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I'm not a mathematician, so my question may be not so clear, sorry.

Let's say we toss up an ideal coin and win 1 dollar on heads and lose 1 dollar on tails. So, expected value is M = 1×0.5 − 1×0.5 = 0. It may be interpreted as: we will gain/lose nothing playing a game for a long time. It is pretty clear.

Now, let's consider bankroll size. It will be changed on every coin flip. For Heads Tails Tails Tails Heads sequence, bankroll size will be 1, 0, -1, -2, -1 dollar. As I understand, bankroll size is a random variable. Is it even possible to evaluate this random variable somehow? How to calculate expected value and variance? May I evaluate minimum and maximum of bankroll size?

Thanks

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  • $\begingroup$ FC et al - remember you can flag questions for moderator attention. We'll close questions more promptly when this happens. Also, once you hit 3000 you can vote to close yourself! $\endgroup$
    – Kim Morrison
    Commented Nov 14, 2009 at 4:59
  • $\begingroup$ @stas, closed, per explantion by FC above. More generally, if you feel the need to preface with "I'm not a mathematician" then you ought to have a particular reason in mind to expect that your question is "of interest to mathematicians". $\endgroup$
    – Kim Morrison
    Commented Nov 14, 2009 at 5:02

2 Answers 2

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The distribution you're looking at is called the binomial distribution.

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  • $\begingroup$ Hmm, I guess expected value must be 0 but for binomial distribution it depends on number of coin flips (E(x) = np)! How it is even possible?! $\endgroup$
    – stas
    Commented Nov 13, 2009 at 23:26
  • $\begingroup$ You need to correct for the number of trials; the binomial distribution counts the total number of times you win, not your bankroll. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 14, 2009 at 1:57
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This is a classic problem. Try googling "Gambler's Ruin" and "Drunkard's Walk",

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