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My question is, if someone tossed a fair coin 100 times, what is the most number of times that a result will likely present itself in a row. Alternatively put, what is the largest string of consecutive flips of the same result that has a probability of occurring >50% in 100 coin flips.

Conversely, I think this question can be answered by giving the equation for the expected number of flips of a fair coin before X number of consecutive flips are the same results (without constraining that the results be heads or tails).

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  • $\begingroup$ gato-docs.its.txstate.edu/mathworks/DistributionOfLongestRun.pdf should answer all your questions. $\endgroup$ – A.S. Dec 7 '15 at 16:26
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The odds of getting two of the same result in a row is $\frac14$. Imagine we have all the coin flips listed out, and we consider each set of adjacent two we can make. In n flips, there are $n - 1$ sets of two possible. The likelyhood that at least one of those sets of flips are the same is $1 - (1 - \frac14)^{n - 1} = 1 - (\frac34)^{n - 1}$. This quickly becomes more likely than not after 4 flips.

In general, the odds of flipping x of the same result in a row is $\frac{1}{2^x}$. We can make $n - x + 1$ sets out of those flips. So, the probability "$p$" of getting at least one set of $x$ flips in a row with $n$ flips becomes:

$$p = 1 - (1 - \frac{1}{2^x})^{n - x + 1}$$

Solving that for $n$ gives:

$$n = x - 1 + \frac{\ln(1 - p)}{ln(1 - \frac{1}{2^x})}$$

Sub in $p = 0.5$ and $x$ based on how many flips in a row you'd like, and $n$ should give the least about of flips to make that above 50%.

After trying a few values, I find that if you flip 100 times, getting 7 in a row goes above 50%.

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  • $\begingroup$ the sets of two adjacent positions are overlapping, so I would think the multiplicative game formula isnt valid. $\endgroup$ – mlu Dec 7 '15 at 13:50
  • $\begingroup$ Darn. Of course, take this at face value. $\endgroup$ – Kaynex Dec 7 '15 at 20:33

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