coin tossing

davieddy · 2010-04-18, 02:05

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

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

David

lfm · 2010-04-18, 02:13

Quote:

Originally Posted by davieddy

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

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

David

a quick guess :

999993/128 ?

davieddy · 2010-04-18, 02:46

Quote:

Originally Posted by lfm

a quick guess :

999993/128 ?

That's what I think.

So counting heads (half a million):

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

Does it?

davieddy · 2010-04-18, 02:52

Yes

And the number of runs of 5 or more heads is

million/64

CRGreathouse · 2010-04-18, 03:13

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

retina · 2010-04-18, 03:19

Did you forget about THHHHHTHHHHHT? And THHHHHTHHHHHTHHHHHT? etc.

edit: nvmd, CRGreathouse got it.

davieddy · 2010-04-18, 03:47

Quote:

Originally Posted by retina

Did you forget about THHHHHTHHHHHT? And THHHHHTHHHHHTHHHHHT? etc.

edit: nvmd, CRGreathouse got it.

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.

davieddy · 2010-04-18, 04:14

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

David

davieddy · 2010-04-18, 11:45

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

wblipp · 2010-04-18, 13:57

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.

davieddy · 2010-04-18, 23:26

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

David

2010-04-18, 02:05	#1
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	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

2010-04-18, 02:52	#4
davieddy "Lucan" Dec 2006 England 14512₈ Posts	Yes And the number of runs of 5 or more heads is million/64 Last fiddled with by davieddy on 2010-04-18 at 03:32

2010-04-18, 03:19	#6
retina Undefined "The unspeakable one" Jun 2006 My evil lair 2²×1,549 Posts	Did you forget about THHHHHTHHHHHT? And THHHHHTHHHHHTHHHHHT? etc. edit: nvmd, CRGreathouse got it. Last fiddled with by retina on 2010-04-18 at 03:22 Reason: CRGreathouse got it.

2010-04-18, 04:14	#8
davieddy "Lucan" Dec 2006 England 2·3·13·83 Posts	Pedantry I think the million tosses should be prefixed and postfixed with half a tail. David Last fiddled with by davieddy on 2010-04-18 at 04:17

2010-04-18, 11:45	#9
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	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 Last fiddled with by davieddy on 2010-04-18 at 12:21

2010-04-18, 03:13	#5
CRGreathouse Aug 2006 5979₁₀ Posts	Nitpick: I would expect 999994/128; am I wrong?

2010-04-18, 13:57	#10
wblipp "William" May 2003 New Haven 2·7·13² Posts	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.

2010-04-18, 23:26	#11
davieddy "Lucan" Dec 2006 England 1100101001010₂ Posts	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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Coin Paradox	petrw1	Puzzles	1	2015-02-06 23:17
Coin tossing 2	henryzz	Puzzles	10	2010-05-14 11:21
Coin Toss Game	davar55	Puzzles	26	2007-11-03 09:46
Coin Problem.	mfgoode	Puzzles	65	2007-01-26 00:25
Infinite Coin-Toss?	eaS	Puzzles	20	2003-09-21 02:48