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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

H H H H H

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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1 Answer 1

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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
1  
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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