# Question regarding Solution to Geometric Probability

I came across the following problem, while browsing my textbook, which was previously answered on this site.

Probability about a geometric distribution

I've attempted to solve the problem with the same reasoning, but with Bayes Rule, and have arrived at the same solution. However, I have trouble conceptualizing what the author meant in saying "If no-one obtains "head", the game continues with the same probabilities as before." If that is the case, why does that affect the probability recursively? Could someone explain why we add the case where no one wins to the probability, and why we multiply it by $p$?

Credit to joriki for the original solution, since the post is over a year old, I didn't want to ask a followup.

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## 1 Answer

The game has three states:

$S$:$\quad\$ So far nobody has thrown heads,

$M$: $\quad$Mary has thrown heads,

$M'$: $\quad\geq1$ of the boys has thrown heads, but Mary hasn't.

$M$ and $M'$ are terminal states with assigned probabilities $1$ and $0$ that Mary wins the game. Let $p$ denote the probability that Mary wins when we are in state $S$. The transition probabilities between the three states are as follows:

\eqalign{S\to M:\quad &p_2\cr S\to M':\quad &(1-p_2)(p_1+p_3-p_1p_3)\cr S\to S:\quad&(1-p_1)(1-p_2)(1-p_3)\cr}

Looking at the flow-chart we see that $p$ satisfies the equation $$p=p_2\cdot 1+ (1-p_1)(1-p_2)(1-p_3)\cdot p\ ,$$ as given in Joriki's answer.

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But I suppose my question should be, why have transition probabilities if the probability of each event is independent? – zhuyxn Nov 20 '12 at 12:18
@zhuyxn: The independence of the three throws at each transition is built into the transition probabilities. The independence of the throws in subsequent transitions is reflected in the fact that only the resulting state is remembered. – Christian Blatter Nov 20 '12 at 12:26