Question: Probability of getting two tails in a row if a coin is flipped 15 times?  Edit

Answer: Let n = number of coin flips.

There are 2^n possible sequences of heads and tails.

The number of those which do NOT have a pair or consecutive tails
is F(n+2) where F(n) = nth Fibonacci number: 0, 1, 1, 2, 3, 5, 8, ...

Hence P(no pair of tails) = (2^n - F(n+2)) / 2^n

For n = 15, we have 2^n = 32768, F(n+2) = 1597,
so P(at least one pair of tails - and probably some triples and multiple pairs, too) =
(32768 - 1597) / 32768 = 95%


Here is a table of results for n = 1 to 16
fail means "NO consecutive pair of tails within n flips"

P(at least one pair of tails in a row) = 100% - P(fail).

n ...all ... fail ... P(fail)
.1: . . 2 . .2 = 100.0%
.2: . . 4 . .3 = 75.0%
.3: . . 8 . .5 = 62.5%
.4: . .16 . .8 = 50.0%
.5: . .32 . 13 = 40.6%
.6: . .64 . 21 = 32.8%
.7: . 128 . 34 = 26.6%
.8: . 256 . 55 = 21.5%
.9: . 512 . 89 = 17.4%
10: .1024 .144 = 14.1%
11: .2048 .233 = 11.4%
12: .4096 .377 = 9.2%
13: .8192 .610 = 7.4%
14: 16384 .987 = 6.0%
15: 32768 1597 = 4.9%
16: 65536 2584 = 3.9%

It makes intuitive sense that no pair would be quite rare in any significant number of trials.
The most tails you could have in that case would be 1/2 the number of flips,
and they would have to alternate H T H T H T H T ....
or there could be fewer tails, spread out among the heads.
That's clearly the minority of all sequences, since the entire half
where there are more tails than heads is eliminated right off the bat.  Edit