Question

On each round, Ann and Bob each simultaneously toss a fair coin. Let Xn be the number of heads tossed in the 2n flips which occur during the first n rounds. For each integer m > 0, let rm denote the probability that there exists an n such that Xn = m.

Answer 1

`P=(2n!)/(m!*(2n-m)!)*0.5^(2n)`

Explanation:

In a coin toss the probability of tossing a head is 0.5 (50% head/50% tails)

If n is the number of rounds and 2n the number of coins tossed (one for each player), the probability of having m heads tossed is:

`R=(2n!)/(m!*(2n-m)!)`

R is the number of cases (combination of coins tossed) that gives a m number of heads. Each case has a probability of
`P_(case)=0.5^(2n)` so:

`P=(2n!)/(m!*(2n-m)!)*0.5^(2n)`

For example, to toss 4 heads in 5 rounds:

• n=5
• 2n=10
• m=4

`P=(10!)/(4!*(10-4)!)*0.5^(10)`

`P=(10*9*8*7*6!)/(4!*6!)*0.5^(10)`

`P=(10*9*8*7)/(4!)*0.5^(10)`

`P=(10*9*8*7)/(4!)*0.5^(10)=0.205`