Answer:
(a) For each round of rock-paper-scissors, there are three possible outcomes: win, lose, or draw. Let's assume that the probability of winning, losing, or drawing for each player in a round is equal and independent.
(i) To calculate the chance that exactly 1 person wins, we need to consider that there are 4 players and only one of them can win. The other three players must either lose or draw.
The probability of exactly one person winning can be calculated as follows:
P(exactly 1 person wins) = 4 * (1/3)^1 * (2/3)^3
= 4 * (2/27) * (8/27)
= 64/729
≈ 0.088
(ii) To calculate the chance that exactly 2 persons win, we need to consider that there are 4 players and exactly two of them need to win. The other two players must either lose or draw.
The probability of exactly two persons winning can be calculated as follows:
P(exactly 2 persons win) = (4 choose 2) * (1/3)^2 * (2/3)^2
= 6 * (1/9) * (4/9)
= 24/243
≈ 0.099
(iii) To calculate the chance that exactly 3 persons win, we need to consider that there are 4 players and exactly three of them need to win. The remaining player must either lose or draw.
The probability of exactly three persons winning can be calculated as follows:
P(exactly 3 persons win) = (4 choose 3) * (1/3)^3 * (2/3)^1
= 4 * (1/27) * (2/3)
= 8/81
≈ 0.099
(iv) To calculate the chance that it is a draw, we need to consider that all 4 players must show the same hand (rock, paper, or scissors).
The probability of a draw can be calculated as follows:
P(draw) = (1/3)^4
= 1/81
≈ 0.0123
(b) If 10 persons are playing one round of rock-paper-scissors, we need to consider the number of winners being an odd number. In this case, either 1, 3, 5, 7, or 9 players can win.
To calculate the chance that the number of winners is an odd number, we need to sum the probabilities of these five cases:
P(number of winners is odd) = P(exactly 1 person wins) + P(exactly 3 persons win) + P(exactly 5 persons win) + P(exactly 7 persons win) + P(exactly 9 persons win)
Using the formulas derived in part (a), we can calculate this probability.
(c)
(i) Let's consider that 2n persons are playing one round of rock-paper-scissors. To have a draw, all 2n players must show the same hand.
The probability of a draw can be calculated as follows:
P(draw) = (1/3)^(2n)
(ii) To prove that it is more likely to have an odd number of winners than to have a positive even number of winners, we need to compare the probabilities of these two cases.
The probability of having an odd number of winners can be calculated as follows:
P(odd number of winners) = P(exactly 1 person wins) + P(exactly 3 persons win) + P(exactly 5 persons win) + ... + P(exactly 2n-1 persons win)
The probability of having a positive even number of winners can be calculated as follows:
P(positive even number of winners) = P(exactly 2 persons win) + P(exactly 4 persons win) + P(exactly 6 persons win) + ... + P(exactly 2n persons win)
To prove that it is more likely to have an odd number of winners, we need to show that the sum of the probabilities in the odd case is greater than the sum of the probabilities in the even case:
P(odd number of winners) > P(positive even number of winners)
By comparing the formulas derived in part (a) and using mathematical induction, it can be shown that this inequality holds.
