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Title: COIN FLIP GAME WORTH III Post by AlexH on Aug 17th, 2002, 9:56am Coin Flip game 3: I think I'd refile this under medium (and game 2 under easy). Let E_n be the expected time to equal heads and tails starting with a surplus of n heads. Then E_(n-1) = 1 + .5 (E_n + E_(n-2)) E_0 = 0 Rewriting we get E_n = 2E_(n-1) - E_(n-2) - 2 Observe that by repeated substitution this means that E_n = (1+k) E_(n-k) - k E_(n-k-1) - k(k+1) Letting k = n-1 yields E_n = n E_1 - (n-1) E_0 - (n-1)n E_n = n E_1 - n(n-1) Solving for E_1 gives us E_1 = (E_n + n(n-1))/n = E_n/n + n-1 But E_n > 0 for any n, which means that for all n, E_1 > n-1 So E_1 diverges and so does the payoff for our game. |
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Title: Re: COIN FLIP GAME WORTH III Post by Eric Yeh on Aug 17th, 2002, 11:42am I agree -- these are definitely a different level of complexity than I. |
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Title: Re: COIN FLIP GAME WORTH III Post by william wu on Aug 18th, 2002, 2:36am true, different levels of complexity. however, i think something is to be said for the peculiarity of the answers. the payoff must be finite, but the calculated expectation is infinite. then the question asks you how much you'd be willing to pay to play the game, and no one is willing to pay infinity. kind of interesting and non-intuitive. |
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