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A biased coin is tossed until a head appears for the first time. Let $p$ denote the probability of a head, $0 < p < 1$. What is the probability that the number of tosses required is odd?

My attempt:

Let $p = $ probability of a head on any given toss and $X = $ number of tosses required to get a head. Then for any toss $x$ \begin{align*} P(X = x) = (1-p)^{x-1}p. \end{align*} We want to know $P(X = 2n + 1) = (1-p)^{2n}p$. Therefore our cumulative distribution function looks like \begin{align*} P(X \leq 2n+1) = \sum_{i = 1}^{2n+1} P(X = i) & = \sum_{i=1}^{2n+1} (1-p)^{2i} p \\ & = \frac{(1-p)^{4n+4} - (p-1)^2}{p-2}. \end{align*} I am stuck what do do from here, and how to exclude possibilities where $X$ is even.

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    $\begingroup$ Have you tried reindexing the sum so it only contains odd $i$? $\endgroup$ – Kajelad Sep 9 '17 at 17:24
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You can sum up the terms with odd $x$, but there is a much more elegant way :

Let $q$ be the probability that the position of the first success is odd.

Then, the probability that the position of the first success is even is $(1-p)\cdot q$ because we have to start with a failure followed by a success in an odd position.

Since one of those events must happen, we have $$q+(1-p)q=(2-p)q=1$$ giving $$q=\frac{1}{2-p}$$

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    $\begingroup$ My method might be a bit overkill compared to yours :) $\endgroup$ – orlp Sep 9 '17 at 17:32
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Let's start by finding function $h(n)$, which is the chance that we get heads for the first time on toss $n$.

If we get our first heads on toss $n$, all previous $n - 1$ tosses must have been tails, and this one heads. In other words, $h(n) = p(1-p)^{n-1}$.

Now we can rephrase our main problem as finding $\sum_{k=0}^\infty h(2k+1)$.

$$\sum_{k=0}^\infty h(2k+1) = \sum_{k=0}^\infty p(1-p)^{2k}$$

Since the geometric series tells us $\displaystyle \sum_{k=0}^\infty ar^k = \frac{a}{1-r}$ for $|r| < 1$, we can note that above we have $a = p$ and $r = (1-p)^2$.

So our solution is

$$\frac{p}{1-(1-p)^2} = \frac{1}{2-p}$$

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Consider the first two tosses. If $H$ doesn't turn up on either toss, the "game restarts"

$P(first\; H\; on\; toss\; 1) = p,\quad P(first\; H\; on\; toss\; 2) = (1-p)p$

Odds in favor of number of tosses required being odd $= p : (1-p)p$

P(number of tosses required are odd) $= \dfrac{p}{p+(1-p)p} = \dfrac{p}{2-p}$

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