Assume a fair coin. What is the expectation number of tosses 𝑋 until getting 100 consecutive heads?

I have looked a similar problem to get 3 or 5 consecutive heads. But I don't know how to apply it to a more common case.

  • $\begingroup$ What have you tried so far? $\endgroup$ – Klangen Sep 2 '19 at 16:40
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    $\begingroup$ Well, you could set it up recursively. Get $E_{100}$ in terms of $E_{99}$ and so on. I expect that's easier than getting a closed form from first principles (of course the recursion isn't hard to solve so you could use it to get a closed form). $\endgroup$ – lulu Sep 2 '19 at 16:56
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    $\begingroup$ But, really, if you have studied the cases for $n=3$ and $n=5$ you ought to know how to solve it for any $n$. $\endgroup$ – lulu Sep 2 '19 at 17:00

Lets take an example: finding the expected number of flips it will take for getting two consecutive heads.

If the expected number of coin flips is denoted by $x$, $x=(\frac{1}{2}$)($x+1$)$+$($\frac{1}{4}$)$(x+2)$+$(\frac{1}{4})(2)$ $= 6$

This is because there are three scenarios. First, there is a $\frac{1}{2}$ probability that the first flip comes up as tails, resulting in 1 being added to the expected number of flips, $x$. This is the first part of the expression, $(\frac{1}{2}$)($x+1$). Next, if the first flip results in heads, but the second flip results in tails, two flips have been wasted, and the probability of this occurrence is $\frac{1}{4}$. This is denoted by $(\frac{1}{4}$)($x+2$). Lastly, if both flips are heads, two flips are used, and the probability of the event is $\frac{1}{4}$. This is shown with "$(\frac{1}{4})(2)$". All these elements are added together, due to the rule of linearity for expected values, to get $x = 6$.

With this example, we can generalize the equation, for the number of consecutive heads required being $n$.

If the $p$th flip in an attempt is a tails, the entire process is reset, and the expression for that attempt is $(\frac{1}{2^p}$)($x+p$). If $n$th flip is tails, the expression would be $(\frac{1}{2^n}$)($x+n$). If $n$ consecutive heads are achieved, the expression would be $(\frac{1}{2^n}$)$(n)$. Combining these results in $x =$ $(\frac{1}{2}$)($x+1$)$+$($\frac{1}{4}$)$(x+2)$+$(\frac{1}{4})(2)$ $...$ $(\frac{1}{2^p}$)($x+p$) $...$ $(\frac{1}{2^n}$)($x+n$) $+$ $(\frac{1}{2^n}$)$(n)$ $=$ $2^{n+1} - 2$

You can find more detailed explanations and more examples here: https://www.codechef.com/wiki/tutorial-expectation

And another way to solve it here: Is my answer correct? Expected number of coin flips to get 5 consecutive heads

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  • $\begingroup$ Thank you so much. Your answer really makes sense to me. $\endgroup$ – yixi zhou Sep 2 '19 at 21:36

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