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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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    @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*).2012-01-22

1 Answers 1

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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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    Thank you very much! It was very helpfull.2012-01-22