Craps - Probability
A casino I played at had the 3,4,5 odds system where you were allowed 3x on the 4 and 10, 4x on the 5 and 9 and 5 x on the 6 and 8. I feel that with this "system" of placing odds, you reduce the fluctuations (with respect to standard 5x odds on all numbers) in your bankroll, and change the distribution of net gain/loss per session, i.e. you would produce a sharper peak located slightly more to the loss side than with 5x odds. Is this so, and could you put some numbers to it?
That is known as 3-4-5X odds, and is now pretty common. The following table shows all the possible outcomes, for the pass and odds combined, with full odds.
Return Table with 3-4-5X Odds
|Pass line win||1||0.222222||0.222222|
|Pass line loss||-1||0.111111||-0.111111|
|Point of 4 or 10 & win||7||0.055556||0.388889|
|Point of 4 or 10 & lose||-4||0.111111||-0.444444|
|Point of 5 or 9 & win||7||0.088889||0.622222|
|Point of 5 or 9 & lose||-5||0.133333||-0.666667|
|Point of 6 or 8 & win||7||0.126263||0.883838|
|Point of 6 or 8 & lose||-6||0.151515||-0.909091|
The standard deviation per pass line bet is 4.915632.
Grshooter from Kansas City, Missouri
The average number of rolls per shooter is 8.525510. For the probability of exactly 2 to 200 rolls, please see my craps probability of survival page.
Jon from Danville, New Hampshire
Of those 100 points established, on average 41.67 would be on a 6 or 8, 33.33 would be on a 5 or 9, and 25.00 would be on a 4 or 10. You could expect on average 18.94 points made on a 6 or 8, 13.33 on a 5 or 9, and 8.33 on a 4 or 10.
DB from New York, USA
Well, anyone can make a mistake, but craps is an easy game to analyze mathematically so I would be very confident my odds on craps are right. Yes, gambling in one way or another is my full time self-employed profession. I have been to Atlantic City many times in the last few years but two months ago I moved to Las Vegas. So, I'm afraid I wouldn't be gracing Atlantic City with my presence much any longer. I prefer a combinatorial approach as opposed to random simulations whenever I can. Either way, I roll my own software with Visual C++. For random numbers I use a Mersenne Twister.
1) Your preference is to count the come out roll of 12 in the calculation of the house edge on the don’t pass. If one was to choose NOT to count it, would the house edge on the pass line combined with full double odds be exactly equal to that of the house edge on the don’t pass line combined with full double odds?
2) Does the overall house edge against player x go up if player x places come bets (which will be backed up with full double odds) after betting the pass line with full double odds? i.e. player x with just a pass line with full double odds = house edge .572%, player x with same bet but places two come bets with full double odds = house edge (.572%)x(3)?
Jay from Hamilton, Ontario
Thanks for your kind words. Here are my answers.
1. If we define the house edge as the expected loss per unresolved bet (not counting ties) then the house edge on the don’t pass would be 1.40%, just barely less than the 1.41% on the pass line bet. If the player can bet more money on the don’t pass side, which is the case in real but not Internet casinos, then the combined house edge favors the don’t side more the greater the multiple of odds allowed.
2. Assuming the player keeps his odds on during a come out roll then the overall house edge does not change if the player adds come bets, backed up with the odds. However if the player keeps the odds off, which is the default rule, then the overall house edge will actually go up slightly by adding come bets.
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks
Stan Abadie from Harahan, Louisiana
You’re welcome, thanks for the kind words. The probability of throwing the dice n times without a 7, and then throwing a 7, is (5/6)n*(1/6). The probability of throwing n non-sevens, without specifying the next throw would be (5/6)n. So the probability of throwing the dice at least four times without a seven would be (5/6)4=625/1296=0.4823.
Jeff from Las Vegas, US
The following are the possible outcomes of the pass/come bet and their associated probabilities:
- Player wins on come out roll: 22.22%
- Player loses on come out roll: 11.11%
- Player wins on a point: 27.07%
- Player loses on a point: 39.60%
So the player will win on a point about 1 in 3.7 rolls.
