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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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    This is why I hate this question. It's like they teach you something and when you try to use it, you get rekt. I get they're trying to teach you to think outside of the box, but come on...2018-02-27

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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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    Could you explain how you obtained the individual probabilities? In particular, $P($Bill and/or Tom obtains "head" and Mary doesn't$) = (1−(1−p_1)(1−p_3))(1−p_2) \cdot 0$ and $P($no-one obtains "head"$) = (1−p_1)(1−p_2)(1−p_3) \cdot p$2018-02-27