You are participating in a raffle where there are 11 baskets and 10 players who take turns, one after another, distributing 5 tickets among the baskets. Assuming perfect play by all parties, what is the expected number of wins for the last player?
To find the expected number of wins for the last player, we can use the concept of linearity of expectation.
Let's analyze the game step by step:
In the first round, the first player randomly distributes 5 tickets among the 11 baskets. The probability of any particular basket ending up with a ticket is 5/11.
In the second round, the second player distributes 5 tickets among the remaining baskets. As there are now tickets in some of the baskets, the overall probability of a particular basket ending up with a ticket is reduced. Let's calculate this probability:
The probability of a specific basket not receiving a ticket in the first round is (11-1)/11 = 10/11. The probability that this basket does not receive a ticket in the second round is then (10/11) * (10-1)/10 = 9/11.
Therefore, the probability of a specific basket receiving a ticket in the second round is 1 - 9/11 = 2/11.
Following the same logic, we can deduce that in each subsequent round, the probability of a specific basket receiving a ticket decreases by 2/11. Hence, the probability of the last player winning a specific basket in the last round is 2/11.
Since there are 11 baskets, the expected number of wins for the last player can be calculated by multiplying the probability of winning a specific basket by the total number of baskets, which in this case is 11:
Expected number of wins = (2/11) * 11 = 2.
Therefore, the expected number of wins for the last player is 2.