LAST EDITED ON Feb-19-01 AT 06:17 PM (EST)

I am pretty sure I am confused. I believe you posed two independent problems.

>Marios tosses 6 fair coins and
>Michael tosses 5 fair coins.
>What is the probability that
>Marios tosses more "heads" than
>Michael?

This is problem #1.

>
>I notice that :
>1/1 + 1/3 = 4/3
>
>4^2 + 3^2
> = 5^2
>1/3 + 1/5 = 8/15
> 8^2
> + 15^2 =
>17^2
>1/5 + 1/7 = 12/35
> 12^2 +
>35^2 = 37^2

This is problem #2.

>i need to prove a generalization
>of the problem.

Of which problem and what generalization?

>there are two cases : either
>Marios tosses more "heads" than
>Michael or tosses more "tails"
>but NOT both. these
>two cases seem symmetric and
>each occurs with probability 1/2
>. please if its possible
>help me out. this is
>an interesting problem that came
>into my life.

The first fellow may have 1, 2, 3, 4, 5, or 6 heads with probabilities 6·2-6, 15·2-6, 20·2-6, 15·2-6, 6·2-6, 1·2-6, respectively.

The second fellow may have 0, 1, 2, 3, 4, or 5 heads with probabilities 1·2-5, 5·2-5, 10·2-5, 10·2-5, 5·2-5, 1·2-5, respectively.

This is means that the second fellow may have less than 1, 2, 3, 4, 5, or 6 heads with probabilities 1·2-5, 6·2-5, 16·2-5, 26·2-5, 31·2-5, 32·2-5 = 1.

You are therefore looking at the sum:

(6·1 + 15·6 + 20·16 + 15·26 + 6·31 + 1·32)·2-6·2-5

which is 210·2-6·2-5 = 1/2. Which shows that you are right indeed.

The same result might have been obtained without actually computing the probabilities but taking symmetry into account as you have suggested.

>p.s. Thx for the best of
>the best pages i ever
>seen in my life on
>math subject. keep on the
>good work!! :)

Thank you for the kind words.