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Let us suppose that a person are throwing a coin. He'll get one dollar if he win, and he'll pays one dollar if he loses. I understand that the winning will trend to zero in the case of unlimited number of throws.

But:

  • It is possible to find the probability of the maximum winning/loss during the experiment?
  • It is possible to find the probability distribution of the zero total winning during the experiment?
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    I am assuming the coin is fair. Then with probability $1$ we will return to the origin at some time. The probability we are at the origin at time $2n$ is easy to calculate.2011-09-14
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    *I understand that the winning will t(r)end to zero in the case of unlimited number of throws*... Actually no, it will not.2011-09-14
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    @Didier Pilau - Why? As I know this is the expected value. P(O1)V(O1)+P(O2)V(O2)=1/2*1 - 1/2*1 = 02011-09-14
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    Our net "win," **divided by the number $n$ of tosses**, will probably be not far from $0$ if $n$ is large. But that does not mean that our net win will probably be close to $0$. For example, $\sqrt{n}/n$, for large $n$, is close to $0$, but $\sqrt{n}$, for large $n$, is not at all close to $0$.2011-09-14
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    You seem to put more in the expected values than there is. True, the expected net win is exactly zero at every time. But the net win itself fluctuates as much as possible. For example, the set of all the net wins after any fixed time is almost surely the whole integer line. Plus what @André said.2011-09-14
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    I don't understand your second question. Do you mean "It is possible to find the probability of zero total winning *if we stop after some fixed number of throws $n$*?" That question makes sense, and I have the answer: **Yes**. :)2011-09-14
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    @cardinal - Yes, it is close to my question. Is there any dependence between 'n' and the probability of zero total winning if we stop after some fixed number of throws 'n'? Is it function or it is constant?2011-09-14
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    Note that you can only have zero winnings after an even number of plays. So, let's say that you play $2n$ times and let $W_{2n}$ be your winnings. Then $\mathbb P(W_{2n} = 0) = {2n \choose n} 2^{-2n}$.2011-09-14

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