That is a good question. It is obviously more fun to go with the crowd than against it. The question is why does the crowd favor the pass line? Perhaps it is just tradition. Maybe when people first started playing craps in private games the don’t pass wasn’t even an option.
Good question. Let’s think of this in units as opposed to $100 bets. You will always have a bet on the pass or come. On any given roll the probability there is an old pass or come bet on the 4 is 3/9. This is the probability that by looking back at old rolls you will find a 4 before a 7. Likewise the probability of having a bet on 5 is 4/10 and on 6 is 5/11. So the average overall bet is 1+pr(4)+pr(5)+pr(6)+pr(8)+pr(9)+pr(10) = 1+3/9 + 4/10 + 5/11 + 5/11 + 4/10 + 3/9 = 3.3758 units. This average will not true at the beginning, while you are getting in to the game. It will only apply after all point numbers and the 7 have already been rolled at least once.
The probability of winning the hard 4 bet is 1/9. So the probability of winning four times in a row is (1/9)4 = 1 in 6561.
Good question. For those who don’t understand the question, unless otherwise requested, odds on come out bets are not active on come out rolls. So if the player rolls a seven on a come out roll any come bets will lose and odds on come bets will be returned. Likewise if the player’s point on the come bet is rolled on the come out roll the come bet will win but the odds will push. The answer depends on how we define the house edge. If we define it as expected loss to total bets made then turning the odds off would not matter. This is because the player is still betting the odds and it still counts as a bet even if it is returned as a push. However if you define the house edge as expected loss to bets resolved then turning the odds off on a come out roll does indeed increase the house edge. I wrote a computer simulation to determine this effect. Assuming the player takes fives times odds then turning the odds off on come out rolls increases the ratio of losses to total bets resolved from 0.326% to 0.377%, or an increase of 0.051%. So if you want to maximize your return on bets resolved then leave those come odds turned on.
Just a coincidence I assure you. The exact house edge in craps is 7/495, which by definition must be a rational number. In fact I would argue the house edge in all casino games must be a rational number because there are a limited number of possible outcomes in all games, resulting in a house edge of a perfect fraction. 2 is not a perfect square thus the square root of 2 must be irrational by definition. Therefore the two numbers can not be equal. To be specific the house edge on a $100 pass line bet would be $1.41414141... The square root of 2 is 1.4142135623731...
Wally from Houston
Thanks for the compliment. I recommend taking the match play. I’m sure the $100 in slot play was on specially designated machines. From anecdotal evidence I believe these free play slots are extremely stingy, set to pay back about 25%. That match play is worth about 48 cents on the dollar. I recommend betting in on the don’t pass in craps. The reason I favor that over blackjack is that blackjack has a lower probability of winning, thus reducing the value of the match play. For further explanation please see my October 30 2001 column.
- 5- place $5
- 6- place $6
- 8- place $6
- field- $5
- total= $22
They claim the house edge is 1.136%. How is that possible if every individual bet made has a higher house edge?
Good question. To confirm their math I made the following table, based on a field bet paying 3 to 1 on a 12. The lower right cell does shows an expected loss of 25 cents over $22 bet. So the house edge is indeed .25/22 = 1.136%.
Mensa Anything but Seven Combo
|Number||Probability||Field||Place 5||Place 6||Place 8||Win||Return|
The reason the overall house edge appears to be less than the house edge of each individual bet is because the house edge on place bets is generally measured as expected player loss per bet resolved.
However, in this case the player is only keeping the place bets up for one roll. This significantly reduces the house edge on the place bets from 4.00% to 1.11% on the 5 and 9, and from 1.52% to 0.46% on the 6 and 8.
For you purists who think I am inconsistent in measuring the house edge on place bets as per bet resolved (or ignoring ties) then I invite you to visit my craps appendix 2 where all craps bets are measured per roll (including ties).
Steve S. from Long Island, NY
First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = 1/4. Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758. Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...
x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551
There are still video poker games that with proper strategy pay over 100%. I’ve also seen a blackjack game at the Fiesta Rancho and Slots-a-Fun in Las Vegas that had a basic strategy advantage. As I argue in my sports betting section betting NFL underdogs at home against the point spread also has resulted in a historical advantage. So 100x odds in craps is still one of the best bets out there, but not the very best. Yes, 0.014% means that per $100 bet you lose 1.4 cents on average.
