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I'm going to post again because I thought of something else that might be an answer to your question. Probability theory is complicated to learn and it's very tedious to calculate something like this. If I were solving this problem, I would write a program to simulate a few thousand card draws and just look at the relative frequency. If you wrote it in ...


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The name of that sort of problem in probability is 'draw without replacement'. When you do this sort of problem, it's easier to first figure out the probability of not getting what you want, then subtracting that from 100% to get the probability you're interested in. First, we'll find probability of finding H alone. We start by finding the probability of ...


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Here is an incredibly simple way to figure this out without graphs or calculators. Simply count one of your soldiers defeated for every other soldier you intend to defeat & one of your soldiers left behind for every territory your conquer. Example: If you wish to eliminate another player that has 12 soldiers spread over 5 connected territories & that ...


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Let's set a lower bound on the likelihood of winning with doubles, by simply ignoring all cases where it is impossible to win without doubles. Assumption: All board positions considered are equally likely. This is probably not true, but will approach truth in longer games. Consider the case of two men only left on the board, both in the home court, and not ...



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