I know that there are some related questions, but they seem to be overkill for this small exercise.

I have 10 (fair) coin tosses and am interested in the probability that I have at least 4 consecutive heads.

So I had a lot of different ideas, but I think many of them do not work too well.

1) Markov chain: But then, somehow, we need to keep track of the number of coins we have already tossed, so one "loses" the Markov property.

2) count all possibilities: we have $2^{10}$ possibilities in total and can subtract the ones that have at most 3 consecutive heads. But seems to be nasty.

3) recursive equation? Let $p_{i,j}$ for $i \leq j$ be the probability that we have at least $j$ consecutive heads. But also this seems to be not that easy..

So this question is asked in a book in the first chapter, so it shouldn't be too hard?

  • $\begingroup$ "so it shouldn't be too hard" - not true. $\endgroup$ – barak manos Sep 27 '14 at 16:09
  • $\begingroup$ ok, thank you. this was helpful, as I really was looking for a simple solution. So your guess is with recursive equations? Or something directly via conditional probabilities? $\endgroup$ – user136457 Sep 27 '14 at 16:11
  • $\begingroup$ I'd actually go for counting all possibilities. You have $1024$ of them, so you'd better get started :) $\endgroup$ – barak manos Sep 27 '14 at 16:14

Another answer is you could take the formula for getting $5$+ consecutive out of $10$ random draws and then just add in the case $4$s ($139$ of those).

The formula for $5$+ consecutive out of 10 random draws is $p^5(6-5p)$ which for p=$0.5$ gives us $7/64$ which is $112/1024$.

Another answer is you could take an "almost correct" formula for getting $4$+ consecutive out of $10$ random draws but you'd have to subtract out the $5$ occurrences where they are either actually case $5$s or "collide" like in my other enumerated answer to this same question.

The "almost" formula for $4$+ consecutive out of 10 random draws is $p^4(7-6p)$ which for p=$0.5$ gives us $4/16$ which is $256/1024$ then subtract out the $5$ we double counted with this formula (see my enumerated answer elsewhere) and we get the correct $251$ (out of $1024$).

  • $\begingroup$ how do you get the formula for 5+ consecutive out of 10 (and 4+ consecutive out of 10)? $\endgroup$ – user103828 Sep 30 '14 at 19:06
  • $\begingroup$ @user103828: I got a bunch of terms for each of the cases such as P(exactly 5), P(exactly 6)... but when you combine them all together into something like P(5+), a lot of the intermediate terms cancel out and leave a nice short formula. For example, for P(exactly 5) I have $6p^5 - 10p^6 + 4p^7$. P(exactly 6) is $5p^6 - 8p^7 + 3p^8$. P(exactly 7) is $4p^7 - 6p^8 + 2p^9$. P(exactly 8) is $3p^8 - 4p^9 + p^{10}$. P(exactly 9) is $2p^9 - 2p^{10}$. P(exactly 10) is $p^{10}$. You can see many terms cancel out for P(5+). Line up similar terms on paper and you can easily see that. $\endgroup$ – David Oct 1 '14 at 21:45
  • $\begingroup$ @user103828: I misplaced some of my notes that I could use to explain to you how I got these terms but I will keep looking for them and post another comment if I find them, explaining the formulas in more detail. I like my placeholder example cuz you can see exactly what is happening in the cases and how many there are of each. Other solutions tend to me more abstract and a number just "pops out" without the person really understanding why. Getting the right answer is important but understanding why is important to. $\endgroup$ – David Oct 1 '14 at 21:51

Let $S_N$ be the set of the strings over the alphabet $\Sigma=\{0,1\}$ with length $N$, avoiding $4$ consecutive $1$'s, and $T_N=|S_N|$. The only possible prefixes of an element of $S_N$ can be: $$0,\quad 10,\quad 110, \quad 1110$$ hence we have: $$ T_N = T_{N-1}+T_{N-2}+T_{N-3}+T_{N-4} $$ and: $$ T_1=2,\quad T_2=4,\quad T_3=8,\quad T_4=15$$ leading to: $$ T_{10}=773.$$ The probability that at least for consecutive heads appear is so: $$ 1-\frac{773}{2^{10}} = \frac{251}{1024}.$$

  • $\begingroup$ thank you for your answer. But I do not really get why prefixes only can be of these forms. It might be because I do not really get what you mean with prefixes. So for example $1011 \in S_4$, so you would say that it has prefix $10$? $\endgroup$ – user136457 Sep 27 '14 at 16:25
  • $\begingroup$ @user136457: exactly, yes. $\endgroup$ – Jack D'Aurizio Sep 27 '14 at 16:36
  • $\begingroup$ So I think now I got it. Thank you. But does this somehow generalize to other settings? For example if I want to calculate the expected value of number of times the sequence 12345 occurs if I draw 100'000 numbers uniformly at random from 1,...,9? So this seems to be a very similar question, but does something similar apply here? $\endgroup$ – user136457 Sep 27 '14 at 17:27
  • $\begingroup$ Can someone please explain why $T_N = T_{N-1}+T_{N-2}+T_{N-3}+T_{N-4}$? For example, why is $T_5 = T_4 + T_3 + T_2 + T_1$ which is $29$? Actually I don't understand the entire solution. $\endgroup$ – David Sep 27 '14 at 22:24
  • $\begingroup$ @David: If the prefix is $0$, it is followed by an element of $S_{N-1}$; if the prefix is $10$, it is followed by an element of $S_{N-2}$ and so on. $\endgroup$ – Jack D'Aurizio Sep 27 '14 at 23:02

