Two Heads After N Tails

Date: 05/20/99 at 10:45:32
From: jack wilson
Subject: Probability of two heads after n tails

Hi Doc,

I'm sure this is probably a simple question but I can't seem to get
past the initial phase. I believe it is similar to the picking the
same ball problem, but I'm interested in a portion of the non-winning
set.

The question is: what is the probability of flipping two heads after n
number of tails? I know there is a 50/50 chance of flipping a head, 
and that there is a .25 probability of two consecutive heads. I can
also assume from the answer in the picking the same ball problem that
the probability of two consecutive heads after n flips is:

1 - (.75)^(n-1)

but the non-winning set contains both heads and tails, so how do I get
the probability after n number of tails? Since both are equally 
likely, can I assume that the number of tails is n/2?

Thanks,
Jack

Date: 05/20/99 at 13:29:49
From: Doctor Mitteldorf
Subject: Re: Probability of two heads after n tails

Dear Jack,

It's easy to get confused in probability problems, because our 
language corresponds to exact mathematical calculations only if we're 
oh-so-careful about how we phrase things.

If what you mean is "The first n flips came out tails. What's the 
probability now that we will get 2 heads in a row?", then the answer 
is 1/4. It doesn't matter how many tails just happened - that's all 
done and in the past. You have a fresh start for your two shots at 
getting heads.

If what you mean is "What is the probability of first getting n tails 
in a row and then 2 heads in a row?" then the answer is (1/2)^n * 1/4.  
The (1/2)^n is for the n tails, and the 1/4 is for the two heads.

The answer that you read about 1 - (.75)^n doesn't have any 
application to this situation.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/20/99 at 14:13:25
From: Doctor Anthony
Subject: Re: Probability of two heads after n tails

The question is slightly ambiguous. It could mean that a mixture of n 
heads and tails takes place and then 2 heads, OR exactly n tails with 
single heads spread amongst them and then 2 heads. Since you say n 
tails I shall assume you mean the second situation.

We could consider the following diagramatic representation:

 *T*T*T.......*THH

where there are n tails followed by 2 heads, and we can insert single 
H's into gaps between the T's.  The gaps are denoted by *'s

There are in fact n *'s and we must consider all the possibilities of 
inserting 1, 2, 3, ....., n H's into the gaps.

If there are no H's inserted between the T's the probability of the 
sequence is   (1/2)^n x (1/2)^2 

The final (1/2)^2 is the probability of the final two H's.

            = (1/2)^(n+2)

If there is only 1 H to be inserted we can choose the gap in C(n,1) 
ways, and the probability of such a sequence will be

      C(n,1) x (1/2)^(n+1) x (1/2)^2

again, the final (1/2)^2 is the probability of the final two H's.

So the actual probability  = C(n,1) x (1/2)^(n+3).

If we insert 2 H's in amongst the T's, the relevant probability is

      C(n,2) x (1/2)^(n+4)

We can continue in this way until we have inserted n H's into the gaps 
between the T's. The total probability is then given by:

  (1/2)^(n+2)[1 + C(n,1)(1/2) + C(n,2)(1/2)^2 + ..... + C(n,n)(1/2)^n]

The series in brackets is the binomial expansion of (1+x)^n with 
x = 1/2.

It follows that the required probability is 

    (1/2)^(n+2).[3/2]^n

This is the probability of EXACTLY n tails before we get two 
consecutive heads.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/