Question

Suppose that when Team A and Team B play basketball, the difference between their scores in the first half (Team A’s score minus Team B’s score) has approximately a normal distribution with mean 0 and standard deviation 8. Likewise, the difference between their scores in the second half has approximately a normal distribution with mean 0 and standard deviation 8. Also, first half and second half scores are independent. If you know that Team A won the game by 10 points, what is the probability that Team A was ahead at the end of the first half?

Answer 1

Let Team A's score in the first half be X1 and Team B's score in the second half be Y1. Similarly, let Team A's score in the second half be X2 and Team B's score in the second half be Y2.

It is given that M = X1-Y1 ~N(0,8) and N = X2-Y2 ~N(0,8).

So, M+N ~N(0,8+8) i.e. N(0,16)

If Team A won the game by 10 points then (X1+X2)-(Y1+Y2)=10 i.e. M+N=10

Probability that team A was ahead in the first half is given by P(M>0)

Standardising the above we have,