So a random piece of information in a video I watched ages ago popped in my head tonight and I started thinking about it. I believe I am remembering this video properly...

They flipped a coin 100 times you saw the ratio of head and tails to be 50/50. They created a diagram of all the flips. There was a lot of flip-flopping between heads and tail. There were even some strings of 4 or 5 heads/tails in a row. At one point in the chart there were 7 heads in a row. They said in a sample this size, that was expected.

That is where my question is, is there a mathematical formula or something that allows us to compute, in a sample size of 100, where the outcome can go 50% one way or the other, the probability of getting 7 of one outcome in a row?

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    $\begingroup$ Assuming you noticed 7 because it was the longest string. In my opinion, you should be asking the probability of getting $\textbf{at least} $ 7 of one outcome in a row, because I think that is what you are actually wondering about. $\endgroup$ – Improve Dec 4 '15 at 18:47

Here's a similar question with pretty good answers:

What are the odds of getting heads 7 times in a row in 40 tries of flipping a coin?

  • $\begingroup$ so maybe im missing something, thats less than 1% chance to get 7 in a row, I wouldnt rate that as very probably $\endgroup$ – Keltari Sep 20 '15 at 5:46
  • $\begingroup$ I'm sorry, I misread your question the first time, I edited my answer accordingly. $\endgroup$ – Nicolás Siplis Sep 20 '15 at 5:46

What do you mean by 'expected' ?

Suppose you flip a fair coin 2 times. The probability of the event of getting heads on the nth (n = 1, 2) toss, denoted by $H_n$, is 1/2. Note that the $H_n$'s are independent and have the same probability as the $H_n^C$'s.

Let $X$ denote the number of heads of in 2 flips.

$P(X = 0) = P(H_1^C \cap H_2^C) = P(H_1^C) P(H_2^C) = 1/4$

$P(X = 1) = P(H_1^C \cap H_2) + P(H_1 \cap H_2^C)= 1/2$

$P(X = 2) = P(H_1 \cap H_2) = P(H_1) P(H_2) = 1/4$

The 'expected' value of $X$ is given by:

$E[X] = \sum_{i=0}^{2} iP(X=i) = 1(1/2) + 2(1/4) = 1$

Mathematically speaking, we 'expect' that if we flip a fair coin twice, we will get one head. 'Expected' value in this sense is 'the long-run average value of repetitions of the experiment it represents'

  • $\begingroup$ I think I am misremembering. Perhaps it wasnt "expected", but not statistically improbable. $\endgroup$ – Keltari Sep 20 '15 at 5:49
  • $\begingroup$ @Keltari What do you mean by statistically im/probable? You want to compute the probability of such an event you mean? $\endgroup$ – BCLC Sep 21 '15 at 3:12

The problem is complicated to solve, so we answer a different but related question which has a relatively simple answer.

Call a number $k$ from $1$ to $94$ good if the $k$-th toss, and the following $6$ tosses, are all the same. Let $Y$ be the number of good numbers. We will find $E(Y)$, the expectation (mean) of $Y$. Note that if we have a run of exactly $8$ of the same kind, then $Y$ counts this run twice.

Let random variable $X_i$ be defined by $X_i=1$ if $i$ is good, and $X_i=0$ otherwise. Then $Y=X_1+X_2+\cdots+X_{94}$, so by the linearity of expectation we have $$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_{94}).$$ We have $\Pr(X_i=1)=\frac{1}{2^6}$ and therefore $$E(Y)=\frac{94}{64}.$$ A little more complicated argument of the same kind could be used to find the expected number of runs of length exactly $7$.


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