think it happened exactly that way---probably not. If you're going to rely on the math, then it's very important to remain consistent---so nobody should say such a result is impossible. It has a non-zero probability, though not much above zero, but any specific string of 18 results has a very close to zero probability of occurrence (by virtue of the fact that so many combinations of events are possible), but the dice must do something.
In fact, the most likely string of 18 dice rolls, which is to say the highest single probability, is 18 sevens in a row:
(1/6)^18 = 9.8464004e-15
Which is to say 0.0000000000000098464004 or 1/0.0000000000000098464004 = 1 in 101,559,960,000,000
Virtually impossible, but technically the most likely specific sequence of 18 rolls of fair dice. If you go to a Craps Table and observe a series of eighteen rolls, then that specific series was (unless all 7's) less likely than the 1 in 101.56 TRILLION chance of them all being sevens.
Of course, if I come to the Forum and report that I observed a Craps Table with the following sequence:
7-7-5-2-3-9-5-7-11-12-3-7-8-4-4-5-9-7
Nobody would bat an eye, despite that specific series being substantially (well, relatively speaking) less likely than all 7's.
Conclusion
The, "Math Community," simply understands how numbers relate to each other and many betting system believers (and I am not saying you specifically) do not fully comprehend it. The math community's belief in a long-term concept does not debunk the math community's opinion of why gambling systems can fail in the short-term and would fail in the long-term on a negative expectation game.
The concept of a long streak does not change the concept of Regression to the Mean. Why not?
Going back to the Craps Table----let's take into consideration every Pass Line bet that has been made in all of history; I can't even estimate how many that might be: How much do you think a string of 30 consecutive Pass Line losses would change the overall winning percentage of all Pass Line bettors in history?
Answer: It wouldn't even be enough to qualify as a rounding error.
The probability of rolling an 11 on any individual trial certainly does not change; nor the probability of winning a Pass Line bet on any one trial.
ADDED COMMENTARY: This is one of those abstract mathematical concepts that sometimes give people trouble.
The notion of Regression to the Mean is that, over a long enough period of time, results will run close to the probabilities of particular events. The most critical aspect of Regression to the Mean is in a part of the quote I said above, which stated, “Critically, his winning percentage can never be 100% again, which is a short-run example of how Regression Towards the Mean works in the long run. In other words, the Gambler's Fallacy is relatively focused on the notion that a short-run phenomenon (usually many consecutive losses) will not happen based on the fallacious position that, after a particular series of losing results---the next result(s) to be losing becomes less probable.”
Not only can the player’s winning percentage never be 100% again, but it (obviously) can never be MORE than 100%. In other words, losing will reduce his win rate from 100% while winning cannot improve it.
The most critical element to understand about Regression to the Mean is that it is something that (depending on the game in question) can take millions of trials to resolve. In other words, an inordinately high winning streak is not necessarily immediately offset by an inordinately long losing streak, though when it comes to thousands (or more) trials, you begin to expect to see long streaks of both.
The difference between the mathematically valid concept of, “Regression to the Mean,” and the invalid concept of, “Gambler’s Fallacy,” is that Gambler’s Fallacy is structured on the notion that results are somehow not independent of one another. In other words, Gambler’s Fallacy believers think that the odds of winning or losing an individual trial go either up or down based upon the results of previous trials. Regression to the Mean already has it built into itself that all results are independent.
If you want to do this with a coin-flip to test it out, then you can. Here’s an easy way to look at it: One Flip: Either heads or tails---one wins 100%.
Five Flips: Usually a mix, but not always.
Ten Flips: Even more likely to be a mix!
100 Flips: Your coin is gaffed if these all land the same!
Anyway, you can flip a coin 100 times for yourself and log them in rows of five. Each individual flip is 100% heads or tails, but then look at your rows of five (the closest possible result is 60% for one and 40% for the other---assuming it’s not 50/50) and you will see that there are wider percentages ranges of heads or tails winning in some of those rows than your overall results.
You can do the same thing by looking at groups of ten and comparing it to the total results of 100 flips. It’s not particularly likely that 100 flips will land exactly fifty heads and exactly fifty tails, but that may happen, but at least one of your groups of ten will almost certainly deviate from that.
