We have a special coin. When we flip it: the coin always lands in heads the first time, the second time - tails. Beginning with the third flip (n+1) the probability of getting heads is $\dfrac{m}{n}$ - where n is the number of flips, and m - number of heads obtained in n flips. (i.e if k denotes the number of tails, k+m=n).

For example:

$n=2$, probability of getting heads in the third flip is always $\frac{1}{2}$.

$n=3, (m=2, k=1)$ probability of getting heads in the $4^{th}$ flip in this case $\frac{2}{3}$

$n=3, (m=1, k=2)$ probability of getting heads in the $4^{th}$ flip in this case $\frac{1}{3}$

1) What is the probability of getting 200 heads from this coin after 300 flips?

2) What is the probability of getting A heads from this coin flip B times? (B>A) times?

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    $\begingroup$ You'll probably want to make sure that you write "getting heads" in the future. The result of a single coin flip is "heads" or "tails" in the plural, and the phrase "getting head" in the singular has a completely different meaning that you may want to google. $\endgroup$ – joriki Jan 22 '12 at 15:16
  • $\begingroup$ @joriki: Rhooo... you should be ashamed... :-) $\endgroup$ – Did Jan 22 '12 at 15:45
  • $\begingroup$ May be I'll take the pain of making this question answerable! I am unsure of the probability of getting .... when a coin is tossed! $\endgroup$ – user21436 Jan 22 '12 at 15:53
  • $\begingroup$ @Didier: Me? :-) I was just trying to help the OP to avoid a faux pas in the future when the audience might be less forgiving than here. About "Rhooo": I get a lot of Google hits for that, most of them French, but none that explain what it means... $\endgroup$ – joriki Jan 22 '12 at 15:54
  • $\begingroup$ @joriki: Das ist nur ein Lautmalerei... The sound is supposed to evoke some reprobation (but I probably should mention ooo is not as in shampoo but as in cope). $\endgroup$ – Did Jan 22 '12 at 16:04

This is a famous problem in probability that is usually described as Pólya's Urn scheme.

Imagine an urn that we put balls in. We start with one ball labelled H and one labelled T to correspond to your first two coin flips. Subsequent coin flips are modelled by drawing a ball from the urn, and returning the drawn ball plus one ball with the same label.

It is well-known that after $d$ draws, the number of balls in the urn labelled H has a uniform distribution on $\{1,2,\dots, d,d+1\}$.

Accounting for the first two coin flips, the total number of heads after $B$ flips is uniform over $\{1,2,\dots, B-1\}$. So the answer to (1) is $1/299$ and the answer to (2) is $1/(B-1)$.

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    $\begingroup$ You can check this result by writing a recurrence relation for the probability $p(n,m)$ of $m$ heads in $n$ flips, $$p(n,m)=p(n-1,m-1)\frac{m-1}{n-1}+p(n-1,m)\left(1-\frac m{n-1}\right)\;,$$ and checking that the uniform distribution Byron gave satisfies it. (@Byron: You used $k$ with a different meaning than the one it was used with in the question.) $\endgroup$ – joriki Jan 22 '12 at 15:47
  • $\begingroup$ @joriki Good eye! I've changed it now. $\endgroup$ – user940 Jan 22 '12 at 15:49
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    $\begingroup$ Thank you very much! It was very helpfull. $\endgroup$ – Mike Jan 22 '12 at 19:20

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