Find the Mistake In My Calculation For the Odds of Hitting The Fire Bet In Craps.

Question

Background Information: Inside the game of craps, one of the most notorious bets, the fire bet, is known for being exceptionally complicated and hard to win. In order to hit the fire bet, every number must be made as a point. This means that the shooter must roll a number, and then continue to roll until he roles that number again; however, he roles a 7 before he roles the number a second time, the bet is lost. The goal of the shooter is to do this for the 4, 5, 6, 8, 9, and 10.

Question: What are the Odds of a shooter making this Fire Bet?

My Attempt at a solution (Wrong): First, I will try to find the odds of rolling a four before rolling a seven. $\sum_{x=0}^\infty P(4) * P(Anything But 4 or 7)^x$ Then, I will find the probability of rolling a 5 before a 7 and a 6 before a 7. (8, 9, 10 share the same exact probabilities as 4, 5, and 6.)
$$\sum_{x=0}^\infty P(4) * P(Anything But 4 or 7)^x$$ $$\sum_{x=0}^\infty P(5) * P(Anything But 5 or 7)^x$$ $$ \sum_{x=0}^\infty P(6) * P(Anything But 6 or 7)^x$$

This is equal to 1/3, 2/5, and 5/11 respectively.

$$ \sum_{x=0}^\infty \frac{1}{12} * \frac{3 }{4}^x = \frac{1}{3} $$ $$ \sum_{x=0}^\infty \frac{1}{9} * \frac{13}{18} ^x = \frac{2}{5} $$ $$ \sum_{x=0}^\infty \frac{5}{36} * \frac{25}{36} ^x = \frac{5}{11} $$

Now, I can start putting together the final answer, which ended being incorrect. The shooter must shoot a 4 ($\frac{1}{12}$) and then shoot the 4 again before a 7 ($\frac{1}{3}$). The probability of making the four is $\frac{1}{12} * \frac{1}{3}$. We can now do the same for the 5, 6, 7, 8, 9, and 10. This leaves us with the final answer, which ended up being incorrect.

$$(\frac{1}{12} * \frac {1}{3}) * (\frac{1}{9} * \frac{2}{5}) * (\frac{5}{36} * \frac{5}{11}) * (\frac {5}{36} * \frac{5}{11}) * (\frac{1}{9} * \frac{2}{5}) * (\frac{1}{12} * \frac{1}{3}) $$

This gives us the final answer of $\frac{5}{8019}$ ; however, this answer is completely wrong. Can someone help me spot my mistake?

(Please note that the first set of computations that yield the answer $\frac{1}{3}$ , $\frac{2}{5}$ , and $\frac{5}{11}$ are correct, but the final answer is not correct. Thank You.)

The answer I got isn't correct.

Answer 1

This looks like a long calculation.

Let's look at the probability just for making the points $4,5,6,8,9,10$ in that order without ever have a seven-out on a point:

• Probability the first point value is $4$ is $\frac{3}{24}$
• Probability of then making that point is $\frac{3}{9}$
• Probability then rolling new point value before seven-out is $\frac{\frac{21}{24}}{1-\frac{3}{24}\frac{3}{9}}$
• Probability then new point value is $5$ is $\frac{4}{21}$
• Probability of then making that point is $\frac{4}{10}$
• Probability then rolling new point value before seven-out is $\frac{\frac{17}{24}}{1-\frac{3}{24}\frac{3}{9}-\frac{4}{24}\frac{4}{10}}$
• Probability then new point value is $6$ is $\frac{5}{17}$
• Probability of then making that point is $\frac{5}{11}$
• Probability then rolling new point value before seven-out is $\frac{\frac{12}{24}}{1-\frac{3}{24}\frac{3}{9}-\frac{4}{24}\frac{4}{10}-\frac{5}{24}\frac{5}{11}}$
• Probability then new point value is $8$ is $\frac{5}{12}$
• Probability of then making that point is $\frac{5}{11}$
• Probability then rolling new point value before seven-out is $\frac{\frac{7}{24}}{1-\frac{3}{24}\frac{3}{9}-\frac{4}{24}\frac{4}{10}-\frac{5}{24}\frac{5}{11}-\frac{5}{24}\frac{5}{11}}$
• Probability then new point value is $9$ is $\frac{4}{7}$
• Probability of then making that point is $\frac{4}{10}$
• Probability then rolling new point value before seven-out is
  $\frac{\frac{3}{24}}{1-\frac{3}{24}\frac{3}{9}-\frac{4}{24}\frac{4}{10}-\frac{5}{24}\frac{5}{11}-\frac{5}{24}\frac{5}{11}-\frac{4}{24}\frac{4}{10}}$
• Probability then new point value is $10$ is $\frac{3}{3}$
• Probability of then making that point is $\frac{3}{9}$

You multiply all those together and get a very small number. Then you do the same for the other $6!-1$ permutations. Then add them all up.

