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As in title: you're flipping a coin indefinitely (let's say heads gives 1, tails gives 0). E.g.:

${0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1\dots}$

How would you try proving that lengths of series of zeros or ones are of geometric distribution, that is of form:

$\mathbb{P}(\text{length} = k) = (1-p)^{k-1}p$.

Thanks for tips.

lkky7
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    Suppose you've just tossed heads, either as the first toss or immediately following a tails, so that it will be the first toss of a run. The run will consist of $k$ heads iff the next $k$ tosses are $k-1$ heads fllowed by tails. What's the probability of that? – saulspatz May 18 '19 at 13:01
  • $\binom{n}{k} \cdot 0.5^k \cdot (1-0.5)^{n-k} \cdot 0.5$? because it's a binomial trial and the last throw has to be the opposite which has probability of 0.5. But I don't know what is $n$ if we throw indefinitely? Or maybe $n=k$ if we want to get $k$ successes in $k$ tosses? – lkky7 May 18 '19 at 13:11
  • There is no ${n\choose k}$ The first $k-1$ tosses must be heads. The next must be tails. The probability is $2^{-k}$ All the runs are identically distributed (since the coin is fair.) You don't need to worry about tossing infinitely many times. – saulspatz May 18 '19 at 13:16

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I will answer my own question correctly (my comments can be unclear), thanks to @saulspatz:

Let's say we tossed heads, and that begins the run. So we already have one heads. To get $k$-tosses-long run of heads finally, we need $k-1$ more heads and one time tails to end that run. So the probability is:

$\mathbb{P}(k) = \underbrace{0.5^{k-1}}_{k-1\text{ heads}} \cdot \underbrace{0.5}_{\text{1 tails}} = (1 - 0.5)^{k-1} \cdot 0.5$.

And because the geometric distribution is of form: $(1-p)^{k-1}p$, we proved that for a fair coin $\mathbb{P}(k)$ is geometric.

lkky7
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