Game: I throw 1 die 4 times, trying to reach at least one 6, you throw 2 dice 24 times and try to reach at least one double 6 (6,6). Who has greater chance of winning

To estimate, compare (5/6)^4 and (35/36)^24 this is 5/6 and (35/36)^6 this is 30/36 and (35/36)^6 notice that 30/36 is missing six 1/6th from 1 (36/36) and taking powers of (35/36)^6 will reduce the number by nearly 1/36th each time, but less than than, so that (35/36)^6 is greater than 1-6/36=30/36. Therefore the probability of not getting any double six is greater than probability of not getting any 6, and you should choose to roll one die. To understand the reasoning, think about taking powers of 0.90, 0.90^1 = 1-1x0.10 0.90^2 > 1-2x0.10 0.90^3 > 1-3x0.10 and so on

It all comes to which is greater: 1-(5/6)^4 or 1-(35/36)^24. They will expect you to calculate this (which is greater, not actual numbers) without a calculator.

First comment is correct, second comment is wrong since it asks for at least one 6 or at least one (6,6). This also includes the outcomes 2 or more sixes or double (6,6). Hence the easiest way of calculating this is by calculating the complementary probabilities P(no six) and P(no double six), respectively, to get P(at least one six) = 1 - P(no six) and P(at least one double 6) = 1 - P(no double six), which gives the result in the first comment.

I agree with the Deen's answer. Just to add up, basically if 1/36 is constantly decreased six times, it is equal to decreasing 1/6 one time. However, each time you decrease 1/36, the total size shrinks, so the next 1/36th decrease is smaller than 1/6. Therefore, 35/36^6 is greater than 5/6.

I agree with the Deen's answer. Just to add up, basically if 1/36 is constantly decreased six times, it is equal to decreasing 1/6 one time. However, each time you decrease 1/36, the total size shrinks, so the next 1/36th decrease is smaller than 1/36. Therefore, 35/36^6 is greater than 5/6.

Bernoulli's inequality: (1-p)^n > 1-np (35/36)^24 = ((1-1/36)^6)^4 > (1-6/36)^4 = (5/6)^4 so 1-(5/6)^4 > 1-(35/36)^4, i.e., choose the first option

@Try again, 1 - (35/36)^24 = .4914 So you get the opposite answer... "1-(5/6)^4 or 1-(35/36)^24. They will expect you to calculate this (which is greater, not actual numbers) without a calculator." Is there a standard way of estimating this?

First answer is correct. Also comparing the too probability is easy. You just need to compare (35/36)^4 and (5/6)^4. Write 35/36 = (7/6)*(5/6) > (5/6) as 7/6 > 1 So, 1-(35/36)^24 < 1-(5/6)^4 Hence I will prefer atleast 6 option in 4 rolls.

The first comment would be correct if you needed to get a string of 6's, probability from 4 rolls of one 6 is 4/6 (4 * 1/6), probability of 6 and a 6 is 1/36, rolled 24 times = 24/36, so they are both the same odds.