In a group of ten, assuming it’s not 50/50, your closest possible set of results is 60% for one and 40% for the other. What you will notice is...even if heads has been flipped more than tails in your set of 100...one of your groups of ten (if not more) will often have more tails than heads.
If you really want to flip a coin 1,000 times, then you could go even beyond that and start comparing groups of 100 to your overall set of 1,000 to see the differences.
The smaller your grouping; the more skewed your percentages can be. The larger your grouping...while in theory you could have 1,000 heads against 0 tails...that’s almost impossible and the percentages of each become more and more likely to fall into tighter and tighter ranges the larger your sample size.
With that, Gambler’s Fallacy thinks that an individual result will be influenced by a particular set of short-term results to precede it. Regression to the Mean doesn’t think that there’s any short-term predictive value whatsoever…and doesn’t EVER believe in exact predictive value...it just analyzes the particular ranges of results (known as standard deviations) in which it states what the ranges of outcomes are likely to be.
Change in RTP due to not changing strategy (video poker)
PREFACE: This was a fun one that I found a creative way to solve! I almost did something that would have taken me about ten hours, until I determined that there was a much easier way to go about it. Technics On WoV wanted to know what the RTP would be using the 9/6 Jacks or Better Video Poker strategy to play 9/5 Jacks or Better:
Technics ASKS: What is the difference in EV, if the 9-6 JOB simple strategy is used on a 9-5 JOB game, instead of using the full 9-5 JOB strategy?
MISSION146 ANSWERS: Are you ready? You're going to love this solution! Because I'm an idiot, I started doing this a much more difficult way.
What we are going to do is get the 9/6 JoB Perfect Strategy Analysis.
What we are really interested in are the probabilities---since we would not be changing our decision making. We're going to want this to be as specific as possible, so we have to do combinations(result)/(total combinations), thus:
Royal: 41,126,022/1,661,102,543,100
Straight Flush: 181,573,608/1,661,102,543,100
Four of a Kind: 3,924,430,647/1,661,102,543,100
Full House: 19,122,956,883/1,661,102,543,100
Flush: 18,296,232,180/1,661,102,543,100
Straight: 18,653,130,482/1,661,102,543,100
Three of a Kind: 123,666,922,527/1,661,102,543,100
Two Pair: 214,745,513,679/1,661,102,543,100
One Pair: 356,447,740,914/1,661,102,543,100
Nothing: 906,022,916,158/1,661,102,543,100
Okay, so now we look at the total return for the 9/6 game of 0.995439 and we see how much of that relates to the flush. Since we are assuming that we are going to play perfect 9/6 JoB strategy on a 9/5 game, we can simply change our return from the flush to five (as opposed to six) and see what happens.
(18,296,232,180/1,661,102,543,100) * 6 = 0.0660870658
(18,296,232,180/1,661,102,543,100) * 5 = 0.05507255484
0.05507255484 - 0.0660870658 = -0.01101451096
With that, we simply subtract this difference from the normal return of 9/6 Jacks: 0.995439 - .0110145 = 0.9844245
Okay, so now all that remains is to do the 9/5 game with perfect strategy for 9/5 and look at what the difference is:
0.9844245-0.984498 = -0.00007349999
With that, we see that we lose about .00007349999 or 0.00735% by way of not deviating from the 9/6 Perfect Strategy playing a 9/5 game.
Not that you asked and not that I am going to figure out how much, but a big portion of this EV lost can be avoided by simply figuring out which Four Flush w/Three Royal Holds (9/6 Jacks) simply become holding Three Royal on 9/5 Jacks.
For example, a hand such as 10s-Js-As-Qh-4s is a Four Flush hold on 9/6, but would become Three Royal hold on 9/5...and by a pretty big margin.
On 9/6, this is actually a really close decision such that holding Three Royal doesn't even really cost you much, in fact, it's the Qh acting as a straight penalty card (in this case) that changes the decision. However, with 9/5 Jacks, the decision to hold Three Royal isn't even close---you drop the Qh and 4s by a huge margin.
NOTE: I did this with 9/6 Optimal (as opposed to simple), but you now know how to do it with Simple...or with any other game!
WRAP-UP: That shortcut ruled! The best part is that you can use it to determine what this would do to your overall RTP playing any strategy as compared to the other strategy...it just has to be the same Video Poker variant. (For example, you couldn’t use this to calculate the effect of playing 9/6 Jacks or Better strategy on Double Double Bonus).
