You and your skiing opponent will complete two runs down the mountain, and the times of your runs will be added together. You and your opponent have the same, identical, independent normal probability distribution of finishing times for each run.
Given that you are ahead after the first run, what’s the probability that you will still be ahead after the second run and earn your gold medal?
Let your two random run completion times be denoted by X1 and X2, and your opponent's Y1 and Y2. Let Z1 and Z2 denote the difference between your run time and your opponent's. Then regardless of the initial mean and standard deviation of X1, X2, Y1, and Y2, Z1 and Z2 will be identical, independent normal variables with mean zero and some irrelevant but identical variance. Without loss of generality, we may in fact assume that the Z1 and Z2 are unit normals (with variance 1).
Then the answer to the original question is given by P{Z1+Z2≥0∣Z1≥0}=P{Z1+Z2≥0,Z1≥0}P{Z1≥0}.
Put altogether, you get a conditional win probability after the first run as P{Z1+Z2≥0∣Z1≥0}=3/81/2=34.
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