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

Answer:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

Step-by-step 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:

$R=\frac{2n!}{m!*(2n-m)!}$

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:

$P=\frac{2n!}{m!*(2n-m)!}*0.5^{2n}$

For example, to toss 4 heads in 5 rounds:

• n=5
• 2n=10
• m=4

$P=\frac{10!}{4!*(10-4)!}*0.5^{10}$

$P=\frac{10*9*8*7*6!}{4!*6!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}$

$P=\frac{10*9*8*7}{4!}*0.5^{10}=0.205$