Simple card game analysis (card math + probabilities)
PREFACE: MrV game up with a new idea for a simple card Table Game on Wizard of Vegas and wants to know if it’s viable. Wizard suggested the House Edge would be too low with MrV’s rules, so I created a modification to get the House Edge close to 2% and created a side bet:
MrV Says: Here's an idea for a casino card game; I don't know if it's be tried before but it is so simple that I suspect it has.
I call it "High Card."
Player antes.
Dealer deals out one card to each player, face up, and one to himself, face up. That's it.
The winner is whoever has the highest card.
For example, a seven beats a three.
In the event of a tie, the higher rank wins, e.g. a spade beats a diamond.
The house edge comes from declaring that the queen of spades ('the bitch") is always a winner for the dealer, whether the dealer or the player holds it.
MISSION146 RESPONDS: Wizard has proclaimed the House Edge of this too low, so here's a simple solution to that: If both player and dealer tie with each having a hand total of five, or less, the player loses. The player probably won't care too much because the player did not expect to win the hand anyway.
With that, let's look at the probability of such an event:
(4/52 * 3/51) * 4 = 0.01809954751
Okay, so ties on the other nine ranks would be a push:
(4/52 * 3/51) * 9 = 0.0407239819
Okay, so any other result would have either the player or dealer winning---and both would be equally likely, so:
(1-0.0407239819-0.01809954751) = 0.94117647059/2 = 0.47058823529 Which leaves an expected return, per unit bet, as follows:
0.47058823529 - (0.47058823529) - (0.0407239819*0) - 0.01809954751 = -0.01809954751
Naturally. Thereby resulting in an expected loss of .01809954751 units per unit bet and House Edge of 1.809954751%.
If the player wants to, "Feel like he has won," then you simply have a side bet on a tie. The probability of a tie is:
0.0407239819+0.01809954751 = 0.05882352941
What we will do is the player gets paid 15 to 1 on any tie, so:
(0.05882352941 * 15) - (1 - 0.05882352941) = -0.05882352944
Thereby, our side bet has an expected loss of .05882352944 units relative to unit bet with a House Edge of 5.882352944%. The draw is that, when the player loses to ties of five, or under, on the base game...the player wins the side bet anyway.
(Analysis based on single deck)
WRAP-UP: There’s not really much to add to this, but if you have any simple games that you think I could help you figure out (or want to create a zero House Edge game to play with friends at home), feel free to PM me your basic idea and I will try to tinker around with it...but only if you don’t mind me putting it in an upcoming, “Ask Mission.”
Drawing odds (casino drawings)
PREFACE: Unfortunately, some questions cannot be precisely answered due to a lack of information. However, you can get a general guesstimate if you’re willing to make a few assumptions. Hunterhill asked a question on WoV that I would have preferred to answer precisely...but it was impossible as too much information was unknown.
Hunterhill ASKS: Question on a drawing?10 names are called you have 5% of total entries which are one million. what are the odds that you don’t get called? Once someone is called they can’t get called again. Also same as above but what if 15 names were called. What if you only had 2% of the entries? Thanks for any help.
MISSION146 ANSWERS: Unfortunately, the solution cannot be determined without knowing precisely how many entries (if different) that everyone else has. The reason why is because someone else being called at anytime (starting with the first person called) effectively eliminates all of their other entries from the pool...but we don't know how many entries those are. That person being called could eliminate 100 entries or 100,000, so you really don't know.
Simply put: Assuming you aren't called the first time, the person with the greatest number of entries being called helps you WAY more than the person with the fewest number of entries being called.
The answer would be easy if you want to assume that everyone has the same number of entries. In this case, we'll just say that you have 5 of 100 (as does everyone else) because that would be the same thing as saying 50,000 of 1,000,000:
Your probability of being called is:
(5/100) + (5/95) + (5/90) + (5/85) + (5/80) = 0.27951066391
Or, roughly 27.951066391%
As you can see, we are simply eliminating five entries for each person called. Why what I suggested above matters is that we can look at the same thing, except for the fourth call, we will assume that the person called either had TEN entries or ONE entry and see what happens to the probabilities based on that---everyone else still has five entires or as close as possible---it really doesn't matter in this example:
(5/100) + (5/95) + (5/90) + (5/85) + (5/84) = 0.27653447343
(5/100) + (5/95) + (5/90) + (5/85) + (5/75) = 0.28367733058
In the most extreme case, only five individual people would have entries at all, in which event, the probability that you would be drawn sooner or later would be 100%.
