The question posed is: "Suppose you flip a coin n times. If you win a dollar every time the coin comes up heads and lose a dollar every time it comes up tails, what is the probability that you will win k dollars?"

The answer given is: "If 2 doesn't divide $n + k$, the probability is 0. If 2 does divide $n + k$, the probability is ${n \choose \frac {n + k}{2} }$"

Does this make sense to anyone?

2. Originally Posted by earachefl
The question posed is: "Suppose you flip a coin n times. If you win a dollar every time the coin comes up heads and lose a dollar every time it comes up tails, what is the probability that you will win k dollars?"

The answer given is: "If 2 doesn't divide $n + k$, the probability is 0. If 2 does divide $n + k$, the probability is ${n \choose \frac {n + k}{2} }$"

Does this make sense to anyone?
To win k dollars requires n1 wins and n2 losses such that n1+n2=n, and
n1-n2=k.

This last imples that 2 n1 - (n1+n2) = k, so:

2 n1 = k+n

Hence k+n is even so the probabilty of winning k dollars in n trials is 0 if k+n is odd.

Also if k+n is even then the number of wins is (k+n)/2, and the probability of this is:

$
p(k, n) = {n \choose \frac{k+n}{2}} (1/2)^n
$

RonL

3. So they did leave out part of the answer:
$
(1/2)^n
$

which does make sense. Thanks!