Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I would like to solve this version of the unfair coin problem.

The coin has a probability p for head and 1-p for tail as usual, but the winner is the person first to reach the number of wins W.

For example, one player choose head, and the other choose tail. The players flip the coin until someone reach W=5 wins.

For W=2 we have the following possible outcomes:

AA - A wins

ABA - A wins

ABB - B wins

BAA - A wins

BAB - B wins

BB - B wins

Note that the case AAA and BBB doesn't make sense because at the second flip the one player already won.

For W =3 we have:

AAA - A wins

AABA - A wins

AABBA - A wins

AABBB - B wins

ABAA - A wins

ABABA - A wins

ABABB - B wins

ABBAA - A wins

ABBAB - B wins

ABBB - B wins

BAAA - A wins

BAABA - B wins

BAABB - B wins

BABAA - A wins

BABAB - B wins

BABB - B wins

BBAAA - A wins

BBAAB - B wins

BBAB - B wins

BBB - B wins

etc...

The maximum number of flips required to win is clearly W + (W-1) = 2 * W-1.

The number of possible outcomes, if I got it right, should be 2 * [sum(i=1,W-1) {W*i)+1], that is W^2*(W-1)+2.

That is 6 for W=2, and 20 for W=3.

However, I couldn't find the probability of winning for A (and B) as a function of W and p: A(W,p).

Do you know of any algorithm for solving this problem? I'm not an expert but I guess this is a typical Markov's chain problem.

Thanks

pagheca

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Flipping unfair coins (harder)

**Physics Forums | Science Articles, Homework Help, Discussion**