I agree that this is a very bad decision and poor advice from the dealers. Once a point of 6 or 8 has been rolled the player edge on a don’t pass or don’t come bet is (6/11)*1 + (5/11)*-1 = 1/11 = 9.09%. Taking "no action" is the same as trading it for a bet with a 1.36% house edge. So this decision costs the player 10.45%. To any dealers encouraging this I say shame on you.
B.L. from NYC
The following table shows the house edge is 5.56%.
Haig from Englewood
The fewer the sevens the greater the odds favor the pass line bet. The following table shows the house edge according to the percentage of sevens, assuming the probability of all other numbers is proportional to the fair probability.
House Edge in Craps According to Seven Probability
|Seven Probability||Pass House Edge||Don’t Pass House Edge|
Andy from Hollywood
I get a lot of questions about combinations of craps bets. Normally I don’t answer them but when you address me as "the great and powerful Wizard" it greatly improves your odds of getting a reply. Your mistake is that both bets are not resolved all of the time. When you win either the 6 or 8 you are taking the other bet down, which brings down the expected loss because you are betting less. So your math is right but you are comparing apples to oranges.
JOE from EUREKA, CA
You’re correct, dice alone can not determine the outcome in craps. There are various ways of using cards in place of dice and still have the odds exactly the same. One way is to use two separate decks, thus there is no effect of removal. Another way is to have a 7-card deck, featuring the numbers 1 to 6, plus a seventh "double" card. The first card drawn can never be the double card. If it is then it is put back in and the process repeats from the beginning. If the double card is drawn second then it counts as whatever the first number drawn was. Regardless of how the casino does it I have never seen hard evidence of a case where the odds were different than if two dice were used. So I think you are omitting something from the rules.
Yes, there was a story taped in which some frat boys at UNLV were trying to parlay $1,000 into $5,000 to buy a high end television. They sought out my advice on how to best achieve this goal quickly. I was limited to the games at the Golden Nugget. The Nugget has 10x odds in craps, which I felt offered the opportunity to achieve the goal. It was my strategy on each come out roll to bet min(bankroll/11, (5000-bankroll)/21), subject to convenient rounding, and take the maximum odds. This way we would never go over $5,000 after a 4 or 10 win, would always have enough to take full odds, and would risk the maximum amount if we didn’t have enough to get to $5,000.
For the first bet, this formula would call for a pass line bet of $90.91, but I rounded it up to $100. Then a point was rolled, I think a 6 or 8. On the second roll the shooter sevened out. So the entire grand was lost in two rolls. It apparently didn’t make for very entertaining television and that story never made the air.
Two questions I can anticipate being asked would be (1) why did I have them bet the pass as opposed to the don’t pass, and (2) why didn’t I bet $91 on the line and $910 on the odds, adding the extra dollar out of my own pocket. To answer the first question, I think that for purposes of going for a quick big win the pass line is better. While the overall house edge is less on the don’t pass, I felt it would have taken more rolls to achieve the $5,000 goal, thus exposing more money to the house edge. To answer the second question, there is not much difference between 9x odds and 10x odds and I thought it would look better on television to be betting only black chips, at least to start.
Greg from Fairfax
As my blackjack section shows, the 2 to 1 on blackjacks is worth 2.27% and doubling on 3 cards is worth 0.23%. Otherwise the rules look standard. All things considered, the house edge in the blackjack game has a player advantage of 2.1%. The probability of winning on a 4 or 10 in craps is (6/36)×(3/9) = 5.56%. Every time this happens you get an extra unit, so it is worth 5.56%. Normally the house edge on the come bet is 1.41%, so overall the player edge under this rule is 4.15%. So I agree that craps was the better game to play.
Will from Rector
I didn’t know they had a buy bet in Crapless Craps. The following table shows the house edge of place and buy bets, assuming there were no rounding of winnings. In your example of a $30 buy bet on 2 or 12 the winnings would be 6*$30-$1=$179. So the expected return is [(1/7)*$179 + (6/7)*-$30] / $30 = -0.0048, so we’re very close.
