game: a coin is tossed and then a dice is rolled

Question

I have the following game:

Person A tosses a coin and then person B rolls a die. This is repeated independently until a head or one of the numbers 1, 2, 3, 4 appears, at which time the game is stopped. Person A wins with the head and B wins with one of the numbers 1, 2, 3, 4. Compute the probability P(A) that person A wins the game.

The book gives the following explanation:

"For an exact derivation, notice that it is implicit in the statement A wins the game that the game is completed. Using abbreviated notation, the game is completed if H or T {1,..., 4} occurs. Using independence, the probability that A wins is thus the conditional probability (1/2)/[(1/2) + (1/2)(4/6)] = 3/5."

I understand what he means by "abbreviated notation" and I also understand it must be something like

$P(\text{{A wins}}|\text{Game ended})=\displaystyle\frac{P(\text{{A wins}}\cap\text{{Game ended}})}{P(\text{game ended})}=\frac{P(\text{{A wins}}\cap\text{{Game ended}})}{P(\text{{A wins}}\cap\text{{Game ended}})+P(\text{{B wins}}\cap\text{{Game ended}})}$

I believe my question is: why are $P(\text{{A wins}}\cap\text{{Game ended}})=\displaystyle\frac{1}{2}$ and $P(\text{{B wins}}\cap\text{{Game ended}})=\displaystyle\frac{1}{2}\cdot\frac{4}{6}$. Shouldn't it be waaay more difficult to calculate these?

Could somebody further explain what is going on?

Answer 1

Let X be the event that H is thrown and Y that T followed by 1, 2, 3 or 4 is thrown.
Given that the game is over, either X or Y had to occur, and note that they are mutually exclusive, so P(X or Y) = P(X) + P(Y).
So we have
P( A won | game is over) = (P( X | X or Y) = P( X or Y | X) $\cdot$ P(X)) / P(X or Y) = $\frac{\frac{1}{2}\cdot1}{\frac{1}{2}+\frac{1}{2}\cdot\frac{2}{3}} = \frac{3}{5}$

Another nice way to solve this is to let p be the probability that A wins when no one has won and it's A's turn (as in the beginning of the game) and solving
$p =$ $\frac{1}{2} + p\cdot\frac{1}{2}\cdot\frac{1}{3}$
To see why this works, condition on A's throw. If A throws H with probability one half, it's a win and if A throws T with probability one half, we need B to throw 5 or 6 with probability one third to be at the start again, where A again has the probability p of winning.

Answer 2

Draw a (discrete) tree graph showing the possible outcomes. Start at the root vertex (at the lower left), representing the beginning of play. There are two edges ("links") from the root vertex representing the outcomes for player $A$: one edge is labeled $H$ (for heads), and terminates. That has a probability of $1/2$.

Consider the other edge, corresponding to $T$ (for tails). At the vertex at end of that edge are two new edges representing player $B$'s results: (1,2,3,4) representing $B$ winning (2/3 probability), and (5,6) representing $B$ losing (1/3) probability. We only care about the case $B$ loses. The vertex at that case has two outcomes for $A$: $H$ (probability 1/2) and $A$ wins, or $T$ (probability 1/2) $A$ hasn't yet won.

Thus the probability $A$ wins (in this short version) is $1/2 + 1/2(1/3 \cdot 1/2)$. Continue to infinity. Add up those probabilities:

$$\sum\limits_{n=0}^\infty \frac{1}{2 \cdot 6^n} = \frac{3}{5}$$