Question

Alice and Bob play a game using some dice providing a random integer from 1 to 6. The rules of their game are: Alice wins if no 4 appears after the dice are thrown, bob wins if exactly one 4 appears after the dice are thrown, if two or more 4 appear, the dice are thrown again. How many dice do Alice and Bob need in their game to have equal probability of winning?

Answer 1

not sure exactly, but if \(n\) dice are thrown the probability that there is exactly one 4 is \(n\times (\frac{1}{6})\times (\frac{5}{6})^n\) and the the probability no 4s are thrown is \((\frac{5}{6})^n\) so maybe we can set them equal and try and solve for \(n\)
or maybe we can do it by trial and error

Answer 2

damn typo there
probability that exactly 1 4 shows is
\[n\times (\frac{1}{6})\times( \frac{5}{6})^{n-1}\]

Answer 3

correct question is below

Alice and Bob play a game using some dice providing a random integer from 1 to 6. The rules of their game are:







- Alice wins if no 4 appears after the dice are thrown
- Bob wins if exactly one 4 appears after the dice are thrown
- If two or more 4 appear, the dice are thrown again

How many dice do Alice and Bob need in their game to have equal probability of winning?

Answer 4

might guess at \(n=5\)

Answer 5

if you throw the dice five times, the the probability you get no 4's is
\[\left(\frac{5}{6}\right)^5\]

Answer 6

the probability you get exactly 1 4 is
\[5\times (\frac{1}{6})\times( \frac{5}{6})^4\]

Answer 7

so i think it is 5 times

Answer 8

Ok thanks but why do you have to multiply the whole expression by 5 if you want to find the probability of getting exactly one 4?

Answer 9

because there are five different combinations that give one 4

Answer 10

(4, not 4, not 4, not 4, not 4)
(not 4, 4 , not 4, not 4, not 4)
(not 4, not 4, 4, not 4, not 4)
(not 4, not 4, not 4, 4, not 4)
(not 4, not 4, not 4, not 4, 4)

each of these has the same probability \((\frac{1}{6})(\frac{5}{6})^4\)