Question: Bob repeatedly rolls a fair 6-sided die. What is the probability that he rolls his first 5 before he rolls his second (not necessarily distinct) even number?
What I've Tried: Let 'n' be the number of rolls in the game. This means that n has to be at least 2. I then created cases:
n = 2: Both rolls would have to be even so \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\)
n = 3: We have to have an odd number (not a 5) and two even numbers (but one of the even numbers has to be last), so \(\frac{1}{2}\cdot\frac{1}{3}\cdot2\cdot\frac{1}{2} = \frac{1}{6}\)
After this, I'm not sure what to do. Help would be appreciated!
+2407 o = rolling a 1 or 3 (1/3 probability)
e = rolling a 2, 4, or 6 (1/2 probability)
f = rolling a 5 (1/6 probability)
So we have 3 cases. Rolling a bunch of "o"s before rolling an f. Rolling a bunch of "o"s and 1 "e" before rolling an f. Rolling an f.
Rolling a bunch of "o"s before rolling an f.
((1/3) * 1/6) + ((1/3)^2 * 1/6) + ((1/3)^3 * 1/6).... + ((1/3)^infinity * 1/6)
This is a geometric sequence.
a = 1/18, r = 1/3, a/(1-r)
(1/18)/(2/3) = 1/12
Rolling a bunch of "o"s and 1 "e" before rolling an f.
2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6)
I'm not actually sure how to calculate this, but hopefully someone else can help with that. #It probably is around 1/5 to 1/4.
Rolling an f.
1/6
1/6 + 1/12 + (2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6))
1/4 + (2((1/3) * 1/6) + 3((1/3)^2 * 1/6) + 4((1/3)^3 * 1/6)... + (infinity - 1)((1/3)^infinity * 1/6))
Sorry I wasn't able to help more. :((
=^._.^=
1 - Probability of rolling a "5" ==1/6. The expected probability of getting a "5" ==n * 1/6, solve for n.
n ==30 rolls.
2 - In 30 rolls, the expected probability of rolling a "2", "4", "6" ==30 * 1/2==15
3 - Then the probability of rolling a "5" before a "2", or a "4", or a "6" ==5 / 15 ==1/3
