Yves and some friends are playing a fair game in which 18 cards are dealt to 6 players. One of the cards is a queen. The player who receives the queen goes first. What is the probability that Yves will go first?
Is it 1/3?
To find the probability that Yves will go first, we need to determine the total number of possible outcomes and the number of favorable outcomes.
There are 6 players, so each player has an equal chance of receiving the queen card. Hence, the number of favorable outcomes is 1, as only one player can receive the queen.
To find the total number of possible outcomes, we need to consider how the 18 cards can be distributed among the 6 players. This can be calculated using the concept of combinations.
There are a total of 18 cards, and we need to distribute them among 6 players. This can be represented as selecting 6 players out of 18. The number of combinations can be calculated using the formula:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of objects, and r is the number of objects selected.
In this case, n = 18 and r = 6. Hence, the total number of possible outcomes is:
C(18, 6) = 18! / (6!(18-6)!)
= 18! / (6!12!)
= (18 * 17 * 16 * 15 * 14 * 13) / (6 * 5 * 4 * 3 * 2 * 1)
= 18564
Now, we can calculate the probability that Yves will go first by dividing the number of favorable outcomes (1) by the total number of possible outcomes (18564):
Probability = 1 / 18564
≈ 0.000054
Therefore, the probability that Yves will go first is approximately 0.000054, or 1 in 18,564. It is very unlikely for Yves to go first in this fair game.