Question

Adam and Eve roll a die. if it shows 1 or 2, Eve wins an apple from Adam and if it shows 3, 4, 5, 6 Adam wins an apple from Eve. Eve has two apples, Adam has only one. What is the probability Adam gets both Eve's apples before he loses his.

Answer 1

The SNAKE WINS!

Answer 2

Okay, lets try this (though I suck at probability):
P(A1) => adam wins apple 1
P(A2|A1) => adam wins apple 2 given that he won apple 1.

Answer 3

oh crap.. there's also before he loses his. Does finding the probability of Adam winning both apples help?

Answer 4

P(A1) = 4/6=2/3

Answer 5

yes it does help to figure out what the probability he wins right away. that will get to the solution

Answer 6

P(A2) = 4/6 = 2/3, so P(A2|A1) = 4/9/2/3 = 2/3 ?

Answer 7

I'm still shaky on P(A2|A1) formula. If you want to find probability of A|B do u add or multiply?

Answer 8

for numerator?

Answer 9

don't worry about conditional probability formula. trials are independent, so think about how adam can win quickest, and then what happens if he doesn't win in two rolls

Answer 10

oh yea i just realized it.

Answer 11

so P(Adam wins quickest) = 4/9
P(Adam loses on first) = P(Adam loses on second) = 1/3

Answer 12

do i just subtract now?

Answer 13

adam can't lose on first roll and still win the game, so ignore that one.

Answer 14

oh wait so that's it? 4/9?

Answer 15

either wins on first two tosses, or then wins and loses. what happens then?

Answer 16

4/9 - 1/3?

Answer 17

2/3

Answer 18

not 2/3

Answer 19

8/9