# PROBABILITY:if two fair die are rolled repeatedly. Find the probability of rolling a double three times before rolling a sum of 7 two times.?

Update:

Probability of rolling any matching pair is 6/6 * 1/6 = 1/6.

Probability of rolling any sum of 7 is 6/36 = 1/6

So I just did 1/6 *1/6 *1/6 *1/6*1/6

What do you think?

Relevance

You have already computed P(double) = P(sum of 7) = 1/6.

However, rolls that are neither have to be ignored, and in the reduced sample space, since they are equiproable, we can take p = q = 1/2

Calling a double as event A and a sum of 7 as event B,

favorable outcomes are

AAA: Pr = 1/8

AABA: Pr = 1/16

ABAA: Pr = 1/16

BAAA: Pr = 1/16

adding up, indicated Pr = 5/16 <--------

edit:

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1. Corrected, see Ian's remarks.

2. Re your last comment, the game can be taken to end as soon as A wins, and because B can at most score once, max # of rounds can at most be 4.

ps:

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On second thought, the simplest way is to compute P(x≥3) for x is binomial(4,1/2), since the "game" must end after 4 rounds

Indicated Pr = (4c3 + 4c4)/2^4 = 5/16 <-------

[ AAA gets replaced by AAAA plus AAAB ]

• Linus5 years agoReport

I am reading the book and I noticed this is related to the "the problem of the points". Which exactly says "x wins before y loses" is derived from formula C(m+n-1,k)p^k q^(m+n-1-k). Which is the binomial

• A = {(x, x)}; P(A) = 1/6

B = {(x1, x2): x1 + x2 = 7}; P(B) = 1/6

A ∩ B = {}

P({A, A, A, B, B}) = 1/6*1/6*1/6*1/6*1/6 = 1/6^5

P({A, A, B, A, B}) = ditto

.

.

P({B, A, A, A, B}) = ditto

So I think 4/36^5. Just a guess... (It could happen in 4 ways not including frivolous intervening rolls.)

Any number of rolls resulting in not(A) and not(B) could occur in the intervening rolls up to the final roll, but the sum of the probabilities of those rolls from 0 to inf. is 1, and have equal opportunity of occurring across all ways to get 3 event A's before 2 event B's, so they have no effect on the outcome???

• M3
Lv 7
5 years agoReport

4/36^5 is an inconceivably low answer for the problem by a reality check.

• M3's solution is very nearly correct. The only issue is that AAAB would already be included in the event AAA, so the AAAB probability, 1/16, should not be counted. So the answer should be 5/16 instead of 3/8.

M3 corrected the error and added a neat alternative solution; the point is that winning 3 rounds (getting doubles 3 times) *before* losing 2 rounds (getting a sum of 7 twice) is equivalent to winning at least 3 of the first 4 rounds.

Have a blessed, wonderful day!

• Linus5 years agoReport

I am reading the book and I noticed this is related to the "the problem of the points". Which exactly says "x wins before y loses" is derived from formula C(m+n-1,k)p^k q^(m+n-1-k). Which is the binomial