Maths Question

Yep.

Yes to the first part, no to the second.

http://mathworld.wolfram.com/Run.html

That page has the mathematical details.

Unless I've miscalculated, the expected number of runs length 16 or greater is 0.5

Also, the "longest expected run" of heads in 65536 flips is 15.

This is from formulae 14 and 15 on that page.

Thanks for replying, I looked at that page and now my brain hurts. It raised another question though, I used the run of heads just to make it easier. I was under the impression that 16 heads have the same expectation as any other sequence.

So, in other words, if I picked a random sequence of 16 heads and tails and then continuously flipped a coin, I could expect to see that sequence once in 65536 flips but for some reason there is less chance of 16 heads?

Thanks again.

It's subtle but there is a difference between different sequences when looking at sub-sequences. It's counter-intuitive (and makes my head hurt too) but I think it's true. I'll try to do some examples to convince myself of that fact.

E.g. if we toss a coin 4 times, all possible (and equally likely) results are:

HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT

Now if we look for the subsequence HT, that's in 11 of the 16 and appears 12 times total. If we look for HH, that's only in 8 of the 11. (but still appearing 12 times in total if we could HHHH as 3 times)

Yeah, Ive got it and its answered another question, I think.

As I see it, the reason for this is because the coin is being continually flipped. If it was only flipped 16 times then every sequence has the same chance of occurring because there would be no sub sequences?

Thanks for your help.

Both Winning and Losing runs can be calculated in terms of reasonable probability.

In this case (Heads and Tails odds = 2 over 65535 flips) the result is:

Winning runs could be: 16
Losing runs could be: 16

I made a calculator ages ago for working this out.
see the output here:

http://www.xpokerbot.co.uk/tools/2@65536.jpg