A fair coin is tossed repeatedly until 5 consecutive heads occurs.

What is the expected number of coin tosses?

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13 Answers13


Let $e$ be the expected number of tosses. It is clear that $e$ is finite.

Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus $$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$ Solve this linear equation for $e$. We get $e=62$.

André Nicolas
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    It is clear that e is finite, but how can you show it properly though ? Thanks. – Dark Jul 03 '15 at 17:39
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    If one wants, let $X$ be the number of tosses. Then $\Pr(X=n)\le (1/2)^{n-5}$. So $E(X)\le \sum n (1/2)^{n-5}$, a convergent series. – André Nicolas Jul 03 '15 at 17:58
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    The same method obviously generalizes to give $e_n$, the expected number of tosses to get $n$ consecutive heads ($n \ge 1$): $$e_n=\frac{1}{2}(e_n+1)+\frac{1}{4}(e_n+2)+\frac{1}{8}(e_n+3)+\frac{1}{16}(e_n+4)+\cdots +\frac{1}{2^n}(e_n+n)+\frac{1}{2^n}(n),$$ the solution of which is easily found to be $$e_n = 2(2^n - 1).$$ – r.e.s. Jul 19 '15 at 23:57
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    Yes, good point. The most common assigned problem is expected number of tosses until two heads in a row, answer $6$. We can use the same idea for $n$ heads in a row, probability of head $p$. – André Nicolas Jul 20 '15 at 00:12
  • Can you help me to find out what is the expected number of tossing if the probability of getting head, $p\neq0.5$? Is it $$E_n=np^n+\sum_{k=1}^n p^k(E_n+k)?$$ – Anastasiya-Romanova 秀 May 16 '16 at 02:37
  • @Anastasiya-Romanova秀: The reasoning I described gives something like $e=(1-p)(e+1)+p(1-p)(e+2)+p^2(1-p)(e+3)+\cdots$. – André Nicolas May 16 '16 at 13:24
  • So $X$ follows a geometric distribution then. Let's say we want to get a pattern $HH$, is the logic as follows: 1st experiment you get $T$ so the expected of tossing will get an extra 1 toss then you repeat the experiment. The 2nd experiment you get $HT$ so the expected of tossing will get an extra 2 tosses then you repeat the experiment again. If you toss twice and you immediately get $HH$, the experiment is over. So the expected is $$E_2=2p^2+\sum_{k=1}^2 p^{k-1}(1-p)(E_2+k)\quad\Longrightarrow\quad E_2=8.75$$ In general, did you mean: $$E_n=np^n+\sum_{k=1}^n p^{k-1}(1-p)(E_n+k)?$$ – Anastasiya-Romanova 秀 May 17 '16 at 03:11
  • @Anastasiya-Romanova秀: Maybe it is best if you ask a separate question. Comments are difficult for any extended typing. The number of tosses until a sequence of $h$ consecutive heads does not have a geometric distribution. I used a conditioning argument to bypass finding the distribution. – André Nicolas May 17 '16 at 03:23
  • @AndréNicolas: Pls tell me why Pr(X=n)≤(1/2)n−5Pr(X=n)≤(1/2)n−5 – anhldbk Mar 20 '17 at 14:00
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    Why are TT, TTT not considered? – Jaydev Jul 24 '17 at 01:24
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    @Jaydev TT and TTT both are covered by the case "if we get a tail immediately". – David K Oct 06 '17 at 13:13
  • It may be worth mentioning that for the general case of $0

    – r.e.s. Dec 06 '17 at 21:38
  • I don't understand expectations. Could anyone please explain in simpler terms? – Alex Apr 07 '20 at 19:50
  • @AndréNicolas (& for anhldbk) $Pr(X=n) \leq (1/2)^{n-5}$ is not correct. We can check that $Pr(X = 12) = (29/32)\cdot (1/2)^6 > (1/2)^7$. However, the main point is correct. The expected time is finite because we can divide the string of draws into chunks of 5, and the probability of a "5 heads" chunk is $(1/2)^5$. The arrival time of 5 heads is drawn according to the exponential distribution, and is finite. Counting strings of 5 draws by $m$, the expected number of such draws until 5 heads is $t = 1/2^5 (1-1/2^5)^{m-1}\cdot m$. Therefore, for $n$, you will have an upper bound for $e$ of $5t$. – Surge Nov 30 '20 at 11:27
  • I feel you didnt explain the answer... i havent leared much from it.. – Simba Aug 10 '21 at 17:04

Lets calculate it for $n$ consecutive tosses the expected number of tosses needed.