WRAP-UP: So, this is a good way that you can estimate your probability of being called in a drawing based on the rules of the drawing and what probabilities that you do know. In this case, Hunterhill at least had some idea of how many entries he had relative to the total.
Calculating specific results in craps
PREFACE: The long story short is that this post came about as a result of some misinformation that was being bandied about when it comes to Craps probabilities. You can read the thread here if you really want to. The first thing that will follow is a breakdown of the probabilities for every possible Pass Line outcome:
MISSION146 SAYS: You want really simple math, here it is:
Pass Line Win, 7 or 11: 8/36
Pass Line Loss, 2, 3, 12: 4/36
Point Established 4 or 10: 6/36
Point Established 5 or 9: 8/36
Point Established: 6 or 8: 10/36
TOTAL: 36/36
Pass Line Win CO Probability: 8/36 = .2222222
Pass Line Point Established 4 or 10 and WIN Probability: (6/36) * (3/9) = 0.05555555555 Pass Line Point Established 5 or 9 and WIN Probability: (8/36) * (4/10) = 0.08888888888 Pass Line Point Established 6 or 10 and WIN Probability: (10/36) * (5/11) = 0.12626262626
Sum of Point Established + WIN: 0.12626262626 + .08888888888 + .05555555555 = 0.27070707069
Sum of ALL WIN Probabilities: 0.27070707069 + .222222222 = 0.49292929289 Pass Line Loss CO Probability: 4/36 = .1111111
Pass Line Point Established 4 or 10 and LOSE Probability: (6/36) * (6/9) = .1111111111 Pass Line Point Established 5 or 9 and LOSE Probability: (8/36) * (6/10) = 0.13333333333 Pass Line Point Established 6 or 8 and LOSE Probability: (10/36) * (6/11) = 0.15151515151
Sum of Point Established + LOSS: 0.15151515151 + .133333333333 + .111111111111 = 0.39595959595
Sum of ALL LOSS Probabilities: 0.39595959595 + .1111111 = 0.50707070706 Difference in Probabilities: 0.50707070706 - 0.49292929289 = 0.01414141417 Ergo, a House Edge of 1.41414141%.
The end. Case closed. That's just how dice probabilities work which is what makes analyzing this bet so trivial. I can't make these terms any simpler above, though I guess I didn't necessarily have to account for equal probability results together and could have done them separately.
MIDDLE: In response to some very wrong and misleading comments (the biggest being that the probability of the House winning being 73%+ when a point is established), I made the following comments:
(You can see the thread and find a link to the blog post in question, but I don’t recommend reading it at all as it contains deliberately misleading statements)
MISSION146 CONTINUES:
1.) Assuming a point has been established, the probability of winning depends on the point in question, but overall, is 27.070707069%. The probability of losing when a point is established is 39.595959595%. As anyone would expect, this lines up with the fact that 24/36 numbers are point numbers.
2.) The probability of winning on the CO is 11.111111% more than of losing on the CO, the difference in probability of winning or losing, a point having been established, is about 12.525252525252% in favor of losing. When you subtract the difference in probability of losing with a point established from the difference in probability of winning without, you will also arrive at the House Edge of roughly 1.4141414141%. Everyone who even understands the basic Rules of Craps would know that is where this comes from.
3.) The House wins approximately 50.70707070--- of ALL hands, regardless of whether or not there is a point established.
4.) The Odds Bet should be viewed as a totally separate bet and has a House Edge of 0% anyway. While you need to have a Point Established to make an Odds bet, making an Odds bet is not required when you have a Point Established.
When you refer to, "Establishment," bettors who have continuous Come Bets or up to a certain number of Come Bets working and you look at all of those losing on a Seven Out, what you are ignoring is that these Come Bets + Odds can also win on the same number multiple times prior to the original Point Established ever being resolved.
THAT is the difference. The Point Established number on the original Pass Line bet can ONLY win or lose once in that hand. Other Point Numbers (assuming continuous Come + Odds betting) can win multiple times before the original Pass Line bet has been resolved.
For example, consider the following hand:
8, 5, 4, 5....
