Player A play 100 coinflip games, The probability of Head/Tails is exact 50%

What is the probability that A loses at least 7 games consecutive out of 100 games?

Extended : What is the probability that A loses at least K consecutive games when he/she plays N games? (k < N)

My though is :

The probability of losing 7 consecutive games out of 100 games (94 cases : 1-7, 2-8, 3-9,.....) is: $94(\frac{1}{2})^{100}$

Do the same thing with 8 consecutive games and above (up to 100) : $93(\frac{1}{2})^{100}$, $92(\frac{1}{2})^{100}$,....., $(\frac {1} {2})^{100}$

The probability of losing at least 7 games consecutive out of 100 games is : $$\sum_{x=6}^{99} (100-x) \frac{1}{2}^{100}$$

but I think it won't be true when increase to K consecutive games out of 1000 games and above.

  • 1
    $\begingroup$ What have you tried? This is a Markov chain with the states the number of losing games at the end. $\endgroup$ Jul 21, 2021 at 5:37
  • $\begingroup$ What have you tried so far ? Where are you stuck ? You need to show some effort to attract help, else it will get closed sooner or later ! Pl. put whatever you have attempted in the main body, not in the comments. $\endgroup$ Jul 21, 2021 at 7:19
  • $\begingroup$ I've edited my question $\endgroup$
    – TennoX
    Jul 21, 2021 at 8:40
  • 1
    $\begingroup$ Does this answer your question? Probability for the length of the longest run in $n$ Bernoulli trials $\endgroup$ Jul 21, 2021 at 17:00

1 Answer 1


This is a binomial probability question. For clarity, let's say that A needs to lose 7 games. Then the probability is actually $\binom{100}{7}(\frac12)^7(\frac12)^{93}$ because there are $\binom{100}{7}$ ways you could re-arrange the ways that A loses those seven games amongst the 100 that A plays. So if she needs to lose at least 7, take the contrapositive case and go 1-p(A losing less than 7), then you'll just have to take the binomial probability for A loses one game, A loses 2 games all the way up to A loses 6 games.

However, for 100 coin flips, since the number of trials is so large and numbers like $\frac12^{100}$ so small you can take the normal distribution approximation, which I assume would be suited to the follow up question of when you're looking for $P(X≤K)$ where K≤N. What you need to do is to find the mean(given by $\mu$), which in the case of binomial, is equal to the probability of success which is half in this case. Then you need to find the $\sigma$, which is given by the formula $\sigma=\sqrt{npq}$ where n=number of times the experiment is conducted, p is the probability of success and q is the probability of failure. So for a coin tossed N times, $\sigma=\frac{\sqrt{N}}{2}$. Then you convert K(the number of losses you want) into a z-score, using the formula $z=\frac{K-\mu}{\sigma}$. So then your z-score becomes $z=\frac{K-\frac12}{\frac{\sqrt{N}}{2}}$ which simplfies down to $z=\frac{2K-1}{\sqrt{N}}$. Then you can google the z-scores table and match your z-score to the probability on the table or you can evaluate the integral $\int\limits_{-\infty}^{z}\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2}dx$ which is the standard normal distribution curve.


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