Question

Individuals A and B begin to play a sequence of chess games. Let S 5 5A wins a game6 , and suppose that outcomes of successive games are independent with P(S) 5 p and P(F) 5 1 2 p (they never draw). They will play until one of them wins ten games. Let X 5 the number of games played (with possible values 10, 11, . . . , 19). a. For x 5 10, 11, c, 19, obtain an expression for p(x) 5 P(X 5 x). b. If a draw is possible, with p 5 P(S), q 5 P(F), 1 2 p 2 q 5 P(draw), what are the possible values of X? What is P(20 # X) ? [Hint: P(20 # X) 5 1 2 P(X , 20).]

Answer 1

a. To obtain the expression for p(x) for x = 10, 11, ..., 19, we need to consider the possible sequences of wins and losses that could result in each number of games played.

For x = 10, it means that one player won all 10 games. This can only happen if player A wins all 10 games or if player B wins all 10 games. So, for x = 10, we have:

p(10) = P(A wins all 10 games) + P(B wins all 10 games)

= p^10 + (1/2 - p)^10

For x = 11, it means that one player won 10 games and the other player won 1 game. This can happen in two ways: either player A wins 10 games and player B wins 1 game, or player B wins 10 games and player A wins 1 game. So, for x = 11, we have:

p(11) = P(A wins 10 games and B wins 1 game) + P(B wins 10 games and A wins 1 game)

= 10 * p^10 * (1/2 - p) + 10 * (1/2 - p)^10 * p

Similarly, you can continue this process for x = 12, 13, ..., 19, considering all possible combinations of wins and losses.

b. If a draw is possible, the possible values of X would range from 10 to 20, as a game can end in a draw after the 10th game. P(20 ≤ X) can be calculated using the hint provided:

P(20 ≤ X) = 1/2 * P(X ≥ 20)

= 1/2 * [1 - P(X < 20)]

= 1/2 * [1 - (p(10) + p(11) + ... + p(19))]

You can substitute the expressions for p(x) obtained in part a to calculate the final result for P(20 ≤ X).

Explanation: