# A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is?

A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is? How do i approach this problem ?

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What's the probability of heads on the first toss? What's the probability of two tails then a heads? What's the probability of four tails and then a heads? What's the probability of six tails and then a heads? Get where this is going? –  blue Jun 14 '14 at 21:53
not exactly. Probability of heads on the first toss is 1/2, on two tails and then a head is 1/8. i get this. But all these are happening on odd numbered toss itself. what exactly are we looking for here? –  user157104 Jun 14 '14 at 22:00
We are looking at the probability the first heads is on toss $n$, for each odd number $n$, then adding all of those probabilities up. –  blue Jun 14 '14 at 22:04
so 2/3 will be the answer? by forming a geometric progression? –  user157104 Jun 14 '14 at 22:09
$1/2+1/8+1/32+\cdots=2/3$, yes. –  blue Jun 15 '14 at 1:06

Let $A$ be the probability of flipping heads first after an odd number of flips, and $B=1-A$ the probability that heads is first after an even number of flips. Then $$A = \frac12 +\frac12 B$$ because the first term is the probability $\frac12$ that heads comes up on the first flip, and if it doesn't (also with probability $\frac 12$), then you need it to come up after an even number of additional flips, which it will do with probability $B$.

Substituting $B=1-A$ we get \begin{align} A & = \frac12 + \frac12(1-A)\\ &= \frac12+\frac12-\frac A2 \\ \frac{3A}2 & = 1 \\ A & = \frac23 \end{align}

We get the same answer as the geometric series approach, but without the geometric series. And this approach works even when the sequence of events is too complicated to write the geometric series explicitly. For example, you can easily use this method to analyze the chance of winning a game of craps, but to sum an explicit series whose terms are the chance of winning on the $n$th throw would be very difficult.

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One way to solve this is by computing essentially a recurrence (write the probability in terms of itself). Let $X$ be the event that we flip heads on an odd numbered flip. By the law of total probability, $Pr(X)=.5+.5Pr(Y)$, where $Y$ is the event that we flip heads on an odd numbered flip that is not the first flip, given we have already failed the first. In particular, what we are doing here is conditioning on whether the first flip is heads. In this case, we got it on an odd flip for sure. In the second case, we get it on an odd flip with probability $Pr(Y)$.

It isn't hard to see that $Pr(Y)=1-Pr(X)$, as after we have already flipped once, this is the same situation as before, except now an even number is the first one, so by symmetry we can do this. By substitution and solving, we see that $Pr(X)=.5+.5-.5Pr(X)$, so $1.5Pr(X)=1$, so $Pr(X)=2/3$.

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The probability of coming up heads on an odd toss is either:

(A) You come up heads on the first toss, or
(B) You come up tails on the first two tosses, then come up with heads on an odd toss

These two cases are mutually exclusive, so the total probability is the sum of the two cases. Call your final result $P$.

$$P = \underbrace{\frac 12}_{\text{Case (A)}} + \underbrace{\frac 12\cdot\frac 12 \cdot P}_{\text{Case (B)}}$$ $$\frac 34 P = \frac 12$$ $$P = \frac 23$$

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