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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: A problem in probability (complex)

**Physics Forums | Science Articles, Homework Help, Discussion**