# Probability of winning a Craps game

A Craps game consists of throwing 2 dices. If the sum is either 7 or 11, you win. Else, if the sum is either 2,3 or 12, then you loose. If the sum is either 4, 5, 6, 8, 9 or 10, then let's call the sum $$X$$. You will throw the two dices again, until the sum 7 (you loose) or the sum is $$X$$ (you win). What is the probability of winning the game?

I made some calculations, yet I want to verify if the result is right.

The probability of winning in the first throw, $$V_1$$ is:

$$P(V_1) = \frac{\{\text{sum is either 7 or 11}\}}{36} = \frac{8}{36}$$

Assume the sum is $$x \in X =\{4,5,6,8,9,10\}$$. The probability of winning in the second throw, knowing the result of your first throw is $$x$$ is given by:

$$P(V_2) = 2\left(\frac{3^2}{36^2}+\frac{4^2}{36^2} +\frac{5^2}{36^2}\right)$$

Because $$P(X=4)=P(X=10)=\frac{3}{36}$$ $$P(X=5) = P(X=9) = \frac{4}{36}$$ $$P(X=6) = P(X=8) = \frac{5}{36}$$

Following the same track of thought, one may have the probability of winnin in the $$\text{3^{rd}}$$ throw is:

$$P(V_3) = 2\left( \frac{3^2}{36^2}\frac{27}{36}+\frac{4^2}{36^2}\frac{26}{36} + \frac{5^2}{36^2}\frac{25}{36}\right)$$

Because you need the second throw not to be $$7$$ neither the sum you had in first round $$X$$.

We can extend the probability of winning in the n-th round as:

$$P(V_n) = \frac{3^2}{36^2} \left(\frac{27}{36}\right)^{n-2} +\frac{4^2}{36^2} \left(\frac{26}{36}\right)^{n-2} + \frac{5^2}{36^2}\left(\frac{25}{36}\right)^{n-2}$$

So we can evaluate the sum as a geometric series, when $$0 < \alpha <1$$:

$$\sum_{n=0}^{\infty} \alpha^n = \frac{1}{1-\alpha}$$

So we have the probability of winning:

$$P(V) = \frac{8}{36} + 2\left( \frac{3^2}{36^2} \frac{36}{9} + \frac{4^2}{36^2}\frac{36}{10} + \frac{5^2}{36^2}\frac{36}{11} \right)$$

$$P(V) = \frac{8}{36} + 2\left( \frac{1}{36} + \frac{25}{36\cdot 5} + \frac{25}{18 \cdot 11} \right) \approx 0,493$$

What do you think? Is it right?

## 2 Answers

I'd attack the problem of whether you successfully score a point before crapping out with a $$7$$ a little more simply. If you have a point, the only issue is whether you roll that point before you roll a $$7$$. All other throws are irrelevant so you don't need to walk through the geometric series. Just use a Bayesian calculation -- given that a throw is either a $$z$$ or a $$7$$ (where $$z$$ is your point), what is the probability that it's a $$z$$?

So if your point is $$4$$ or $$10$$, your chance of winning is $$\frac{1}{3}.$$

If your point is $$5$$ or $$9$$, your chance of winning is $$\frac{2}{5}$$.

If your point is $$6$$ or $$8$$, your chance of winning is $$\frac{5}{11}$$.

So by my calculation

$$P(V)=\frac{2}{9}+\frac{1}{6}\frac{1}{3}+\frac{2}{9}\frac{2}{5}+\frac{5}{18}\frac{5}{11} = \frac{244}{495} \approx 49.29\%.$$

• Thank you Robert! It's an elegant approach. I tend to overcomplicate things :) – M.Gonzalez Mar 22 at 0:50

Let's solve the problem in a much easier manner. Denote the probability of winning the game after $$n$$ times playing Craps by $$p_n$$. Therefore$$p_n{=\Pr(\text{Winning at the n-th chance})\\=\Pr(\text{Postponing the first n-1 times and} \\\ \ \ \text{ winning the last one})\\=\Big(\Pr(\text{The sum takes on values 4,5,6,8,9,10})\Big)^{n-1}\\\ \ \ \cdot \Pr(\text{The sum possesses the values 7,11})}$$From the other side$$\Pr(\text{The sum takes on values 4,5,6,8,9,10})=1-{8\over 36}-{4\over 36}={2\over 3}\\\Pr(\text{The sum possesses the values 7,11})={2\over 9}$$therefore $$p_n=\left({2\over 3}\right)^{n-1}\cdot {2\over 9}$$so we can write $$\Pr(\text{Winning at some time})=\sum_{n=1}^\infty p_n={2\over 9}\cdot{1\over 1-{2\over 3}}={2\over 3}$$