Place and Buy Bets in Crapless Crapspass and buying oddsin Crapless Craps
|Bet||Pays||Prob. Win||House Edge|
|Place 2, 12||11 to 2||0.142857||0.071429|
|Place 3,11||11 to 4||0.25||0.0625|
|Buy 2, 12 (commision only on wins)||119 to 20||0.142857||0.007143|
|Buy 3,11 (commision only on wins)||59 to 20||0.25||0.0125|
|Buy 2, 12 (commision always)||119 to 21||0.142857||0.047619|
|Buy 3,11 (commision always)||59 to 21||0.25||0.047619|
Donald from Las Vegas
That is very tight to limit the dealers like that. On a $2 bet the house edge goes up to 29.02%, and a $5 bet it is 41.94%.
John B. from Riverside, Illinois
We can see from my analysis of the Fire Bet that the probability of a shooter making all six points is 0.000162435. So, the value of the promotion per shooter is $4,000 × 0.000162435 = 0.649739.
The next question to be asked is what is the expected loss per shooter. The house edge on the pass line bet is 7/495 = 1.414141%. The tricky part is how many pass line bets will a shooter make, on average.
There are four possible states the shooter can be in. Let's define each one as the expected number of future pass line bets for that shooter.
- A = Come out roll
- B = Point of 4 or 10 made
- C = Point of 5 or 9 made
- D = Point of 6 or 8 made
Here are the equations showing the probability of each state leading to the next state.
A = 1 + (12/36)*A + (6/36)*B + (8/36)*C + (10/36)*D
B = (1/3)*A
C = (2/5)*A
D = (5/11)*A
A little algebra results in A = 2.525510, the number of pass line bets made per shooter.
So, the expected loss per $5 shooter is $5*2.525510*0.0141414 = 0.178571.
The expected amount bet by the shooter is $5*2.525510=$12.627551.
Finally, the expected return is the expected win divided by the expected bet: (0.649739-0.178571)/12.627551 = 3.73127%. So the house edge is -3.73%.
James from Santa Cruz
Yes, the probability of each double is 1/36. However you have to compare that to the probability of rolling a losing combination. For a hard four, there are 8 losing rolls (two each of 1-6, 2-5, 3-4, and 1-3), so the probability of winning is 1/9. For a hard six, there are ten losing rolls (two each of 1-6, 2-5, 3-4, 1-5 and 2-4), so the probability of winning is 1/11. The hard six pays more because the probability of winning is less.
Ron from Collinsville, IL
Crapless Craps offers those two bets too. There is one way to roll a 2, and six ways to roll a 7, so the probability of winning a place bet on the 2 is 1/7. Same probability is the same for the 12. As explained in the baccarat question, if the probability of something is p, then fair odds are (1/p)-1 to 1. In this case fair odds would be 6 to 1. The house edge can be expressed as (t-a)/(t+1), where t is the true odds, and a is the actual odds. In Crapless Craps the place bet on the 2 and 12 pays 11 to 2. Using this formula, the house edge on the 2 and 12 is (6-5.5)/(6+1) = 0.5/7 = 7.14%.
In Crapless Craps the 3 and 11 pay 11 to 4. Using the same formula, t=3, and a=2.75, so the house edge is 0.25/4 = 6.25%.
Ben from Austin, TX
If we ignore the house edge (which is very low in craps if played properly), the probability of winning $500, as opposed to losing $1,000, is 2/3. The probability of 4 out of 5 winning sessions would be 5×(2/3)4×(1/3) = 32.9%.
The probability of rolling a 7 is 1/6, and the probability of rolling a 12 is 1/36. The probability of rolling a 7, given that a roll is a 7 or 12 is (1/6)/((1/6)+(1/36)) = 6/7. So the probability that the first six times a 6 or 12 is rolled it is a 6 every time is (6/7)6 = 39.66%.
If you rephrase the question to be what is the probability of rolling five 6’s before a 12, then the answer is (6/7)5 = 46.27%. With four rolls it is (6/7)4 = 53.98%. So there is no number of 7’s before a 12 that is exactly 50/50. If you’re looking for a good sucker bet, suggest you can either roll four 7’s before a 12, or a 12 before five 7’s.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
On the come out roll there are three possible outcomes at this point.
