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 A Coin Game (Posted on 2004-05-20)
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 Bert wins?

 See The Solution Submitted by Brian Smith Rating: 3.5000 (4 votes)

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 Puzzle Solution | Comment 11 of 12 |

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:
H(A)>H(B)
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.

 Posted by K Sengupta on 2009-02-26 15:31:58

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