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Topic: The Toad and the Water Lilies (Read 1294 times) 

ThudnBlunder
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The Toad and the Water Lilies
« on: Jun 6^{th}, 2008, 9:40am » 
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A toad hops down a long line of water lilies. Before each hop it flips a fair coin to decide whether to hop two lilies forward or one lily back. What is the expected fraction of the water lilies it will land on?


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Eigenray
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Re: The Toad and the Water Lilies
« Reply #1 on: Jun 10^{th}, 2008, 12:15pm » 
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Bit tricky, this one, but nice: hidden:  Let p_{n} be the probability that the toad ever lands on the pad n units to the right. We have p_{0} = 1, and for n != 0, p_{n} = (p_{n2}+p_{n+1})/2, or p_{n+1} = 2p_{n}  p_{n2}. The characteristic polynomial of this recurrence is x^{3}  2x^{2} + 1 = (x1)(x^{2}x1), which has roots 1, r, and s, where r = (1+sqrt 5)/2, s = (1sqrt 5)/2. But because the recurrence only holds for n!=0, we have to solve it twice: p_{n} = A + Br^{n} + Cs^{n}, n >= 1; p_{n} = X + Yr^{n} + Zs^{n}, n <= 0. This has 6 unknowns, but we also know the following: (a) p_{n} is bounded for all n. This gives B=0 and Z=0. (b) p_{n} converges to 0 as n goes to infinity. This gives X=0. (c) p_{0} = 1. This gives A+C=1, Y=1. Equating the two formula for p_{1} gives finally 1/r = A+C/s = A+(1A)/s, and therefore A = (rs)/[r(1s)] = (3sqrt5  5)/2, which is the limit of p_{n} as n goes to infinity.  We can also say that the expected value of the farthest left the toad ever goes is (1+sqrt5)/2.

« Last Edit: Jun 10^{th}, 2008, 12:17pm by Eigenray » 
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Barukh
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Re: The Toad and the Water Lilies
« Reply #2 on: Jun 12^{th}, 2008, 6:14am » 
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Nicely done, Eigenray!


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ThudnBlunder
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Re: The Toad and the Water Lilies
« Reply #3 on: Jun 12^{th}, 2008, 7:23am » 
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on Jun 12^{th}, 2008, 6:14am, Barukh wrote:Nicely done, Eigenray! 
 Yes, indeed! A less classical, more handwavy sort of solution is as follows: Let p = the probability that a toad starting at 1 regresses to 0 at some point. To never regress it must jump forward (probability = 1/2) and not subsequently regress three times (probability = 1  p^{3}) So we have 1  p = (1  p^{3})/2 giving p = (5  1)/2 Now consider the probability that during a particular jump the toad leaps over a lily that it never has and never will land on. For this to happen it must a) be jumping forward (probability = 1/2) b) never regress from where it lands (probability = 1  p) c) not have reached in the past the lily it is jumping over (probability = 1  p) So the probability that a particular jump carries the toad over a skipped lily = (1  p)^{2}/2 Since the toad travels at an average rate of 1/2 lily per jump, the fraction of skipped lilies = (1  p)^{2} Hence the fraction of water lilies it will land on = 1  (1  p)^{2} = p(2  p) = (35  5)/2 0.854102...


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jollytall
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Re: The Toad and the Water Lilies
« Reply #4 on: Aug 9^{th}, 2008, 12:40pm » 
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Shouldn't it be devided by two, for he is only jumping  practically  into one direction? The question talks about a long line, not a halfline.


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