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Calculate the probability of getting exactly 50 heads and 50 tails after flipping a fair coin 100 times. for this question we can easily apply the the binomial distribution formula, as $100 \choose 50$ $\frac{1}{2}^{100}$

What if we are asked

Calculate the probability of getting exactly consecutive 10 heads or 10 tails after flipping a fair coin 100 times? would binomial distribution still work. thank you

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    $\begingroup$ Do you mean 50 heads followed by 50 tails? If you really mean 10 heads and 10 tails, you might need to be more specific: are you allowing this to occur more than once in the 100 flips, are you allowing 11 heads followed by 12 tails to count, etc. $\endgroup$ – angryavian Nov 28 '17 at 18:34
  • $\begingroup$ This is not at all clear. Since $10+10<100$ you can not get "exactly" $10$ heads and $10$ tails in $100$ tosses. What are you asking? $\endgroup$ – lulu Nov 28 '17 at 18:40
  • $\begingroup$ When you say exactly 10 heads and 10 tails, do you mean after obtaining 10 heads and then 10 tails in the 100 trials, you then get a head? $\endgroup$ – Remy Nov 28 '17 at 19:41
  • $\begingroup$ @lulu exactly consecutive. Doesn't that mean 11 consecutive wont't count ? if that have different meaning, apologize. still improve my english still as well, hope you don't mind $\endgroup$ – Dmomo Nov 28 '17 at 19:46
  • $\begingroup$ I'd say the phrasing is very ambiguous. I could make a guess as to your meaning. Maybe something like "the longest consecutive streak of Heads has length $10$ and similarly for Tails" but even that isn't clear. Suppose there are three strings of length $10$ which are all Heads, and one such string that's all Tails. Is that acceptable? $\endgroup$ – lulu Nov 28 '17 at 19:52
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The probability of getting exactly consecutive 10 heads and 10 tails is $\frac{1}{2}^{20}$.

Then you have to consider that it can happen from zero to five times. So you have to use the Bernoulli’s Formula to calculate the total probability: $$\sum_{k=0}^{5} \binom{5}{k} \left( \frac{1}{2}^{20} \right)^{5} \left(1 - \frac{1}{2}^{20} \right)^{5-k}$$

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  • $\begingroup$ Sorry. I meant Calculate the probability of getting exactly consecutive 10 heads or 10 tails after flipping a fair coin 100 times? would binomial distribution still work. changing " 10 heads and 10 tails" to " 10 heads or 10 tails" thank you $\endgroup$ – Dmomo Nov 28 '17 at 19:49

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