Four people(North, South, East, West) are each dealt 13 cards from an ordinary deck of 52 cards.
a. If South has exactly one ace, what is the probability that his partner North has the other three aces?
b. If North and South together have 10 hearts, what is the probability that either East or West has the other 3 hearts?
To solve these probability questions, we need to use the concept of combinations. Combinations are a way to count the number of ways to select a group of items from a set without concern for the order of selection.
a. If South has exactly one ace, what is the probability that his partner North has the other three aces?
First, let's consider the total number of ways to distribute the remaining 12 aces among the three players (North, East, and West). We can think of this as selecting 3 aces out of the 12 available. This can be calculated using the combination formula:
C(12, 3) = 12! / (3!(12 - 3)!) = 220
Therefore, there are 220 ways to distribute the remaining 3 aces among the three players.
Now, let's focus on the specific scenario where South has one ace and North has the other three aces. There are 3 aces remaining after South receives one, and we want North to have all three of them.
The probability of North having all three aces can be calculated as:
P(North has 3 aces) = 1 / (number of ways to distribute the remaining 3 aces)
P(North has 3 aces) = 1 / 3 = 1/3 ≈ 0.333
Therefore, the probability that North has the other three aces given that South has exactly one ace is approximately 0.333 or 33.3%.
b. If North and South together have 10 hearts, what is the probability that either East or West has the other 3 hearts?
This problem involves a similar concept of combinations. We need to calculate the number of ways to distribute the remaining 3 hearts among the two players (East and West).
Let's first consider the total number of ways to distribute the remaining 3 hearts among the two players:
C(39, 3) = 39! / (3!(39 - 3)!) = 9,139
Therefore, there are 9,139 ways to distribute the remaining 3 hearts among East and West.
Now, let's focus on the scenario where North and South have 10 hearts, and we want either East or West to have the other 3 hearts.
The probability of either East or West having the other 3 hearts can be calculated as the sum of the probabilities of each case:
P(Either East or West has 3 hearts) = P(East has 3 hearts) + P(West has 3 hearts)
To calculate these individual probabilities, we need to consider the distribution of the remaining 3 hearts among the two players (East and West). The possible combinations are:
1. East has 3 hearts, West has 0 hearts: C(10, 3) * C(29, 0) = 120
2. East has 2 hearts, West has 1 heart: C(10, 2) * C(29, 1) = 870
3. East has 1 heart, West has 2 hearts: C(10, 1) * C(29, 2) = 4060
4. East has 0 hearts, West has 3 hearts: C(10, 0) * C(29, 3) = 3654
Adding up these individual cases:
P(East has 3 hearts) = 120 / 9,139 ≈ 0.013
P(West has 3 hearts) = 3654 / 9,139 ≈ 0.4
P(Either East or West has 3 hearts) = 0.013 + 0.4 ≈ 0.413
Therefore, the probability that either East or West has the other 3 hearts, given that North and South have 10 hearts, is approximately 0.413 or 41.3%.