# Thread: Variation of a coin flip problem

1. ## Variation of a coin flip problem

Question:
A, B, C take turns fliping a coin. Calculate the probability that A wins, B wins, C wins; the first to flip a head. Assume A flips, then B, then C.

I understand it follows a sequeance {1, 01, 001, 0001.....etc)
so theres can be infinity amount of turns before 1 gets a head.
So how would you add the probability of all the A's wining {1, 0001, 0000001......}

2. First, what is the probability that the first head is flipped at time $T=n$ (for $n\geq 1$) ? This is the same as saying that the $n-1$ first throws are tails, and the $n$-th one is a head. So this probability is, using independence: $P(T=n)=\left(\frac{1}{2}\right)^{n-1}\times\frac{1}{2}$.

Then, for instance, the probability that A wins is $P(T=1)+P(T=4)+\cdots+P(T=3k+1)+\cdots=\sum_{k=0}^\ infty P(T=3k+1)$ $=\sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(3k+1)-1}\times\frac{1}{2}$. Remembering that $\sum_{k=0}^\infty a^k=\frac{1}{1-a}$ when $|a|<1$, the probability finally becomes: $\frac{1}{1-\left(\frac{1}{2}\right)^3}\times\frac{1}{2}=\frac {4}{7}$.

Changing $3k+1$ into $3k+2$ and $3k+3$, you get the B and C cases.

3. Here is another proof.

Let $p_A,p_B,p_C$ denote the probabilities that A, B or C wins. We have of course $p_A+p_B+p_C=1$.

In addition, for B to win, the first coin must be a tail, and then it is the same condition as for A (for a shifted sequence of throws), so that $p_B=\frac{1}{2}p_A$. The same way, for C to win, the first two throws must be tails, and then the same condition as that for A to win must be checked for the following throws, so that $p_C=\frac{1}{4}p_A$.

Coming back to the first equation, plugging in the last two relations, you get that $\frac{8}{7}p_A=1$, hence $p_A=\frac{7}{8}$.

4. i got confused now, how come the two probabilities aren't the same??
one is 4/7, the other 7/8?

5. Originally Posted by weakmath
i got confused now, how come the two probabilities aren't the same??
one is 4/7, the other 7/8?
Oh yes, thank you ! This is a typo in the second solution : $1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}$, so we get $\frac{7}{4}p_A=1$ and $p_A=\frac{4}{7}$.