Re: The old coin-flip debate (long including quotes)
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I just can't see how you can have an equal chance of finishing 2nd or 3rd. You would have to be more likely to finish 2nd.
Let's assume the one of the shortstacks is going to win. What are the chances that you finished 2nd? Well, you have 4000 and the other shortstack has 2000....so I'd say it's 67% (4000/6000) that you "win" 2nd place.
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I'm not sure what you mean when you say that you're assuming that one of the short stacks is 'going to win'. Would you be more specific? It would help me respond. Using Sklansky's method for three people, the chances of the first player finishing in each position is determined by figuring out the chances of the second and third players finishing in each position and subtracting those from 100%.
Brett
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