- Sevening out.
- Repeating a point already made (4 to 9).
- Rolling a 10 on the come out roll, and then making it.
We need to quantify the second and third probabilities only. The shooter will eventually make a point, and then eventually make it or seven out. The probability that the point established and then made is 4 to 9 is:
(3/24)×(3/9) + (4/24)×(4/10) + (5/24)×(5/11) + (5/24)×(5/11) + (4/24)×(4/10) = 0.364394.
The probability of establishing a 10 point and then making it is (3/24)*(1/3) = 0.041667.
Let p be the probability of making a 10 point before sevening out. If the player makes any other point, he is right back to where he started from. So...
p = 0.364394 × p + 0.041667
p × (1-0.364394) = 0.041667
p = 0.041667/(1-0.364394)
p = 0.065554
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
Given that a point is established, the probability that the shooter makes the point is pr(point is 4 or 10) ×pr(making 4 or 10) + pr(point is 5 or 9) × pr(making 5 or 9) + pr(point is 6 or 8) ×pr(making 6 or 8) = (6/24) × (3/9) + (8/24) × (4/10) + (10/24) × (5/11) = 201/495 = 0.406061.
If the probability of an event is p, then the expected number of times it will happen before failure is p/(1-p). So, the expected number of points per shooter is 0.406061/(1-0.406061) = 0.683673.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.
First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.
Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.
The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.
So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.
Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.
What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:
pr(A or B) = pr(A) + pr(B) - pr(A and B)
You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,
pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3) - pr(4 before 2 and 3) = (3/4)+(3/5)-(3/6) = 0.85.
The probability of not getting the four along the way to the two and three is 1.0 - 0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15*12 = 43.8.
Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.
What is the probability of getting the five before achieving the two, three, or four? The general rule is:
pr (A or B or C) = pr(A) + pr(B) + pr(C) - pr(A and B) - pr(A and C) - pr(B and C) + pr(A and B and C)
So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)-pr(5 before 2 and 3)-pr(5 before 2 and 4)-pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)-(4/7)-(4/8)-(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1 - 83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)*9 = 44.5.
Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.
Here is the general rule for pr(A or B or C or ... or Z)
pr(A or B or C or ... or Z) =
pr(A) + pr(B) + ... + pr(Z)
- pr (A and B) - pr(A and C) - ... - pr(Y and Z) Subtract the probability of every combination of two events
+ pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
- pr (A and B and C and D) - pr(A and B and C and E) - ... - pr(W and X and Y and Z) Subtract the probability of every combination of four events
Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.
The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.
Expected Number of Rolls Problem
|Highest Number Needed||Probability||Expected Rolls if Needed||Probability not Needed||Probability Needed||Expected Total Rolls|
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
I was playing craps at one of your advertisers and got 38% too many sevens. I suspect they are cheating. Here is my full roll history: 7,5,7,2,4,6,8,7,9,4,9,6,6,6,5,12,7,11,8,4,7,7,9,5,12,5,11,5,8,1,7,7,6,6,6,5,5,9,8,10,9,7,7,11,8,9,3,7,6,10,6,7,8,7,8,6,6,5,5,9,6,7. I think you should quit endorsing this cheating casino!
In 61 rolls the expected number of sevens is 61×(1/6) = 10.17. You had 14. The probability of exactly 14 sevens is 7.96% and the probability of 14 or more is 12.77%. So, nothing unusual there. I also did a chi-squared test on every roll. I know that it isn't very kosher to do a chi-squared test on such a small sample, so take the results with a grain of salt. Here are the results:
Chi-Squared Test on 61 Dice Rolls.
The bottom right cell shows a chi-squared statistic of 8.70. The probability of a statistic that high or higher with ten degrees of freedom is 56.09%. These results were close to the peak of the bell curve, so the casino easily passes the chi-squared randomness test.
The answer is 219.149467.
There are two ways I can think of to solve this. The first is with a Markov Chain. The following table shows the expected rolls needed from any given state of the 128 possible.
