coin tossing
 

A run of 5 "heads" is defined as the sequence THHHHHT.

How many such runs would you expect in a million tosses?

David

[QUOTE=davieddy;212228]A run of 5 "heads" is defined as the sequence THHHHHT.

How many such runs would you expect in a million tosses?

David[/QUOTE]

a quick guess :

[spoiler]999993/128[/spoiler] ?

[quote=lfm;212229]a quick guess :

[spoiler]999993/128[/spoiler] ?[/quote]

That's what I think.

So counting heads (half a million):

1/8 + 2/16 + 3/32 + 4/64 +.... should = 1/2

Does it?

[spoiler]Yes[/spoiler]:smile:

And the number of runs of 5 or more heads is

[spoiler]million/64[/spoiler]

Nitpick: I would expect [spoiler]999994/128[/spoiler]; am I wrong?

Did you forget about THHHHHTHHHHHT? And THHHHHTHHHHHTHHHHHT? etc.

edit: nvmd, CRGreathouse got it.

[quote=retina;212238]Did you forget about THHHHHTHHHHHT? And THHHHHTHHHHHTHHHHHT? etc.

edit: nvmd, CRGreathouse got it.[/quote]
We didn't "forget" it.

My line of thinking was that each T had a probability of 1/64 of
being followed by HHHHHT and I expected 500,000 tails.

Pedantry
 

I think the million tosses should be prefixed and
postfixed with half a tail.

David

GIMPS Lucky Streak
 

As you know, Wagstaffe expects a ratio of 1.48 between
exponents of successive Mersenne primes.
The probabiliy of the "gap" being greater than this is 1/e.
The "Half Gap" when the probability is 50-50 is smaller (say1.3).
The mean of gaps smaller than the halfgap is less than sqrt(1.3).

A neat way of describing the lucky streak M40-M47 is that we have hit
a run of 7 gaps smaller than the "Half Gap".

From our previous deliberations, we expect this to occur once in
512 Mersenne primes.
We expect runs of 7 or more gaps less/greater than the Halfgap
to occur once in 128 Mersenne primes.

Our find is not that freakish!

David

I agree with CRGreathouse. davieddy's analysis method works with the additional observation that is doesn't apply to tails in the last 6 tosses. An alternative method is to treat it as an 8 state Markov process on the states
Not-Started
Tail
One Head
Two Heads
Three Heads
Four Heads
Five Heads
Finished

With Intial probability "not started = 1". On the seventh step this reaches steady state probability. For that and all following steps, the probability of being in the "Finished" state is 1/128. For the first six steps it is 0.

We make a great "double act"
 

Who was it who said that there was no problem
so complicated that with a little ingenuity
couldn't be made more complicated?

David