Lets denote $E_n$ for $n$ consecutive heads. Now if we get one more head after $E_{n-1}$, then we have $n$ consecutive heads or if it is a tail then again we have to repeat the procedure.

So for the two scenarios:

  1. $E_{n-1}+1$
  2. $E_{n}{+1}$ ($1$ for a tail)

So, $E_n=\frac12(E_{n-1} +1)+\frac12(E_{n-1}+ E_n+ 1)$, so $E_n= 2E_{n-1}+2$.

We have the general recurrence relation. Define $f(n)=E_n+2$ with $f(0)=2$. So,

\begin{align} f(n)&=2f(n-1) \\ \implies f(n)&=2^{n+1} \end{align}

Therefore, $E_n = 2^{n+1}-2 = 2(2^n-1)$

For $n=5$, it will give us $2(2^5-1)=62$.

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ravi pradeep
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Here is a generating function approach.

Consider the following toss strings, probabilities, and terms

$$ \color{#00A000}{ \begin{array}{llc} T&\frac12&\qquad\frac12x\\ HT&\frac14&\qquad\frac14x^2\\ HHT&\frac18&\qquad\frac18x^3\\ HHHT&\frac1{16}&\qquad\frac1{16}x^4\\ HHHHT&\frac1{32}&\qquad\frac1{32}x^5\\ \color{#C00000}{HHHHH}&\color{#C00000}{\frac1{32}}&\color{#C00000}{\qquad\frac1{32}x^5} \end{array} } $$ Each term has the probability as its coefficient and the length of the string as its exponent.

Possible outcomes are any combination of the green strings followed by the red string. We get the generating function of the probability of ending after $n$ tosses to be $$ \begin{align} f(x)&=\sum_{k=0}^\infty\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)^k\frac1{32}x^5\\ &=\frac{\frac1{32}x^5}{1-\left(\frac12x+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)}\\ &=\frac{\frac1{32}x^5}{1-\frac{\frac12x-\frac1{64}x^6}{1-\frac12x}}\\ &=\frac{\frac1{32}x^5-\frac1{64}x^6}{1-x+\frac1{64}x^6} \end{align} $$ The average duration is then $$ \begin{align} f'(1) &=\left.\frac{\left(\frac5{32}x^4-\frac6{64}x^5\right)\left(1-x+\frac1{64}x^6\right)-\left(\frac1{32}x^5-\frac1{64}x^6\right)\left(-1+\frac6{64}x^5\right)}{\left(1-x+\frac1{64}x^6\right)^2}\right|_{\large x=1}\\ &=\frac{\frac4{64}\frac1{64}+\frac1{64}\frac{58}{64}}{\left(\frac1{64}\right)^2}\\[12pt] &=62 \end{align} $$

