Which one has a higher probability? Getting at least 1 six with 6 dice or getting at least 2 six with 12 dice?

Answer 1

Make sure to use the prob function for binomial distribution.

p(at lease 1 six with 6 dice) = 1 - p(no six with 6 dice)
= 1 - (5/6)^6 = 0.665 or so.

p(at least 2 sixes with 12 die) = 1 - p(no six) - p(one six)
= 1 - (5/6)^12 - (12C1)(1/6)(5/6)^11 where 12C1 = 12!/11!*1! = 12
= 0.619.

So the first one is higher.

Answer 2

Prob(at least 1 six with 6 dice) = 1 - Prob( no six with 6 dice) =
1 - (5/6)^6 = (6^6 - 5^6)/ 6^6 = 0.6651.

Prob(at least 2 six with 12 dice) = 1 - Prob(no six with 12 dice) - Prob( 1 six with 12 dice) =
1 - (5/6)^12 - 12(1/6)(5/6)^11 = 0.6187.

The first one has higher probability.

Answer 3

Probability of getting at least 1 six with 6 dice = 1 - (probability of getting no sixes)
= 1 - (5/6)^6
= 1 - 0.33489797668038408779149519890261
= 0.66510202331961591220850480109739

Probability of getting at least 2 sixes with 12 dice = 1 - (probability of zero or one six)
= 1 - (5/6)^12 - (5/6)^12
= 1 - 0.22431330956923016854175722234453
= 0.77568669043076983145824277765547

So, the probability of getting at least two sixes with 12 dice is greater than getting at least one six with six dice. QED

Answer 4

probability of getting 1 six with 6 dice =1/(6^6)
probability of getting 2 six with 12 dice=2/(6^12)
so the first option has more probability.

Answer 5

both