Question 558903: player 1 chooses a number between 1 and 10. Then player 2 chooses a number between 1 and 10. What is the probability of player 2 choosing a higher number than player 1?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Let N1 and N2 be the numbers chosen by players 1 and 2.
Note that we have three probabilities: P(N2 > N1), P(N1 > N2) and P(N1 = N2). Exactly one of these happens, so the sum of these three probabilities is 1.
By symmetry, we can say that P(N2 > N1) = P(N1 > N2) because the players' numbers are independent of each other, and {N1, N2} = {9,4} and {N1, N2} = {4,9} have the same probability of occurring, for example.
To find P(N1 = N2), simply fix N1 (e.g. suppose N1 = 6). Then P(N1 = N2) = 1/10 (one number out of ten is equal to P1).
Hence we have P(N2 > N1) + P(N1 > N2) + 1/10 = 1 --> P(N2 > N1) + P(N1 > N2) = 9/10. Since these two probabilities are equal, P(N2 > N1) equals half of 9/10, or 9/20.
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