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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.

User Squillman
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Answer:


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

User Lil Ari
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