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    Could you elaborate briefly on why the derivative gives the expected number of flips? – Austin Mohr Aug 20 '13 at 02:37
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    @AustinMohr: If $f(x)$ is the generating function of the probability $p_n$ of the ending after $n$ tosses $$ f(x)=\sum_{n=0}^\infty p_nx^n $$ then, because the probability of lasting an infinite number of tosses is $0$, we have $$ \begin{align} f(1) &=\sum_{n=0}^\infty p_n\\ &=1 \end{align} $$ Furthermore, $$ \begin{align} f'(1) &=\sum_{n=0}^\infty n\,p_n\\ &=\mathrm{E}(n) \end{align} $$ – robjohn Aug 20 '13 at 05:13
  • Do you know how to find the distribution (or expectation and variance) for the number of tosses until either 5 consecutive heads or 5 consecutive tails? (Or 5 consecutive equal results from rolling dice.) Is there a question on math.se about this? – ShreevatsaR Dec 15 '15 at 14:17
  • I found an answer using martingales here: https://www.quora.com/Independent-trials-each-of-which-are-equally-likely-to-have-any-of-m-possible-outcomes-are-performed-until-the-same-outcome-occurs-k-consecutive-times-What-is-the-expected-number-of-trials/answer/Douglas-Zare but I'm curious if there is a generating functions way (also about the distribution, say variance or number of trials until 90% probability of seeing what we want). – ShreevatsaR Dec 15 '15 at 14:31
  • @ShreevatsaR: The generating function for the probability of ending on $n$ tosses for that problem is similar: $$\frac{\frac1{16}x^5}{1-\left(\frac12x+\frac14x^2 +\frac18x^3+\frac1{16}x^4\right)}$$ From that, we can compute the expectation and variance. Looking at the coefficients of the series for the generating function we can also find out that in $66 $ tosses, we will have a $90.0761\%$ chance of seeing $5$ heads or $5$ tails in a row. – robjohn Dec 15 '15 at 17:32
  • Why is TT, TTT not considered? – Jaydev Jul 24 '17 at 01:24
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    @Jaydev: note that the substrings listed in the table are components of the full string. the full string is composed of a combination of the green substrings followed by the red substring. Thus, $TT$ is represented by $$\overbrace{\left(\tfrac12x\right)}^{\color{#0A0}{T}}\overbrace{\left(\tfrac12x\right)}^{\color{#0A0}{T}}=\tfrac14x^2$$ – robjohn Jul 24 '17 at 01:53
  • @user211337: the first equality is defining the generating function. The factor $$\left(\vphantom{\frac12}\right.\overbrace{\quad\frac12x\quad}^T+\overbrace{\quad\frac14x^2\quad}^{HT}+\overbrace{\quad\frac18x^3\quad}^{HHT}+\overbrace{\ \ \frac1{16}x^4\ \ }^{HHHT}+\overbrace{\ \ \frac1{32}x^5\ \ }^{HHHHT}\left.\vphantom{\frac12}\right)$$ represents an atom of one of $T,HT,HHT,HHHT,HHHHT$. There can be any number of these atoms (thus the sum) preceding the final $HHHHH$, which is represented by $\frac1{32}x^5$... – robjohn Oct 15 '21 at 17:07
  • ...We can map any sequence ending in $HHHHH$ to exactly one product in one of the terms of the sum. For example, $\color{#C00}{T}\color{#F80}{T}\color{#090}{HHT}\color{#00F}{HHHHH}$ is mapped to a product in the $k=3$ term $$\left(\color{#C00}{\frac12x}+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)\\\left(\color{#F80}{\frac12x}+\frac14x^2+\frac18x^3+\frac1{16}x^4+\frac1{32}x^5\right)\\\left(\frac12x+\frac14x^2+\color{#090}{\frac18x^3}+\frac1{16}x^4+\frac1{32}x^5\right)\\\color{#00F}{\frac1{32}x^5}$$ ... – robjohn Oct 15 '21 at 17:08
  • ...The coefficient of that product is the probability of that sequence and the exponent of $x$ is the length of the sequence. When all are added up, we have the coefficient of $x^n$ is the probability that the sequence has length $n$. – robjohn Oct 15 '21 at 17:08
  • @robjohn many thanks for the lovely explanation, I've posted the original comment as a question [here](https://math.stackexchange.com/questions/4277625/generating-function-for-length-of-coin-tosses-until-5-consecutive-heads) because I thought that was more appropriate. In particular, I am unsure why my method of conditioning on the green strings fails to give the correct answer – user211337 Oct 15 '21 at 19:19

This problem is solvable with the next step conditioning method. Let $\mu_k$ denote the mean number of tosses until 5 consecutive heads occurs, given that $k$ consecutive heads just occured. Obviously $\mu_5=0$. Conditioning on the outcome of the next coin throw: $$ \mu_k = 1 + \frac{1}{2} \mu_{k+1} + \frac{1}{2} \mu_0 $$ Solving the resulting linear system:

In[28]:= Solve[Table[mu[k] == 1 + 1/2 mu[k + 1] + mu[0]/2, {k, 0, 4}],
   Table[mu[k], {k, 0, 4}]] /. mu[5] -> 0

Out[28]= {{mu[0] -> 62, mu[1] -> 60, mu[2] -> 56, mu[3] -> 48, 
  mu[4] -> 32}}

Hence the expected number of coin flips $\mu_0$ equals 62.

