2

In craps, say I roll a 5 for the place bet. What is the probability that I roll another 5 before rolling a 7?

Is this correct?: $P(5 \text{ before 7}) = P(5) + P(\neg 7 \neg 5, 5) + P(\neg 7 \neg 5, \neg 7 \neg 5, 5) + ....$. This becomes $\frac{4}{36}+\frac{26}{36}\frac{4}{36} + ...\left( \frac{26}{36} \right)^n\frac{4}{36} = \frac{1}{9} \sum \limits_{i=0}^n \left( \frac{26}{36} \right)^i = \frac{1}{9} \frac{36}{10} = \frac{2}{5}$ ?

Dilip Sarwate
  • 25,197
adfasdf
  • 71
  • two dice in craps. im talking about their sum – adfasdf Jan 19 '14 at 04:38
  • 1
    Please accept this answer and answers to a couple of other questions you asked by pressing the tick if you are happy with the answer. It is unlikely you will receive future help on math.stackexchange if people notice you never accept any answers. – Lost1 Jan 21 '14 at 16:33

2 Answers2

1

Your notation $\neg 7 \neg 5$ confused me a bit and so I composed a long answer, but now that I understand what you wrote a little better, yes, what you have written is correct.

When you roll a 5, that becomes your point and then you repeatedly roll the dice until either your point shows up and you win, or you roll a 7 and you lose. Thus, having established a point of 5, your (conditional) win probability is $$P(5)+P(N,5) + P(N,N,5) + \cdots = \frac{1}{9} + \frac{13}{18}\times \frac{1}{9} + \left(\frac{13}{18}\right)^2\times \frac{1}{9} + \cdots = \frac{1}{9}\times \frac{1}{1-\frac{13}{18}} = \frac{2}{5}$$ where $N$ is the event that the roll is neither 5 nor 7 (what you have written as $\neg 7 \neg 5$).

More generally, if $A$ and $B$ are mutually exclusive events, then on a sequence of independent trials, the probability that $A$ occurs before $B$ is $\displaystyle \frac{P(A)}{P(A)+P(B)}.$

See, for example, this answer.

Community
  • 1
Dilip Sarwate
  • 25,197
  • Funny that this is almost identical to the question i just answered. – Lost1 Jan 21 '14 at 16:25
  • @Lost1 I expect that essentially the same question has been asked and answered on this site hundreds of times. Unfortunately, the search mechanisms are not that great in finding the duplicates and closing a question as a duplicate of one that has been asked before. – Dilip Sarwate Jan 21 '14 at 16:28
  • A thought just occurred to me: there is a nice way of formulating it as a Markov chain hitting probability with 3 states, 0 5 7 and ask what is the hitting prob of 5 before 7. – Lost1 Jan 21 '14 at 16:31
  • @Lost1 Markov chains might be overkill. Since the "rest" state 0 is "neither 5 nor 7", we are wanting the probability that the first departure from 0 state is to state 5 instead of state 7 and this is just $P(5)/(P(5)+P(7))$ without needing to sum series and the like. See my other answer cited above for a solution via this conditional probability viewpoint. – Dilip Sarwate Jan 21 '14 at 16:37
  • Yes but it allows us to arrive at the result by solving 2 simultaneous equations, instead summing a geometric discussion. Anyhow, i dunno why need we discuss this trivial question. – Lost1 Jan 21 '14 at 16:44
0

Here is an alternative way of using conditional probability, beside the answer via conditional probability by @Dilip Sarwate in another question. (This answer is adapted from the book "An Introduction to Stochastic Modeling" [Section 2.2 The Dice Game Craps] by Pinsky and Karlin)

Consider repeated rolls of the pair of dice and let $Z_n$ ($n \in \mathbb{N}$) be the sum observed on the $n$-th roll. Denote the event of "$\textrm{5 before 7}$" by $B$ and its probability $\alpha$.

According to the law of total probability, we obtain

$$\alpha = P \{ B\} = \sum_{k=2}^{12} P \{B \mid Z_1 = k \} P \{ Z_1 = k\}$$

Now we have

$$P \{ B \mid Z_1 = 5 \} = 1, \quad P \{ B \mid Z_1 = 7 \} = 0.$$

If the first roll results in anything other than 5 or 7, the problem is repeated in a statistically identical setting. That is,

$$P \{ B \mid Z_1 = k \} = \alpha, \forall k \neq 5, 7.$$

Therefore, we have

$$\alpha = P \{ Z_1 = 5 \} \times 1 + P \{ Z_1 = 7 \} \times 0 + \sum_{k \neq 5,7} P \{ Z_1 = k \} \times \alpha \\ = P \{ Z_1 = 5 \} + [1 - P \{ Z_1 = 5 \} - P \{ Z_1 = 7 \}] \alpha$$

Solving this equation, we get

$$\alpha = \frac{P \{ Z_1 = 5 \}}{P \{ Z_1 = 5 \} + P \{ Z_1 = 7 \}} = \frac{\frac{4}{36}}{\frac{4}{36} + \frac{6}{36}} = \frac{2}{5}.$$

Community
  • 1
hengxin
  • 3,687
  • 26
  • 49