Should you take the bet?
A gambler is willing to bet at even odds that using two ordinary 6-sided dice, he'll throw a 6 and an 8 before throwing two 7's. Should you take the bet?
Note that the rolls need not be in consecutive order.
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1 solution
You should clarify what "throw a 6 and an 8" means.
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It means exactly what you think it means. Roll a combination of 2 dice totaling 6 (1-5, 2-4, 3-3) and a combination of 2 dice totaling 8 (4-4, 5-3, 6-2), getting both such combinations before twice rolling a combination totaling 7 (4-3, 5-2, 6-1).
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What I was going for was "Do they have to be consecutive rolls?"
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@Calvin Lin – Nope. If you roll a 4, a 6, a 9, a 10, a 7, a 2, an 11, a 5, and an 8, in that order, you win, because you rolled a 6 and an 8 before rolling two 7s.
I enjoyed looking at the problem using the expectation values for how long each outcome would take in numbers of throws. For example, the expectation value for the number of throws it would take to throw two sevens is given by k = 2 ∑ ∞ k ( k − 1 ) ( 3 6 6 ) 2 ( 3 6 3 0 ) k − 2 = 1 2 whereas the expectation value for the number of throws it would take to throw both a six and an eight is k = 1 ∑ ∞ k ( 3 6 1 0 ) ( 3 6 2 6 ) k − 1 + k = 1 ∑ ∞ k ( 3 6 5 ) ( 3 6 3 1 ) k − 1 = 5 1 8 + 5 3 6 = 1 0 . 8 We can find the values of these sums a few different ways, but I think the most powerful is using polylogarithms .
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That is the wrong way to approach this problem. See here to understand why lower expected value doesn't necessarily imply will occur first. You are not accounting for the positive / negative covariance.
But he has to throw both 6 and 8 before a 7 as mentioned in the question 6"and"8
The chances of throwing a 6 or an 8 with 2 dice are 10 in 36, while the chances of throwing a 7 are 6 in 36. So there are 10 chances out of 16 that he'll throw a 6 or an 8 before a 7, and 6 chances in 16 he'll throw a 7 first.
Once he throws either a 6 or an 8, the chances of him throwing the other one are 5 in 36, while the chances of him throwing a 7 are 6 in 36. So the chances of throwing the 7 first is 6 in 11.
Combining these facts, his chances of winning are (10/16) * (5/11) [throws them both before a 7] + (10/16) * (6/11) * (5/11) [throws one, than a 7, than the other] + (6/16) * (10/16) * (5/11) [throws a 7, then both a 6 and an 8.] This adds up to 54.558%, making him the favorite at even odds.