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  • What tool did you use for solving? – hola May 11 '15 at 17:19
  • @pushpen.paul I used Mathematica – Sasha May 11 '15 at 17:20
  • Can you please explain the original equation? – BOS Sep 20 '16 at 15:58
  • @BOS Since $\mu_k$ is the conditional expectation, consider the next coin toss. Because a new toss was made, we add 1, in the next state, with equal probabilities we either get next head, in which case we gonna get $k+1$ heads, hence $\mu_{k+1}$, or the tail, in which case we break the streak of consecutive heads, hence $\mu_0$. – Sasha Sep 23 '16 at 03:03
  • Thank you @Sasha, I think I get it now. – BOS Sep 24 '16 at 09:53
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    @Sasha This is the Markov way of solving it, right? This seems most intuitive to me of all methods presented here. – user3496060 Dec 22 '17 at 08:06

I’m slightly surprised that no one has suggested this solution yet in such a highly active question.

Call a stretch of $5$ tosses all of which are heads a “success”. Divide a long series of tosses into segments after each tails that follows a success. The expected number of (overlapping) successes in each resulting segment is $1+\frac12+\frac14+\cdots=2$. Denote by $x$ the expected length of these segments up to the end of the first success. We must have $\frac{x+2}2=2^5$, since each segment on average contains $x+2$ tosses and $2$ successes, and the average number of successes per toss is $2^{-5}$. Thus $x=2^6-2=62$. Our initial state, in which we haven’t counted any heads yet, is equivalent to the state after a tails. Thus in this case, too, $62$ is the expected number of tosses up to the end of the first success.

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Use Markov chains. The nice part of Markov Chains is that they can be applied to a huge class of similar problems with relatively little thought (it's almost formulaic in application). It's also the most intuitive way to handle these problems.

(in Matlab code notation below)

%% setup full transition matrix with states from zero heads to 5 heads

T = $[ones(5,1)*.5,eye(5)*.5];$

$T = [T;zeros(1,6)]$

%%Take subset "Q" comprised of just transient states (5 heads is absorbing state) $Q = T(1:end-1,1:end-1);$

$M = inv(eye(5)-Q)$

absorbing Markov Chain has a similar example as this question BTW...

ans =


Where each row is the expected number of steps before being absorbed when starting in that transient state (0 through 4 heads, top to bottom).

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We can solve this without equations. Ask the following (auxiliary) question: how many flips you need to get either $N$ heads or $N$ tails. Then to get $N$ heads only you need twice as many flips. Start with the question of how many flips you need to get either $H$ or $T$. The answer is $$x = \frac12 (1) + \frac12 (1) = 1.$$ The reason is that there is $\frac12$ probability to get $H$, and after $1$ flip you are done. The same for $T$. To get only one $H$ you then need two flips. OK. Now we ask what it takes to get $HH$ or $TT$. The result is $$x=\frac12(1+2) + \frac12(1+2).$$ The number $2$ appears because, say you flip $H$ first, then you need on average $2$ flips to get another $H$, as we learned earlier. The same for $T$. So you need $3$ flips to get $TT$ or $HH$, and you need $6$ flips to get $HH$ only. And so on. You need $\frac12 (1+6) + \frac12(1+6) = 7$ flips to get either $HHH$ or $TTT$, and $14$ to get $HHH$ only. If you need $HHHH$ or $TTTT$, then flip $\frac12(1+14) + \frac12(1+14) = 15$ times, or $30$ times to get just $HHHH$. The sequence is $1, 3, 7, 15, \ldots$ to get either heads or tails. The formula is easy to extract: you need $2^N-1$ flips to get either $N$ heads or $N$ tails, or $2^{N+1}-2$ to get $N$ heads only. If $N=5$ we get the answer: $62$.

