The game of craps and dice

Question

The game of craps involves the repeated tossing of a pair of dice. In the game of craps, Nancy throws a five on the first toss of a pair of dice. ["Five" means that the sum of the number of dots on both dice is equal to five.] The rules call for her to keep on tossing the dice until she gets a five again, in which case she wins, or until she gets a seven, in which case she loses. What is the probability that she will win.

My attempt:

$P$ $($Sum of five occurs$)=$ $(2/6)+ (1/6)(2/6)+(1/6)^2(2/6)+...$

$P$ $($Sum of five occurs$)$ $=$ $\sum_{x=1}^\infty (2/6)(1/6)^{x-1}$

$P$ $($Sum of five occurs$)=$ $(2/6)/1-(1/6)= 2/5$

I know the answer is correct. However I am not sure if this is the way how you would show the working. Is there any other way we can show how to get this probability with a formula?

Answer 1

On each roll, the probability of total 5 is $4/36$, and the probability of total 7 is $6/36$.

Notice that if Nancy rolls any total other than 5 or 7, her situation has not changed at all: she still needs to keep rolling until the dice show 5 or 7, and will get the same results. So her probability of a win after that roll is the same as it was before.

$P(\mbox{win}) = P(\mbox{roll 5}) P(\mbox{win} \mid \mbox{roll 5}) +P(\mbox{roll 7}) P(\mbox{win} \mid \mbox{roll 7}) + P(\mbox{roll other}) P(\mbox{win} \mid \mbox{roll other})$

$P(\mbox{win}) = \frac{4}{36} \cdot 1 + \frac{6}{36} \cdot 0 + \frac{26}{36} P(\mbox{win})$

$\frac{10}{36} P(\mbox{win}) = \frac{4}{36}$

$P(\mbox{win}) = \frac{2}{5}$.

Answer 2

Under the stated rules, the only sums that matter are $5$ and $7$, which have probabilities $\frac{4}{36}$ and $\frac{6}{36}$ respectively.

The probability she got a $5$ given she got a $5$ or a $7$ is $$\frac{\frac{4}{36}}{\frac{4}{36}+\frac{6}{36}}.$$ The formula is the standard conditional probability formula.

Answer 3

At the $n$th roll, the probability is simply the probability of a 5 hitting, while the $n-1$ previous rolls were not 5 or 7.

There are 6 ways to roll a 7, and 4 ways to roll a 5, so the probability that the dice show neither on any given roll is $\frac{26}{36}$.

So the probability of winning on the $n$th roll is $\left(\frac{26}{36}\right)^{n-1} \frac{4}{36}$.

The overall probability of winning is the probability of winning on the first roll, plus the second, plus the third, and so on:

$$P[win] = \sum_{k=1}^\infty \left(\frac{26}{36}\right)^{k-1} \frac{4}{36} = \frac{\frac{4}{36}}{1-\frac{26}{36}} = 0.4.$$

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Gambler's note: Probability of $0.4$ corresponds to odds of $2:3$, or "two dollars to win three," which is exactly the payoff of odds bets on the pass/come lines for 5s and 9s on the craps table. The pass/come lines, however, only pay even money, so it's clear that odds bets are better. Of course, to play odds, you have to play the point as well.