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Topic: COIN FLIP GAME WORTH III (Read 1800 times) |
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AlexH
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Posts: 156
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COIN FLIP GAME WORTH III
« on: Aug 17th, 2002, 9:56am » |
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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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Eric Yeh
Senior Riddler
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Re: COIN FLIP GAME WORTH III
« Reply #1 on: Aug 17th, 2002, 11:42am » |
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I agree -- these are definitely a different level of complexity than I.
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william wu
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Re: COIN FLIP GAME WORTH III
« Reply #2 on: Aug 18th, 2002, 2:36am » |
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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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