# Probability about a geometric distribution

Bill, Mary and Tom have coins with respective probabilities $p_1,p_2,p_3$ of turning up heads. They toss their coins independently at the same times.

What is the probability that neither Bill nor Tom get a head before Mary?

I tried to write it down as a geometric distribution but the equation became extremely long. I believe there must be an easier way to look at this problem.

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## 1 Answer

Each group of simultaneous tosses has three possible outcomes. If Mary obtains "head", the game is over with positive outcome. If Bill and/or Tom obtains "head" and Mary doesn't, the game is over with negative outcome. If no-one obtains "head", the game continues with the same probabilities as before. Thus, the probability of a positive outcome is the sum of three contributions:

$$\begin{eqnarray} p&=&p_2\cdot1+(1-(1-p_1)(1-p_3))(1-p_2)\cdot0+(1-p_1)(1-p_2)(1-p_3)\cdot p\\ &=&p_2\cdot1+(1-p_1)(1-p_2)(1-p_3)\cdot p\;. \end{eqnarray}$$

Solving for $p$ yields

$$p=\frac{p_2}{1-(1-p_1)(1-p_2)(1-p_3)}\;.$$

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