Alex flips a fair coin 20 times. Bert spins a fair coin 21
times. Bert wins if he gets more heads than Alex, else Alex wins.
Note that Alex wins if there is a tie. What is the probability that
(In reply to answer
by K Sengupta)
In general, let us suppose that Bert spins the coin (m+1) times and Alex flips the coin m times.
Let the respective number of heads produced by Bert and Alex be H(A) and H(B); while respective number of tails produced by Bert and Alex be T(A) and T(B)
Then, by the problem:
(i) H(A) + T(A) = m+1; and: (II) T(A) + T(B) = m
Now, if Bert produced more heads than Alex; then:
Or, H(A) >= H(B) + 1
Or, m+1 – T(A) > = m – T(B) + 1
Or, T(A)<= T(B)
Accordingly, H(A) > H(B) implies T(A)<= T(B) while H(A) < H(B) implies T(A)>= T(B)
Therefore, if Bert produced more heads than Alex, then it follows that Bert could not have come up more tails than Alex and vice versa. ……………(#)
Denoting the events that Bert produced more heads than Alex and Bert produced more tails than Alex respectively by E1 and E2, it follows by (#) that E1 and E2 are mutually exclusive. So, E1 and E2 are the only possible events and it follows from symmetry that Prob.(E1) = Prob.(E2); giving :
P(E1) = P(E2) = 1/2; since E1 and E2 are independent, disjoint and mutually exclusive ; and so, their probabilities must sum to 1.
Accordingly, the probability that Bert will produce more heads with his coin toss than Alex is 1/2, and consequently, the required probability that Bert wins is 1/2.
Note: It may be observed that the probability of Bert's win is independent of the value of m.