Here is some R code that does that:

possiblepoints <- c(4:6, 8:10)
numpoints <- length(possiblepoints)
numperms <- factorial(numpoints)
ways <- 6 - abs(7 - (1:12)) 

permutations <- function(n){
    ## from https://stackoverflow.com/a/20199902/672221
    ## by user Museful 
    if(n==1){
        return(matrix(1))
    } else {
        sp <- permutations(n-1)
        p <- nrow(sp)
        A <- matrix(nrow=n*p,ncol=n)
        for(i in 1:n){
            A[(i-1)*p+1:p,] <- cbind(i,sp+(sp>=i))
        }
        return(A)
    }
}

pointperms<- matrix(possiblepoints[permutations(numpoints)],ncol=numpoints)

probabilities <- rep(1, numperms)
notcomeout <- 6^2 - ways[2] - ways[3] - ways[12] - ways[7] - ways[11]
remainways <- rep(notcomeout, numperms)
newdenom <- rep(1, numperms)

then going through the six points

for (k in 1:numpoints){
  probnewpointbeforesevenout <- remainways / notcomeout / newdenom 
  probabilities <- probabilities * probnewpointbeforesevenout 
  probthisnewpoint <- ways[pointperms[,k]] / remainways 
  probabilities <- probabilities * probthisnewpoint 
  probmakingpoint <- ways[pointperms[,k]] / (ways[pointperms[,k]]+ways[7])
  probabilities <- probabilities * probmakingpoint
  remainways <- remainways - ways[pointperms[,k]] 
  newdenom <- newdenom - probmakingpoint*ways[pointperms[,k]]/notcomeout   
}

sum(probabilities)
# 0.0001624347
1/sum(probabilities)
# 6156.318

and those final figures seem to be the same as values published by Mark Bollman and Michael Shackleford

Answer 2

The following solution uses exponential generating functions to show that the probability of winning the fire bet is $0.000162435$.

Let $p_n$ be the probability of rolling $n$ with a pair of dice: $$\begin{matrix} n &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12\\ p_n &1/36 &2/36 &3/36 &4/36 &5/36 &6/36 &5/36 &4/36 &3/36 &2/36 &1/36\\ \end{matrix}$$ If the point (the first roll in a game) is $n$, then the probability of rolling the point before rolling a $7$ is $$\frac{p_n}{p_n + p_7}$$ The rules require that the point is one of $\{4,5,6,8,9,10\}$. Consequently, the probability of rolling a point of $n$ and also winning the game is $$s_n =\frac{p_n^2}{p_n + p_7}$$

As a first step, we would like to find the probability of winning the fire bet when the last point made is $4$, so we have a sequence of games in which the player never "sevens out", all of the points $5,6,8,9$ and $10$ are rolled with success at least once, and the final point of $4$ (which is won) is the first occurrence of $4$ as a point. Any number of initial "points" of $2,3,7,11$ or $12$ may appear in the sequence of points; these have no effect on the fire bet. Let $a_n$ be the probability of a sequence of $n$ games in which all the points $5,6,8,9$ and $10$ are rolled with success at least once and $4$ is never a point--in other words, a sequence of games suitable to win the fire bet with final point of $4$ up to just before rolling $4$ as a point. The exponential generating function of $a_n$ is $$f_4(x) = e^{(p_2+p_3+s_7+p_{11}+p_{12})x} \prod_{n \in \{5,6,8,9,10\}} (e^{s_n x}-1)$$

The probabiity of winning the fire bet when the last point made is $4$ is then $$w_4 = s_4 \sum_{n=0}^{\infty} a_n = s_4 \int_0^{\infty} e^{-x} f_4(x) \; dx = 0.0000299565 \tag{*}$$ (I used Mathematica to evaluate the integral, a task which would otherwise be tedious and error-prone.)

Proceeding analogously for the cases in which the final points in a winning sequence are $5$ or $6$, we find $w_5=0.0000270913 $ and $w_6=0.0000241696$. By symmetry $w_4=w_{10}$, $w_5=w_9$, and $w_6=w_8$, s0 the probability of winning the fire bet is $$2(w_4+w_5+w_6) = \boxed{0.000162435}$$

$(*)$ Note: If $$f_4(x) = \sum_{n=0}^{\infty} a_n \frac{1}{n!}x^n$$ then $$\sum_{n=0}^{\infty} a_n = \int_0^{\infty} e^{-x} f_4(x) \; dx$$ provided the integral converges, because of the identity $\int_0^{\infty} e^{-x} x^n \; dx = n!$.