Siong Thye Goh
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I would simplify the problem as follows:

Let $e$ = Expected number of flips until $5$ consecutive $H$, i.e., $E[5H]$
Let $f$ = Expected number of flips until $5$ consecutive $H$ when we have seen one $H$, i.e., $E[5H|H]$
Let $g$ = Expected number of flips until $5$ consecutive $H$ when we have seen two $H$, i.e., $E[5H|2H]$
Let $h$ = Expected number of flips until $5$ consecutive $H$ when we have seen three $H$, i.e., $E[5H|3H]$
Let $i$ = Expected number of flips until $5$ consecutive $H$ when we have seen four $H$, i.e., $E[5H|4H]$

Now Start flipping coin, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ then expected number of flips until 5 consecutive $H$ is $(f+1)$. Alternatively if $T$, we wasted 1 flip and expected number is still $(e+1)$ $$ e=\frac12(e+1)+\frac12(f+1)\; $$

We now need $f$ to solve above to get $e$. Now we start with 1 $H$ and seeking 4 more $H$ to get total 5 $H$. Again, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ (total $2H$ so far) then expected number of flips until 5 consecutive $H$ is $(g+1)$. Alternatively if $T$, we wasted this flip and expected number is back to $(e+1)$

$$ f=\frac12(g+1)+\frac12(e+1)\; $$

Continuing this way... $$ g=\frac12(h+1)+\frac12(e+1)\; $$ $$ h=\frac12(i+1)+\frac12(e+1)\; $$

Finally, Now we have 4 $H$ and seeking last $H$ to get total 5 $H$. Still, there is $\frac{1}{2}$ probability of getting $H$ or $T$. So if we get $H$ (total $5H$) then we need just $1$ flip. Alternatively if $T$ is observed, we wasted this flip and expected number is back to $(e+1)$ $$ i=\frac12(1)+\frac12(e+1)\; $$

Solving these equations, $e=62$, $f=60$, $g=56$, $h=48$, $i=32$
This solution offers some insight into conditional expectations of number of flips needed till 5 consecutive $H$ given 1, 2, 3 and 4 consecutive $H$.

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No one else seems to have suggested the following approach. Suppose we keep flipping a coin until we get five heads in a row. Define a "run" as either five consecutive heads or a tails flip plus the preceding streak of heads flips. (A run could be a single tails flip.) The number of coin flips is equal to the number of runs with at least four heads ($R_{4+}$), plus the number of runs with at least three heads ($R_{3+}$), and so on down to the number of runs with at least zero heads ($R_{0+}$). We can see this by expanding the terms:

$R_{0+}$ = # runs with 0 heads + # runs with 1 head + ... + # runs with 5 heads

$R_{1+}$ = # runs with 1 head + # runs with 2 heads + ... + # runs with 5 heads


$R_{4+}$ = # runs with 4 heads + # runs with 5 heads

# flips = # flips in runs with 0 heads + # flips in runs with 1 head + ... + # flips in runs with 5 heads

# flips in runs with 0 heads = # runs with 0 heads

# flips in runs with 1 head = 2 x # runs with 1 head


# flips in runs with 4 heads = 5 x # runs with 4 heads

# flips in runs with 5 heads = 5 x # runs with 5 heads

By linearity of expectation, the expected number of coin flips is $E(R_{0+}) + E(R_{1+}) + \ldots + E(R_{4+})$. $E(R_{4+})$ is $2 E(R_{5+}) = 2$, because one half of the time we flip at least four heads in a row, we go on to flip five heads in a row, i.e. the following coin flip is heads. In other words, in expectation, it takes two runs that start with four heads to achieve one run of five heads. Likewise, $E(R_{3+}) = 2 E(R_{4+}) = 4$, $E(R_{2+}) = 2 E(R_{3+}) = 8$, $E(R_{1+}) = 2 E(R_{2+}) = 16$, and $E(R_{0+}) = 2 E(R_{1+}) = 32$. The expected number of coin flips is $32 + 16 + 8 + 4 + 2 = 62$.

