This is the solution to problem 9 from Fifty Challenging Problems in Probability by Frederick Mosteller (1987). The problem is paraphrased below; for reference, it is inspired by the original book.
In the gambling game of craps, we roll two fair six-sided dice and observe the total. If we roll a 7 or an 11, we win! We can also win in a more elaborate way:
- Throw anything except 2, 3, or 12. Our throw is called a point.
- Keep throwing. If we throw the point, we win! If we throw a 7, we lose. Otherwise, we just keep throwing.
What is the probability of winning a game of craps?
Let’s dive in!
Looking at some winning paths
When solving this, I first wrote down some winning sequences of totals:
- 7;
- 11;
- 4, 4;
- 5, 2, 6, 2, 8, 12, 11, 5;
- and so on.
Assuming we win, we can either do it on the first throw or with a point.
Probability of winning on the first throw
The probability of winning on the first throw is the same as the probability of rolling a 7 or an 11 immediately. How do we find those? Let’s draw a 6 by 6 table where each row represents the outcome of die D1, each column represents the outcome of die D2, and each cell represents their total.
| D1/D2 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Since there are 36 cells, the probability of rolling a total of \(x\) is equal to the count of \(x\) in the table, divided by 36. In more standard notation:
\[ X \sim \begin{pmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \end{pmatrix}. \]The probability of winning in the first throw is thus \(P(X = 7) + P(X = 11) = \frac{8}{36}.\)
Probability of winning with a point
To win this way, we first roll to get a point. Then we roll until we get the point again – or 7. A valid sequence can be described as \(n, m_1, m_2, \dots, m_{k}, n\), where \(m_i \notin \{7, n\}\) for all \(i = 1, \dots, k\).
What is the probability of such a sequence? Obviously, it depends on \(n\). We note that \(n\) must be part of a valid set of totals \(V = \{4,5, 6,8,9, 10\}\). The probability of winning with a point is the sum of probabilities of winning with each \(v \in V\). We can summarize this as follows:
\[ P_\mathrm{point-win} = \sum_{v \in V} P(v) P(\textrm{roll many times without seeing 7, nor } v) P(v). \]The two \(P(v)\) terms describe the probability of rolling \(v\) initially and as the final roll of the sequence.
How do we compute the middle term? Let’s call it \(P_S(v)\). We have to roll zero or more totals, none of which are in \(\{7, v\}\). We can write:
\[ P_S(v) = \sum_{i=0}^\infty P(\textrm{roll neither 7, nor } v)^i. \]Each term in the sum is the probability of a roll sequence of length \(i\). The sum checks all sequence lengths \(i\). We notice that \(P(\textrm{roll neither 7, nor } v)\) is always the same for a given \(v\) regardless of \(i\). Since probabilities are between 0 and 1, the entire sum is a geometric series of the form:
\[ \sum_{i=0}^\infty q^i = \frac{1}{1 - q}. \]Let’s compute this individually for all values \(v\in V\). I’ll use #\(v\) to denote the number of times \(v\) appears in the totals table. Let \(q(v) = P(\textrm{roll neither 7, nor } v) = \frac{36 - 6 - \#v}{36}\).
| \(v\) | #\(v\) | \(q(v)\) | \(P_S(v) = \sum_{i=0}^\infty q(v)^i\) |
|---|---|---|---|
| 4 | 3 | \(\frac{3}{4}\) | \(4\) |
| 5 | 4 | \(\frac{13}{18}\) | \(\frac{18}{5}\) |
| 6 | 5 | \(\frac{25}{36}\) | \(\frac{36}{11}\) |
| 8 | 5 | \(\frac{25}{36}\) | \(\frac{36}{11}\) |
| 9 | 4 | \(\frac{13}{18}\) | \(\frac{18}{5}\) |
| 10 | 3 | \(\frac{3}{4}\) | \(4\) |
The probabilities are the same for \(\{4, 10\}\), \(\{5, 9\}\), and \(\{6, 8\}\) due to symmetry in the totals table. What’s left is to compute \(P_\mathrm{point-win}\):
\[ P_\mathrm{point-win} = 2 \cdot \left( \left(\frac{3}{36}\right)^2 \cdot 4 + \left(\frac{4}{36}\right)^2 \cdot \frac{18}{5} + \left(\frac{5}{36}\right)^2 \cdot \frac{36}{11}\right) = \frac{134}{495}. \]Add the two together and we get our final win probability:
\[ P(\mathrm{Win}) = P(X = 7) + P(X = 11) + P_\mathrm{point-win} = \frac{8}{36} + \frac{134}{495} \approx 0.49293. \]We’re at a slight disadvantage, but it’s almost even odds!
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