February 1, 2005

The first step in the analysis of craps should be finding the probability of any given total. The following table shows the total of each possible outcome of both dice.

The next table shows the total number of times each total occurs in the table above.

The following graph shows the frequency of each total. It is important to note that the frequency decreases on a linear basis as the total moves away from 7.

The craps 2 bet pays 30 to 1 (often stated as 31 for 1 on the felt) if the player rolls a total of 2 on the next roll. The probability of rolling a 2 is 1/36. For bets that have only two outcomes, win or lose, the expected value can be expressed as pr(win)×win – pr(loss).

So the expected value of the craps 2 bet is (1/36)×30 – (35/36) = -5/36 = -13.89%.

The yo 11 bet pays 15 to 1 if the player rolls a total of 11 on the next roll. The probability of rolling an 11 is 2/36. So the expected value of the yo 11 bet is (2/36)×15 – (34/36) = -4/36 = -11.11%.

The any 7 bet pays 4 to 1 if the player rolls a total of 7 on the next roll. The probability of rolling a 7 is 6/36. So the expected value of the any 7 bet is (6/36) ×4 - (30/36) = -6/36 = -16.67%.

The any craps bet wins if a 2, 3, or 12 is rolled on the next throw. There is 1 way to roll a 2, 2 ways to roll a 3, and 1 way to roll a 12. So out of 36 possible outcomes 4 win, thus the probability of winning is 4/36 = 1/9. The bet pays 7 to 1. The expected return is thus (4/36)×7 – (32/36) = -4/36 = -1/9 = -11.11%.

Many bets in craps take more than one role to evaluate. For example the Place 4 bets wins if a 4 is rolled before a 7. Let’s consider the general case of two possible outcomes x and y of an event. The probability of x shall be denoted as pr(x) and the probability of y as pr(y). Please take the following on faith.

Let’s consider the Place 4 bet in craps. The bet wins if a 4 is thrown before a 7. The dice will be rolled indefinitely until a 4 or 7 is rolled. From the table above we know the probability of a 4 is 3/36 and the probability of a 7 is 6/36. In this case pr(x) = 3/36 and pr(y) = 6/36.

So the probability of winning this bet is (3/36)/((3/36)+(6/36)) = 3/(3+6) = 1/3.

Note that we can eliminate the 36’s in both the number and denominator to express the probability as (frequency of 4)/(frequency of 4 or 7) = 3/(3+6) = 1/3.

The bet pays 9 to 5. The expected value is

(1/3)×(9/5) – (2/3) = 9/15 – 10/15 = -1/15 = -6.67%. So the house edge is +6.67%. The odds are exactly the same for the Place 10 bet.

The Place 5 bet, which pays 7 to 5 if a 5 is rolled before a 7.

The probability of winning is the number of ways to roll a 5 divided by the number of ways to roll a 5 or 7, or 4/(4+6) = 0.4.

The expected value is 0.4×(7/5) – 0.6 = -0.04 = 4%. The odds are the same for the place 9 bet.

The Place 6 bet, which pays 7 to 6 if a 6 is rolled before a 7.

The probability of winning is the number of ways to roll a 6 divided by the number of ways to roll a 6 or 7, or 5/(5+6) = 0.4545.

The expected value is (5/11)×(7/6) – (6/11) = -1/66 = -0.0152 = -1.52%. The odds are the same for the place 8 bet.

The buy 4 bet is the same as the place 4 bet, except in payoff. The bet pays true odds but the bettor must pay a 5% commission (based on the bet, not the win) to make the bet. The probability of winning is 3/(3+6) = 1/3. If the probability of a bet winning is p then the fair odds are (1-p)/p to 1. In this case, (2/3)/(1/3) = 2 to 1. So after deducting a 5% commission the rest of the bet pays 2 to 1.

Normally this bet is made in increments of $20 so that the commission rounds to an even dollar. So the player is risking $21 to make a $20 bet, so truthfully the player is betting $21. If the player wins the $20 portion of the bet pays 2 to 1, or $40. However his $1 commission is not returned, so his net win is only $39. So another way of stating the odds is 39 to 21, because the player stands to win $39 on a $21 bet.

So the expected value is (1/3)×(39/21) – (2/3) = -1/21 = -4.76%.

Another way to analyze this bet is to break it down into two bets. Buy bets are supposed to pay fair odds, before factoring in the 5% commission. So 20/21 of the bet has a 0% expected value and 1/21 of the bet has a –100% expected value. The average expected value is (20/21)×0 + (1/21)×-1 = -1/21 = -4.76%.

Bet pays fair odds of 3 to 2 if a 5 is rolled before a 7, but the player must pay a non-refundable 5% commission (based on the bet). So on a $21 bet the player stands to have a net win of $20×(3/2)-1 = $29. After adjusting for the 5% commission the win is 29 to 21.

