Infinite Series ProblemDate: 01/15/2002 at 14:41:02 From: Flori Subject: Infinite series and geometric sequences Hi, I have been given the task of finding the correct answer to an infinite series question: You have three people, a, b, and c. Each throws a die once, in turn. If a gets a 6 on his first throw, he wins. If a doesn't get a 6, it's b's turn, but b has to get a 4 or 5 to win. If b doesn't get a 4 or 5, it's c's turn, but c has to get a 1, 2, or 3 to win. If c doesn't get a 1, 2, or 3, then it goes back to a. I have gotten this far: p(a wins) = 1/6 + 1/6*5/18 + 1/6*(5/18)2 + 1/6*(5/18)3 etc.... This has the same kind of thing for p(b wins) and c(wins), but I don't know how to get a final answer for the problem. Date: 01/15/2002 at 17:17:57 From: Doctor Schwa Subject: Re: Infinite series and geometric sequences Neat formula, and nice work! To sum the infinite series like that, you'll need the formula 1 + x + x^2 + x^3 + ... = 1/(1-x) Factor out the 1/6 from your formula and you should be able to make it work. A good check will be if p(a wins) + p(b wins) + p(c wins) = 1. If you'd like to know where that formula comes from, just ask. By the way, an alternate shortcut (which also incidentally proves that formula) would be to say that p(a wins) = 1/6 + (probability that a gets another turn) * p(a wins). I'm thinking that p(a gets another turn) is 5/18, because that's what happened in your formula ... the neat shortcut is that instead of going around again, as you did, you can just use the p(a wins) again. So p(a wins) = 1/6 + (5/18)*p(a wins), (13/18)*p(a wins) = 1/6, p(a wins) = 3/13. Neat! Thanks for the fun problem, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 01/16/2002 at 14:40:31 From: Flori Subject: Cheers Thanks so much for the quick response. It really helped! Floz |
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