Darla and Bill play a game where they pick 2 different numbers from the set {1, 2, 3, 4}. If the sum is larger than the product Darla wins. Otherwise Bill wins. What is the probability that Darla wins?
Yes you are right mathematician
Darla and Bill play a game where they pick 2 different numbers from the set {1, 2, 3, 4}. If the sum is larger than the product Darla wins. Otherwise Bill wins. What is the probability that Darla wins?
1,2 s=3 P=2 D
1,3 S=4 p=3 D
1,4 S=5 P=4 D
2,3 S=5 p=6 B
2,4 S=6 P=8 B
3,4 S=7 P=12 B
THE PROBABILTY THAT DARLA WILL WIN IS 1/2 OR 0.5 OR 50%
I'm going to assume that the choices are random choices, otherwise, Bill could always choose to win.
I'm also going to assume that the game is played "with replacement", meaning that both players choose from all four posibilities, not that one player gets two of the numbers, while the other player gets the two that are left.
So, what are the possibilities for Darla?
1 + 2 = 3 1 + 3 = 4 1 + 4 = 5 2 + 3 = 5 2 + 4 = 6 3 + 4 = 7
There are 6 possibilities for Darla (we must count the two ways for Darla to get 5 as separate possibilities).
Now, what are the possibilities for Bill?
1 x 2 = 2 1 x 3 = 3 1 x 4 = 4 2 x 3 = 6 2 x 4 = 8 3 x 4 = 12
Each of the possibilities for Bill has a 1/6th chance of occuring.
In how many different ways can Darla beat Bill when
Bill gets 2? 6 ways So Carla has a 6/6 chance to win.
Bill gets 3? 5 ways So Carla has a 5/6 chance to win.
Bill gets 4? 4 ways So Carla has a 4/6 chance to win.
Bill gets 6? 1 way So Carla has a 1/6 chance to win.
Bill gets 8? 0 ways So Carla has a 0/6 chance to win.
Bill ets 12? 0 ways So Carla has a 0/6 chance to win.
So now we need to multiply the probability that Bill gets a certain result times the probability that Darla beats him and add all those results together:
(1/6)(6/6) + (1/6)(5/6) + (1/6)(4/6) + (1/6)(1/6) + (1/6)(0/6) + (1/6)(0/6) = 16/36 = 4/9
The probability that Darla beats bill is 4/9.
If the choice is "without replacement" (one of them gets two of the numbers, the other gets the other two numbers) needs a different analysis. Re-post if that is the problem.
Isn't the question asking the probability if only two numbers were picked? As in: Two numbers are picked in total, not two by each?
Yes you are right mathematician
Darla and Bill play a game where they pick 2 different numbers from the set {1, 2, 3, 4}. If the sum is larger than the product Darla wins. Otherwise Bill wins. What is the probability that Darla wins?
1,2 s=3 P=2 D
1,3 S=4 p=3 D
1,4 S=5 P=4 D
2,3 S=5 p=6 B
2,4 S=6 P=8 B
3,4 S=7 P=12 B
THE PROBABILTY THAT DARLA WILL WIN IS 1/2 OR 0.5 OR 50%