The probability of winning is 4/(4+6) = 0.4. So the expected value is 0.4×(29/21) – 0.6 = -1/21 = -4.76%.

Bet pays fair odds of 6 to 5 if a 6 is rolled before a 7, but the player must pay a non-refundable 5% commission (based on the bet). So on a $21 bet the player stands to have a net win of $20×(6/5)-1 = $23. After adjusting for the 5% commission the win is 23 to 21.

The probability of winning is 5/(5+6) = 5/11. So the expected value is (5/11)×(23/21) – (6/11) = -1/21 = -4.76%.

The Big 6 is yet another bet that pays if a 6 is rolled before a 7. However it pays only 1 to 1. The probability of winning is 5/11. The expected return is (5/11)×1 – (6/11) = -1/11 = -9.09%.

Note that there are three ways to make the exact same bet: the place 6, buy 6, and big 6. However they all carry a different house edge, 1.52%, 4.76%, and 9.09% respectively. You would think players would only bet the place 6 of the three, but alas players still bet the other two.

A few casinos only charge the 5% commission if the bet wins on buy bets of 4 and 10. So if the player makes a $20 buy bet on the 4 he does not need to add a dollar for the commission. However if he wins it will be deducted from the win. At fair odds of 2 to 1 the actual win on a $20 but bet would be 2×$20 - $1 = $39. So overall it pays 39 to 20. The probability of winning is 3/(3+6) = 1/3. So the expected value is (1/3)×(39/20) – (2/3) = -1/60 = -1.67%.

In gambling terminology "lay" means to risk more than you stand to win. For example betting against the spread or on a total in sports the bettor must generally lay (or risk) 11 to win 10. In craps the player can lay odds that a 7 will appear before a 4 (or any other "place number"). A 5% commission is charged based on the potential win amount and otherwise pays true odds. The probability of rolling a 7 before a 4 is 6/(3+6) = 2/3. The commission on a $20 win is $1. If the probability of a bet winning is p then the true odds are (1-p)/p to 1. In this case p=2/3 so the true odds are ½ to 1, or 1 to 2.

So, the player must risk $40 to win $20, before factoring in the commission. On a $40 bet the player must add $1 for the commission (5% of the win). If the player wins he will get back $19. Thus the actual odds are 19 to 41.

The expected return is this (2/3)×(19/41) – (1/3) = 38/123 – 41/123 = -3/123 = -1/41 = -2.44%. The odds are the same on the lay 10 bet.

If you wish to calculate the expected return by the averaging method as show for the Buy 4 bet then we must calculate the ratio of "true odds" portion of the bet to the commission. In this case $40 of a $41 bet has a 0% expected value and $1 has a –100% expected value. So the overall expected return is (40/41)×0 + (1/41)×-1 = -1/41 = -2.44%.

The Lay 5 bet wins if a 7 is rolled before a 5. The probability of rolling a 7 before a 5 is 6/(4+6) = 3/5. With a probability of p the true odds are (1-(3/5))/(3/5) = (2/5)/(3/5) = 2/3 to 1, or 2 to 3.

So, the player must risk $30 to win $20, before factoring in the commission. On a $30 bet the player must add $1 for the commission (5% of the win). If the player wins he will get back $19. Thus the actual odds are 19 to 31.

The expected return is this (3/5)×(19/31) – (2/5) = 57/155 – 62/155 = -5/155 = -1/31 = -3.23%. The odds are the same on the lay 9 bet.

Using the averaging method the expected return is (30/31)×0 + (1/31)×-1 = -1/31.

The Lay 6 bet wins if a 7 is rolled before a 6. The probability of rolling a 7 before a 6 is 6/(5+6) = 6/11. With a probability of p the true odds are (1-(6/11))/(6/11) = (5/11)/(6/11) = 5/6 to 1, or 5 to 6.

So, the player must risk $24 to win $20, before factoring in the commission. On a $24 bet the player must add $1 for the commission (5% of the win). If the player wins he will get back $19. Thus the actual odds are 19 to 25.

The expected return is this (6/11)×(19/25) – (5/11) = 114/275 – 125/275 = -11/275 = -1/25 = -4%. The odds are the same on the lay 8 bet.

Using the averaging method the expected return is (24/25)×0 + (1/25)×-1 = -1/25.

There are 3 ways to roll a 4: 2 ways to roll a 1 and 3, and 1 way to roll a 2 and 2. Because the 1 and 3 is more likely it is called the "easy" way and the 2 and 2 is the "hard" way.

The Hard 4 bet wins if a hard 4 is rolled before a 7 or an easy 4. There is only 1 way to roll a hard 4. There are 2 ways to roll an easy 4 and 6 ways to roll a 7. Thus the probability of winning is 1/(1+8) = 1/9. The bet pays 7 to 1. The expected return is (1/9)×7 – (8/9) = -1/9 = -11.11%. The odds are the same on the hard 10 bet.

