the probability for exactly two consecutive successs

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 $100 \choose 50$ $\frac{1}{2}^{100}$

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

• 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. – angryavian Nov 28 '17 at 18:34
• 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? – lulu Nov 28 '17 at 18:40
• 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? – Remy Nov 28 '17 at 19:41
• @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 – Dmomo Nov 28 '17 at 19:46
• 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? – lulu Nov 28 '17 at 19:52

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: $$\sum_{k=0}^{5} \binom{5}{k} \left( \frac{1}{2}^{20} \right)^{5} \left(1 - \frac{1}{2}^{20} \right)^{5-k}$$