More generally, given a biased coin that comes up heads $p$ portion of the time, the expected number of flips to get $n$ heads in a row is $\frac{1}{p} + \frac{1}{p^2} + \ldots + \frac{1}{p^n} = \frac{1 - p^n}{p^n(1 - p)}$.

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Here is an answer that uses martingale. We think in terms of a game: at every time step we toss a coin, you can bet any amount $x$; if the coin toss comes up head, you gain $x$, otherwise you lose $x$. Now let's construct a strategy such as if the $n$th coin toss is tail, then you position (cumulative gain) will be $-n$. So clearly, we should bet 1 on the first toss. If we get a tail, our position is -1, and we will bet 1 again; if we get a head, our position is 1, and we should bet 3 next, since we want to get to -2 if the second toss is a tail. How much should we bet after 2 heads in a row? Let's say the next toss is the $k+1$th toss, then $k-2$th toss was a tail, and by our strategy, our position after $k-2$th toss was $-(k-2)$; after that we bet 1, a head showed up, and then we bet 3, again a head, so our position now is $-(k-2)+4=-k+6$. We want to get to $-(k+1)=-k-1$, so we should bet 7 on the $k+1$th toss. By the same reasoning, we should bet 15 after the third head in a row, and 31 after the fourth head in a row. Since the up and down are symmetric, our position $X$ is a martingale. Let $\tau$ be the first time we have 5 consecutive heads. $\tau$ has finite expectation, so by optional sampling theorem, $X$ stopped at $\tau$ is also a martingale. What is our position at $\tau$? According to our strategy, our position will be $-(\tau-5)+1+3+7+15+31=-\tau+62$. Since our position stopped at $\tau$ is a martingale, and our position starts at 0, this means $$\mathbb{E}[-\tau+62]=0$$ and $$\mathbb{E}[\tau]=62$$

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The question can be generalized to what is the expected number of tosses before we get x heads.Let's call this E(x). We can easily derive a recursive formula for E(x). Now, there are a total of two possibilities, first is that we fail to get the xth consecutive heads in xth attempt and second, we succeed. Probability of success is 1/(2^x) and probability of failure is 1-(1/(2^x)).

Now, if we were to fail to get xth consecutive heads in xth toss (i.e. case 1), the we will have to use a total of (E(x)+1) moves, because one move has been wasted.

On the other hand if we were to succeed in getting xth consecutive head in xth toss (i.e. case 2), the total moves is E(x-1)+1 , because we now take one move more than that was required to get x-1 consecutive heads.


E(x) = P(failure) * (E(x)+1) + P(success) * (E(x-1)+1)
E(x) = [1-(1/(2^x))] * (E(x)+1) + [1/(2^x)] * (E(x-1)+1)

Also E(0) = 0 , because expected number of tosses to get 0 heads is zero, duh


E(1) = (1-0.5) * (E(1)+1) + (0.5) * (E(0)+1) => E(1) = 2

E(2) = (1-0.125) * (E(1)+1) + (0.125) * (E(1)+1) => E(2) = 6


E(3) = 14

E(4) = 30

E(5) = 62


A recursive programming solution is also possible. Below is the solution in Python

# Expected number of tosses to get n heads
def Expectation(n):
    if n == 0:
        return 0

    return 2**n + Expectation(n-1)

>>> Expectation(1)
>>> Expectation(2)
>>> Expectation(3)
>>> Expectation(5)
>>> Expectation(10)

Peter Winkler included this problem in his puzzle collection Mathematical Puzzles (2020), and gave the following very elegant solution, which I quote below:

Since the probability of seeing HHHHH in a particular series of five coin flips is $1/32$, you might think it would take $32$ flips on average to get HHHHH. Indeed, $32$ flips is the average wait between occurrences of HHHHH, but this includes, for example, a wait of length 1 between the first five heads in HHHHHH [six H's] and the last five. Waits of length 1 can’t help us, because we have no “head start” (OK, pun intended) when we begin flipping.

The real answer is much greater. Between runs, half the time you get the wait of $1$ and the rest of the time $1+x$, where $x$ is the desired quantity. Hence it is not $x$ but the average of $1$ and $1+x$ that is equal to $32$, which gives us $x = 62$.

Mike Earnest
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