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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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Add up the probabilities of the coin coming up heads for the first time on toss 1, 3, 5...

$p_o = 1/2 + 1/2^3 + 1/2^5 + ...$

  • The $1/2$ term is pretty obvious, it's the probability of the first toss being heads.

  • The $1/2^3$ term is the probability of getting heads for the first time on the third toss, or the sequence TTH. That sequence has a probability of $1/2 * 1/2 * 1/2$.

  • The $1/2^5$ term is the probability of getting heads for the first time on the fifth toss, or the sequence TTTTH. That sequence has a probability of $1/2 * 1/2 * 1/2 * 1/2 * 1/2$.

Now we can rewrite the series above as

$p_o = 1/2 + 1/8 + 1/32 + ...$

This is a geometric series that sums to $2/3$. The easiest way to show this is with a visual example. Start with the series

$p = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ...$

This is a geometric series that sums to $1$.

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ... = 1

If we sum just the even terms of that series, we can see that they sum to $1/3$.

$1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3$

1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3

If you eliminate the even terms from the full sequence, you're left with just the odd terms, which must add up to $2/3$.

$p_o = 1/2 + 1/8 + 1/32 + ... = 2/3$

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Think recursively - let $p_o$ be the probability of the first head on an odd toss, and let $p_e$ be the probability of the first head on an even toss. Now $p_o+p_e=1$, and we also have that $p_e$ equals the probability of first toss tails times $p_o$. Thus $p_e = 1/2\cdot p_o$; $p_o+1/2\cdot p_o = 1$; $p_o = 2/3$.

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  • $\begingroup$ Could you please expand upon "...we also have that pe equals the probability of first toss tails times po."? $\endgroup$ – MackM May 13 '15 at 17:43

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