Okay, so now you have a five that has been rolled twice and the Come Bet + Odds has already won once on the Five, though the eight is unresolved. Prior to the five being rolled, the player could have lost the equivalent of three PL + Odds bet (two of these are COME + Odds) had a seven been the next number. The third Come Bet would have won.
So, it is true that the seven can cause multiple bets to lose. However, a seven cannot cause the same number to lose twice...but during Continuous Come Bet action, the same number can win multiple times in a hand and can only lose once in a hand.
Tuttigym Said (In the Blog): Mr. O: Further down in this blog, I believe your calculations show an overall point conversion rate at 27+%. Does that mean the House wins approximately 73% of all hands where there is a point established? I am using simple math here. Are my assumptions correct? If so, how does the House's 73% winning advantage on point establishments square with a 1.41% HA.? Can you tell me, approximately, the percentage of outcomes of Come Outs vs Point establishments?
MISSION146 RESPONDS: That is the dumbest Craps-related question that I have ever seen asked.
The Point Conversion Probability of 0.27070707069 does NOT mean that the House wins approximately 73% of all hands where a point is established.
Start with the probability of the House winning on a Point of either Four or Ten is (6/9) = 66.666666% against (3/9) = 33.333333% probability of losing. This is the second-worst case scenario for the Pass Line with the only thing worse being to lose immediately.
Now, let's look at this section of what I said:
Okay, so what you want to do is simply index all of these results to Points Established. That's easy, and this is how you would do it:
.27070707069/(.27070707069+.39595959595) = 0.40606060605 or 40.6061% .39595959595/(.27070707069+.39595959595) = 0.59393939394 or 59.3939%
Assuming a point is established, irrespective of what the individual point actually is, above are the probabilities of winning and losing.
Did it escape your attention that .22222222 + .27070707 = .4929292929? Does fourth grade math not get that far?
CLARIFICATION: If you guys ask me questions, I promise not to be mean to you or call any questions, “Stupid.” For reasons that I will never understand, Tuttigym is determined to make it appear as though Craps math/probabilities/expectations are something other than what they are and often references, “Simple arithmetic,” and, “Fourth Grade Math,” as sufficient to prove what every analyst has ever said about Craps wrong.
The problem is that everyone to analyze Craps mathematically IS GOING TO AGREE 100% OF THE TIME!!! We are not going to agree 100% of the time because there is some sort of conspiracy...we’re going to agree 100% of the time because (for us) Craps Math is really basic stuff. It’s closed form math as it’s straight probabilities with no conditionals and results that are almost binary. In fact, MOST craps bets are binary, but the line bets have multiple resolution points, which I solved individually above.
Craps system analysis (craps)
Tuttigym (of course) want to know about a Craps System: Mr.Odiuosgambit and Mr.ChumpChange are the inspiration for this strategy, so if this works, the credit is theirs, and if it fails, my shoulders are broad enough to take the attacks.
Mr.Change had previously stated that 73% of all point play in craps are winners for the House. He has since revised that figure somewhat, but the overall "probability" for the House is 73%. I am going to assume that percentile number would increase if one were to calculate the House wins on just the 4, 5, 9, & 10. (Perhaps, Mr.Change could do those calculations and provide that
number or percentile.)
This strategy will require only a very modest buy-in ($200) so a vast majority of players might like to try it if they believe there is a validity to the plan.
All wagers are one unit. First, a one unit each simultaneous PL/DP bet at CO. Only possible loss is a 12; a one unit loss easily overcome in point play and by definition occurs once in 36 CO rolls.
If the point is a 6 or 8, no further action by the player; the point plays out; the player breaks even.
If the point is a 4, 5, 9, or 10, place one unit on the DP odds. Let the hand play out. Worst case scenario; the point is converted and the player loses one unit otherwise 73+% of the time ---- 7 out and the player wins the reverse odds $$$ of the DP odds bet.
This type of play would take great discipline especially if the table action is "hot" because the other players might be capitalizing on their own action and play.
Comments, additions, subtractions?
Tuttigym
MISSION146 ANALYZES: I am NOT ANALYZING this for the benefit of Tuttigym. In fact, without Tuttigym being here, I wouldn't have to analyze this garbage.
I am analyzing it for anyone who might be foolish enough to read this system and think it's winning.
