# Probability question on tossing a coin

Two people, A and B, toss a coin which has two possible outcomes, T and H. The probability to get H when A tosses the coin is $\;p_A\;$, and the probability to get H when B tosses the coin is $\;p_B\;$. The appearance of either H or T in each toss of both players A and B is independent of the other tosses' outcomes.

In each person's turn he tosses the coin until T appears and then it is the other player's turn, and the winner of the game is the first player who gets in his turn at least two H's.

I am asked to calculate the probability A wins the game.

I am very confused: first, why do they say the winner is the one who gets " at least " two H's in his turn? I would say that it is exactly at the second appearance of H in one person's turn that the game ends and that person wins, and this can happen only in the second toss, as otherwise there already appeared T and the turn ended.

Second: by independency, I calculated that the probability A wins in his turn equals the probability he gets H exactly in that turn's first two tosses (otherwise he gets T and his turn is over!), so the probability is $\;p_b^2\;$ ...but I can't figure out what happens if A doesn't win in his first turn: then it must be that B does not win in his first turn, so that he either gets T in his first toss ( probability: $\;1-p_B\;$), or in his second one ( probability: $\;(1-p_B)p_B\;$), as any player that completes two tosses and has not lost his turn has already win...is this right? Anyway, I thought the answer could be $\;p_A^2\;$ as it wouldn't matter what happened before...but I really am not sure at all.

Any input will be appreciated.

You are right on your first point: Practically speaking, the game ends as soon as someone tosses two heads in a row on their turn. There is no point in continuing to flip (unless there are further questions about the distribution of the number of heads).

On your second point: I would proceed as follows. $A$ wins if he tosses heads twice; this happens with probability $p_A^2$. If that does not happen, then $B$ wins if he tosses heads twice; the joint probability of $A$ not winning on his first turn and then $B$ winning on his first turn is $(1-p_A^2)p_B^2$.

If neither player wins on their first turn, then $A$ tries again, and wins with joint probability $(1-p_A^2)(1-p_B^2)p_A^2$. If you continue along in this vein, you should obtain a series of probabilities with a clear pattern, and by summing all of the terms where $A$ wins, you should get the overall probability that $A$ wins.

Or, you may see another pattern and avoid most of the algebra...

• Thank you, I appreciate the answer. Other than in the first turn, A can win only in an odd one, say $\;2n+1\;$ , so the probability for that is $$(1-p_A^2)^n(1-p_B^2)^np_A^2$$ Yet if a make $\;n\to\infty\;$ I get $\;0\cdot0\cdot p_A^2=0\;$...am I missing something? – user312943 Apr 6 '16 at 18:13
• @AntoineNemesioParras: Well, sure, the individual terms go to $0$, but the sum doesn't. They form a geometric series, in fact. – Brian Tung Apr 6 '16 at 19:44
• Thank you. I just don't see it. And what is, please, that "other pattern" that'd could save most of the algebra? – user312943 Apr 6 '16 at 21:33
• Hint. Write out the probabilities of winning on the $k$th turn for both $A$ and $B$ in a table, with the $k$th row representing the $k$th turn, and two columns, one each for $A$ and $B$. How are the probabilities for $A$ and $B$ related to each other in each row? – Brian Tung Apr 6 '16 at 23:44

P($A$ wins on first round) $=p_A^2$

P($A$ does not win on first round and $B$ wins on second round)$=(1-p_A^2)(p_B^2)$

Odds in favor of $A = \dfrac{p_A^2}{(1-p_A^2)(p_B^2)}$

[Subsequent cycles of $2$ rounds will only add some common multiplier, odds won't change]

Thus P(A wins) = $\dfrac{p_A^2}{p_A^2 + (1-p_A^2)(p_B^2)}$

• In particular, if $p_A = p_B = \frac12, P(A\;\;wins) = \frac47$ – true blue anil Apr 7 '16 at 4:20