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Old 11-30-2007, 09:46 PM
pzhon pzhon is offline
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Default Re: probability of n ties in a race to three (roshambo)

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thanks, sometimes I think phzon forgets that he isn't talking to grad students.


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I hope it doesn't take a graduate student to evaluate (n+2) choose 2, the most complicated part of the formula I gave.

When I give an answer, I often expect an interested reader to take some time to think about what I said. The amount of time it takes depends on the reader's interest and background. If you expect to get everything at the first reading, you will be disappointed in mathematics.

Your complaints made me regret helping you again. I'm not inclined to spoonfeed you, and if I were, what I would need to say to make everything clear to you would depend on your background, which I don't know.

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when you look at problems like this are you like "duh.... so obviously geometric distribution, and so obvious that it is equally like to end in two decisive throws as three."

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No. Here is a rough transcript of my thought processes: "Is there anything ifu could have meant other than best of 3 decisive results? I don't think so. After 2 decisions, there is a 50% chance it's over, and a 50% chance it will take one more. The number of ties before each decisive result is geometric. What is the convolution of geometric distributions called again? "Hypergeometric" would make sense, but I think that's something else. Yes, Wikipedia says hypergeometric is something else. Oh well. It's easy to read off the explicit formula from basic combinatorics, but that's prone to producing errors. Let me put the formula into Mathematica to check that the total probability is 1. Yup. Check a few values for plausibility. Hmm, the denominators for P(0) and P(1) are equal to 27, but that pattern can't continue. The expected value is 5/4. That makes sense, as that is the average of 2/2 and 3/2. Post."
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