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Here is the problem: There is a fair coin. We flip the coin 10 times, and we look for the probability to have equal appearences of [H]eads and [T]ails.

Here is my solution: The probability will be P=|A|/|W| where

A={all sequences of 5H and 5T} W={all possible outcomes from flipping a coin 10 times}

=> Since every time we flip a coin we have two possible outcomes and we repeat that 10 times, then |W|=2^10 .

=> Now the outcomes that we want, are those that will have 5H and 5T (since we look after for probability of equals appearences of H and T, and since the experiment is repeated 10 times the equality means 5H and 5T). Let's imagine:

H H H H H T T T T T [outcome]

= = = = = = = = = =

1 2 3 4 5 6 7 8 9 10 [experiment]

so i look for: what are the permutations of the above ordering? I can find in how many ways i can assign the numbers to the 5H (and the rest empty slots will be assigned to Ts)=>

10 9 8 7 6

= = = = = --> 30240


and the rest empty slots are filled with the Ts.

So P = 30240/1024 > 1 !!

What is wrong here?

p.s: Intuitively, since the coin is fair, and there is a symmetry (5H, 5T) the P has to be 1/2.

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up vote 0 down vote accepted

Your intuition, unfortunately, is misguided. The fact the coin is fair indicates that if we continue to toss this coin indefinitely, the ratio of heads to tails will tend to $1.$ The fairness of the coin does not say this ratio will occur on any given finite number of tosses however.

Take a special case: What is the probability of tossing 5 heads, then 5 tails, in that order? Hopefully you agree it is $1/(2^{10}).$ How about any other scenario where we must end up with 5 heads and 5 tails? Each of those scenarios have probabliity $1/(2^{10})$ as well, so it remains to see how many such scenarios there are.

We need to toss 5 heads and 5 tails in any order. All the possible orders correspond to the number of ways we can place 5 "H"s in 10 adajacent boxes, and then we there's no more choices involved, the other 5 spaces are automatically tails.

The number of ways to pick 5 boxes from 10 is $ \binom{10}{5} $. Each of these specific scenarios are disjoint events, each with probability $1/(2^{10})$. Thus the probablity of tossing 5 heads and 5 tails in any order is $$ \frac{ \binom{10}{5} }{2^{10} } = \frac{63}{256}.$$

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Thanks.Just a small arithmetic correction: P = 252 / 1024 – Ponty Oct 15 '11 at 14:17
I cancelled a common factor of $4$ : $$ \frac{ 252}{1024} = \frac{ 4 \cdot 63 }{4 \cdot 256 } = \frac{63}{256}.$$ – Ragib Zaman Oct 15 '11 at 14:21

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