Okay, we start off we total bets of $30, which reflect $15 on Pass Line and $15 on Don't Pass. As was correctly pointed out, the Come Out Roll only suffers a loss of one unit on Come Out Rolls of 12. All other Come Out deciding rolls become a push:
1/36 * -$15 = -0.416666666667
Come Out Roll pushes would be on Come Out Rolls of 2, 3, 7 and 11...because of the rules of this system, Come Out rolls of 6 and 8 effectively become a push.
21/36 * 0 = 0 (If you don't know where 21/36 comes from, too bad, I don't care.)
That leaves 14/36 results that resolve either as a win, or as a loss, given the rules of this system.
6/36 (Points of 4 or 10) will result in Laying $15 Odds to try to win $7.50.
(7.5 * 6/9) - (15 * 3/9) = 0
Because the Odds Bet has no House Advantage.
8/36 (Points of 5 or 9) will result in Laying $15 Odds to try to win 15 * 2/3 = $10 With that:
(10 * 6/10) - (15 * 4/10) = 0
Because the Odds Bet has no House Advantage.
With that, we have an expected loss of -0.416666666667 which reflects the loss of $15 for every 1/36 total bets we make. When we look at the expected loss (per attempt) relative to the total we bet each attempt, we get:
-0.416666666667/30 = -0.01388888888
Which reflects a House Edge of 1.3888888889%, which is approximately what happens when you combine the, "Bet Made," house edge of the Pass Line and the Don't Pass.
It is impossible for the House to win 73% of all Point Play because the most favorable possible condition for the House (or the Don't Pass bet), when it comes to a Point being Established, is a Point of 4 or 10. When the point is Four or Ten, the probability of Don't Pass going on to win is 66.666666667%...so given that is the best possible scenario, the overall probability cannot be more than this. That's just rudimentary logic.
If anyone is interested what the overall probability of Don't Pass winning is when we isolate Points of 4, 5, 9 and 10, here is how you would do it.
Points of 4 or 10: Don't Pass---.66666666667, Pass---.33333333333333 Points of 5 or 9: Don't Pass---.6, Pass---.4
Okay, we are ignoring Points of 6 and 8 for the purpose of this system, so with that, we have to compare how frequently points of 4 and 10 happen to points of 5 and 9:
Total Possibilities: 14
Point 4 or 10: (6/14) = 0.42857142857
Point 5 or 9: (8/14) = 0.57142857142
Okay, so with that, we can now multiply the probabilities of the Don't Pass winning based on these points and simply add them together:
(0.57142857142) * .6 = 0.34285714285
(0.42857142857) * .6666666667 = 0.28571428542
0.28571428542 + 0.34285714285 = 0.62857142827
Having isolated only results in which a point of 4, 5, 9 or 10 is established, we see that the Don't Pass will win 0.62857142827 or 62.857142827% of the time.
That means the Pass Line will win 1- 0.62857142827 = 0.37142857173 or 37.142857173% of the time.
With that, we can reanalyze the expected loss (there won't be one) on Odds Bets overall. If the DP Odds bet of $15 loses, it always loses $15, thus:
0.37142857173 * 15 = 5.57142857595
In order to determine how much it wins, we simply look at how much it wins in each individual case and multiply by the probability of that happening, then add the two together:
(10 * 0.34285714285) + (7.5 * 0.28571428542) = 5.57142856915
As you can see, the two numbers are the same with small differences due to rounding. System Conclusion
1.) Is it a winning system?
-Of course not.
2.) Is the probability of a win (assuming a point is established) playing according to these rules, "73%+?"
-Absolutely not.
3.) What is 'good' about this system?
-While you can only lose money on the Come Out Roll (1/36), at least this system calls for only the three lowest House Edge bets (Pass, Don't Pass + Odds) to be made.
WRAP-UP: As I suspected, Tuttigym was dissatisfied with this analysis and posted to the effect that he thinks it’s wrong. Why would Tuttigym do this? Because he hates being told that his Craps systems are losing systems...which all Craps systems are...because Betting Systems DO NOT Work and cannot change the House Edge.
If you think betting systems are fun and you’re playing with money that you can afford to lose---then go for it and have a great time! I’m not here to tell anyone how to gamble. I’m here to tell the truth, to prove the truth with math and to make sure that people are making informed gambling decisions based on facts.