Let $P_N$ be the probability of at least $4$ consecutive heads in $N$ tosses and $p$ be the probability of a head. Then conditioning on the first toss, \begin{align*} P_4 &=p^4 \\ P_5 &=(1-p)P_4+p^4= (2-p)p^4\\ P_6 &=(1-p)P_5+p(1-p)P_4+p^4=(3-2p)p^4 \\ P_7 &=(1-p)P_6+p(1-p)P_5+p^2(1-p)P_4+p^4=(4-3p)p^4 \end{align*} and for $N\geq 8$ $$ P_N = (1-p)P_{N-1}+p(1-p)P_{N-2}+p^2(1-p)P_{N-3}+p^3(1-p)P_{N-4}+p^4 $$ so $$ P_8=(5-4p)p^4 \qquad P_9=(1-p)(5-p^4)p^4+p^4 \\ \qquad P_{10}=(1-p)(6+p^5-3p^4)p^4+p^4 $$ so when $p=1/2$, $$ P_4=0.0625 \qquad P_5=0.09375 \qquad P_6=0.125 \qquad P_7=0.15625 \qquad P_8=0.1875 \qquad P_9=0.216797 \qquad P_{10}=0.245117 $$

Edit: As @Byron Schmuland pointed out in the comments, $p^4$ was missing from $P_6$ and $P_7$.

  • 1
    $\begingroup$ There should be a $+p^4$ at the end of your formulas for $P_6$ and $P_7$, just like at the end of your formula for $P_5$. $\endgroup$ – user940 Sep 28 '14 at 16:23

Just count them up using placeholders. Notice I start with the heads in the leftmost slots and then gradually work them to the right one placeholder at a time (case $8$ is a simple example to illustrate this).

Let H = Head, T = Tail, - = don't care (could be Head or Tail), (nH) = n consecutive Heads.

$10$ in a row max: ($10$H)
(only $1$ occurrence possible)

$9$ in a row max: ($9$H)T or T($9$H)
($2$ occurrences)

$8$ in a row max: ($8$H)T- or T($8$H)T or -T($8$H)
($5$ occurrences)

$7$ in a row max: ($7$H)T-- or T($7$H)T- or -T($7$H)T or --T($7$H)
($12$ occurrences)

$6$ in a row max: ($6$H)T--- or T($6$H)T-- or -T($6$H)T- or --T($6$H)T or ---T($6$H)
($28$ occurrences)

$5$ in a row max: ($5$H)T---- or T($5$H)T--- or -T($5$H)T-- or --T($5$H)T- or ---T($5$H)T or ----T($5$H)
($64$ occurrences)

$4$ in a row max: ($4$H)T----- or T($4$H)T---- or -T($4$H)T--- or --T($4$H)T-- or ---T($4$H)T- or ----T($4$H)T or -----T($4$H)
($144$ occurrences)

Case $4$ is special so it needs extra care. The following patterns match more than once so we have to make sure we only count that trial once as a winner.

($4$H)T----- and -----T($4$H) are case $5$s, not case $4$s, if all - are heads, so subtract these $2$ cases from $144$ to get $142$.

Next we have to make sure we don't double count cases where a string of $4$ consecutive heads can appear twice in the string of length 10. There are $3$ main patterns for that, namely: ($4$H)T($4$H)T, ($4$H)TT($4$H), and T($4$H)T($4$H).

($4$H)T----- collides with ----T($4$H)T if we have ($4$H)T($4$H)T so subtract $1$ from $142$ to get $141$.
($4$H)T----- collides with -----T($4$H) if we have ($4$H)TT($4$H) so subtract $1$ from $141$ to get $140$.
T($4$H)T---- collides with -----T($4$H) if we have T($4$H)T($4$H) so subtract $1$ from $140$ to get $139$.

So we have $144 - 2 - 1 - 1 - 1 = 139$ "corrected" occurrences of case $4$.

Total number of good outcomes is $1 + 2 + 5 + 12 + 28 + 64 + 139 = 251$.

$251 / 1024$ is about $24.5$%.

A slight advantage of this method is you can get a visual of what is happening and you can see how many of each case there are so for example, if you wanted to know $5$+ heads max in a row, just add up cases $5$ thru $10$ which would be $112$ total. A disadvantage is it is more work and is only practical for small numbers of flips such as $10$. If you were looking for $20$+ heads out of $100$ coin flips instead, then don't use this tedious method. Also, if you had asked for $3$+ heads, there would be more special cases to handle so this is not the best method for that situation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.