The question is simple, but the solution is a monster (or at least(adsbygoogle = window.adsbygoogle || []).push({}); mineis). It says,

A playerAposesses n+1 normal coins, and a playerBposesses n coins of the same kind. Each player throws all of his coins. Show that the probability ofAhaving thrown tails (T) more often thanBis ½.

To do this, condition according to wheterAorBhas T more often than the other when both players have thrown n coins. (there are 3 possibilities)

Solution:

I identified the fondamental set as [itex]\Omega[/itex]: The result of n throwing of a coin for the two players A and B."

[tex]\Omega=\left\{((x_1^A,x_2^A,...,x_{n+1}^A),(x_1^B,x_2^B,...,x_{n}^B)): x_i^A,x_i^B\in \left\{T,H\right\}\right\}[/tex]

#[tex]\Omega=2^{n+1}2^n = 2^{2n+1}[/tex]

I then define the event of interest as E: Player A has thrown more T than H after all players have thrown all their coins. And I partition the fondamental set like so,

[itex]F_1[/itex]: Player A has more T than player B after n throws.

[itex]F_2[/itex]: Player A has less T than player B after n throws.

[itex]F_3[/itex]: Player A and player B both have the same amount of T after n throws.

According to the "total probability formula", then,

[tex]P(E) = \sum_{i=1}^3P(E\vert F_i)P(F_i) = \sum_{i=1}^3P(E F_i)= \sum_{i=1}^3 \frac{\mbox{card}(EF_i)}{\mbox{card}(\Omega)}[/tex]

So if I find the cardinality of the [itex]EF_i[/itex], I will have won.

Obviously, #[itex]EF_2[/itex]=0, for even if B only has one more T than A after the nth thrown and A gets T at his n+1th thrown, A and B are only at equality. So A cannot win in this case, hence the intersection is empty.

To find #[itex]EF_3[/itex], I said, supposekis the number of T each player has thrown after their nth thrown. There are [itex]\binom{n}{k}[/itex] ways for each player to thrownkT in n thrown. So there are [itex]\binom{n}{k}^2[/itex] ways for the both of them combined. And, for each of these ways to realize the event "each player throwns k T", there is only one way that A can win, and it is by throwing a T at his n+1th thrown. And sincekcan be any value btw 0 and n, we must sum onkfrom 0 to n:

[tex]\mbox{card}(EF_3)=1\cdot\sum_{k=0}^n\binom{n}{k}^2[/tex]

For #[itex]EF_1[/itex], we first note that for each ways to realize [itex]F_1[/itex], there are two ways to realise E, since player A can either thrown a T or a H at his n+1th thrown and still win (hence the factor 2 in the final expression for #[itex]EF_1[/itex]). To find the number of ways [itex]F_1[/itex] can be realized, we proceed similarly to above. We say, supposekis the number of T thrown by A after n thrown. There are [itex]\binom{n}{k}[/itex] ways that this can happen. And for each of them, there are

[tex]\sum_{j=0}^{k-1}\binom{k-1}{j}[/tex]

ways for B to throw less than k T. So,

[tex]\mbox{card}(EF_1) = 2\cdot\sum_{k=1}^n\sum_{j=0}^{k-1}\binom{n}{k}\binom{k-1}{j}[/tex]

All this together yields,

[tex]P(E)=\frac{1}{2^{2n+1}}\left[2\sum_{k=1}^n\sum_{j=0}^{k-1}\binom{n}{k}\binom{k-1}{j}+\sum_{k=0}^n\binom{n}{k}^2 \right][/tex]

This is not equivalent to ½ as can be seen by computing it for n=1. It give P(E) = 2/8=1/4.

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# A problem in probability (complex)

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