The Hard 6 bet wins if a hard 6 is rolled before a 7 or an easy 6. There is only 1 way to roll a hard 6. There are 4 ways to roll an easy 6 (2 ways of 1 and 5, and 2 ways of 2 and 4) and 6 ways to roll a 7, for a total of 10 ways to lose. Thus the probability of winning is 1/(1+10) = 1/11. The bet pays 9 to 1. The expected return is (1/11)×9 – (10/11) = -1/11 = -9.09%. The odds are the same on the hard 8 bet.

The pass bet is the most fundamental bet in craps. Almost every player makes it. The bet wins if the shooter throws a 7 or 11 on the first roll, known as the come out roll. The bet loses if the shooter throws a 2, 3, or 12 on the come out roll. Otherwise whatever number is thrown becomes known as the point and a dealer will put a big white puck on that number so players remember what the point is. Then the shooter will keep rolling until he either rolls a 7 or the point again. If he rolls the point first the pass bet wins, if he rolls a 7 first the pass loses.

The next table shows all the possible outcomes, the probability of each, what it pays, and the return.

The lower right cell of this table shows an expected return of –1.41%.

The don’t pass is the opposite of the pass, except that a 12 on the come out roll pushes instead of wins. Following is a return table for the don’t pass.

The lower right cell shows an expected return of -1.36%, or a house edge of 1.36%. Some gambling books state the house edge as 1.40%. This is because they incorrectly define the house edge as the expected loss per bet resolved, thus ignoring the push on 12.

If the player makes a pass or don’t pass bet and a point is rolled he may bet additional money that the point will be rolled before a 7 after a pass line bet, or vise versa on a don’t pass bet. Odds bets by definition always pay true odds. If the player is betting the odds after a pass bet he is said to "take" the odds, and after a don’t pass bet is "laying" the odds. The following table shows the probability of winning and true odds of each odds bet.

It is often asked what is the overall house edge for the player who only makes a pass line bet with the maximum odds allowed. Let’s assume the casino lets the player bet 2 times his pass line bet on the odds. The probability of the bet resolving on the come out roll is (1+2+6+2+1)/36 = 1/3. So only 2/3 of the time does the player get to make an odds bet. The average odds bet will thus be (2/3)×2 = 4/3. The player will always bet 1 unit on the pass line. So overall 3/7 of money bet will be on the pass line and 4/7 on the odds. The weighted house edge is thus (3/7)×1.41% + (4/7)×0% = 0.61%.

The next table shows the combined house edge according to various amounts of odds allowed.

Sometimes casinos allow different amounts on different numbers. For example "3-4-5" times odds allows the bettor it bet 3 times the pass line bet on points of 4 or 10, 4 times on 5 or 9, and 5 times on 6 or 8. Note that if the player takes the maximum allowed odds he always stands to win six times his pass line bet on the odds.

The general formula if you can take x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10 is (-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ]

As I write this it is one week before the 2005 Super Bowl. The same method of analyzing many craps bets can also arguably be used to analyze Super Bowl proposition bets. For example Caesars Tahoe has the following bet.

For those unfamiliar with this notation the bettor must risk $170 to win $100 on the TD pass is first bet, or will win $140 on a $100 bet on the interception is first bet. In Europe they use decimal notation, showing how much the player gets back, including his original wager, for a 1 unit bet. So in Europe the odds would look like this:

Converting again to "to one" odds, as used in table games we get the following.

In my opinion I think the European notation makes more sense and I hope, but doubt, the U.S. will eventually adopt it.

If we assume (perhaps incorrectly) that any given pass by McNabb is likely to result in a touchdown pass or interception with constant probabilities then we can look at some past data to estimate the probability of each. From USA Today (http://www.usatoday.com/sports/football/nfl/eagles/stats.htm) we see that in the 2004 season McNabb threw 32 passing touchdowns and threw (as opposed to caught) 11 interceptions. Assuming this ratio of passing touchdowns to interceptions continues in the Super Bowl then the probability of a touchdown pass before an interception would be 32/(32+11) = 74.42%. The probability of interception first is 11/(32+11) = 25.58%.

As we see from the decimal conversion the player will get back 1.5882 units if he bets on a TD pass first and wins. One unit is his original bet back and the 0.5882 is his winnings.

The expected value of betting on a TD first would be pr(TD pass first)×win - pr(loss) = 74.72%×0.5882 - 25.58% = +18.37%. In other words, if my assumptions are correct (a big if), the player would have an 18.37% advantage.

The expected value of the interception bet first is pr(interception first)×win - pr(loss) = 25.58%×1.4 - 74.72% = -38.91%. In other words the house edge on the interception first is 38.91%.