Fire Bet — Markov Chain
Briefly, the expected rolls from any given state is the expected rolls until point is either made or lost (5.063636) plus the expected number of rolls if the player advances to a further state, divided by the probability of not advancing in state.
The other method uses integral calculus. First calculate the expected rolls for each possible outcome to happen. Then take the dot product of the probability of each event and average rolls to get the average rolls to resolve a pass line bet, which the lower right corner shows is 3.375758 = 557/165.
Fire Bet — Expected Rolls
|Event||Probability||Average Rolls||Expected Rolls|
|Point 4 win||0.027778||5||0.138889|
|pt 5 win||0.044444||4.6||0.204444|
|pt 6 win||0.063131||4.272727||0.269743|
|pt 8 win||0.063131||4.272727||0.269743|
|pt 9 win||0.044444||4.6||0.204444|
|pt 10 win||0.027778||5||0.138889|
|pt 4 loss||0.055556||5||0.277778|
|pt 5 loss||0.066667||4.6||0.306667|
|pt 6 loss||0.075758||4.272727273||0.323691|
|pt 8 loss||0.075758||4.272727273||0.323691|
|pt 9 loss||0.066667||4.6||0.306667|
|pt 10 loss||0.055556||5||0.277778|
|Come out roll win||0.222222||1||0.222222|
|Come out roll loss||0.111111||1||0.111111|
From there we can get the expected rolls between any given point winning:
- Rolls between a point of 4 winning = (3/36)*(3/9)*5*(557/165) = 6684/55 = apx 121.527273.
- Rolls between a point of 5 winning = (4/36)*(4/10)*4.6*(557/165) = 1671/21 = apx 75.954545.
- Rolls between a point of 6 winning = (5/36)*(5/11)*(47/11)*(557/165) = 6684/125 = apx 53.472.
The expected rolls for a 10, 9, and 8 point winner are the same as for 4, 5, and 6, respectively.
Let's say that instead of a point-4 winner happening on a discrete basis, it follows an exponential distribution with a mean of 6684/55. The probability such a random variable lasts x units of time without happening is exp(-x/(6684/55)) = exp(-55x/6684).
The probability it has happened within x units of time, at least once, is 1-exp(-55x/6684).
If we represent all six points as continuous variables, then the probability all six have happened within x units of time is (1-exp(-55x/6684))^2 * (1-exp(-22x/1671))^2 * (1-exp(-125x/6684))^2.
The probability at least one of the six events not happening within x units of time is 1 - (1-exp(-55x/6684))^2 * (1-exp(-22x/1671))^2 * (1-exp(-125x/6684))^2.
We can get the expected time for all six events to happen by integrating the above from 0 to infinity.
Using this integral calculator gives an answer of 8706865474775503638338329687/39730260732259873692189000 = apx 219.1494672902.
Why this works is harder to explain, so please take that part on faith.
That dreadful rule would increase the house edge from 1.41% to 33.26%.
I know a dice influencer who claims to have recorded the following rolls in craps. The roller claims he objective was to hit the inside numbers (4, 5, 6, 8, 9, and 10). Can you analyze his results?
First, let's add a column to the table to show the expected tally of each total, assuming a fully random roll.
Craps Data with Expectations
You didn't ask me how to analyze the data, so I'll do it a few different ways.
A chi-squared test has a chi-squared statistic of 21.43009, with 10 degrees of freedom. The probability of data this skewed, or more, is 1.83%.
Looking at just the inside numbers, which you mentioned was the goal, the total achieved was 12,802, while the expected total would be 25,227 × (2/3) = 12613.5. This excess of inside numbers is 2.52 standard deviations above expectations. The probability of such an excess number, or more, is 0.59%.
I couldn't help but notice the lack of sevens. In 25,227 rolls, the expected sevens are 25,227 × (1/6) = 4204.5. The shooter had 3,963. That is 4.08 standard deviations away from expectations. The probability of such a shortfall is 0.0000225, or one in 44,392.
However, I must say it's usually easy to look at historic data and find something fishy about it. Then again, avoiding sevens is an intrinsic goal to the dice influencer.
The scientific way to test whether the shooter can influence the dice is to state the goal BEFORE data is collected.