Under probability, there was an example given, i.e. what is the probability of rolling three dice, and getting at least one six?

MGMAT's strategy is to calculate the probability of not getting a 6 (i.e. 5/6 x 5/6 x 5/6 = 125/216) and reducing 1 from this figure to get the answer. (The answer is 91/216)

However, I calculated it as follows:

(probability of getting one six) + (probability of getting 2 sixes) + (probability of getting 3 sixes)
= 3(1/6*5/6*5/6) + 3(1/6*1/6*5/6) + 3(1/6*1/6*1/6)
= 93/216

I multiplied each probability by 3 in order to take into account the separate scenarios of dice 1, dice 2 or dice 3 rolling a 6 individually.

I understand the logic of MGMAT's approach but I'm uncertain as to why my approach doesn't seem to work. Hope you can help clarify this. Thanks!

hi -

the approach doesn't work because you shouldn't multiply by 3 in all three cases.

to wit:

there are 3 ways to roll one six: [die 1, die 2, die 3]
not-6 ... not-6 ... 6
not-6 ... 6 ... not-6
6 ... not-6 ... not-6
therefore, you are correct in multiplying this probability by 3.

there are also 3 ways to roll two sixes: [die 1, die 2, die 3]
not-6 ... 6 ... 6
6 ... not-6 ... 6
6 ... 6 ... not-6
therefore, you should (and did) also multiply this probability by 3.

however,
there is only one way to roll all sixes (6 ... 6 ... 6).
therefore, you don't multiply that probability by 3. fix this error and you'll arrive at 91/216, the same probability arising from our calculation.

by the way, this post is a strong endorsement for our solution, using the (1 - x) method: why would you want to go to all this trouble, anyway...?

hope no one reading this post believes in all that 'mark of the beast' stuff.

Thanks Ron. Tricky stuff!

I'll definitely use MGMAT's technique, just wanted to see if my concepts were on right. Thanks again!

good times.

it's good to have both solutions at your disposal, so that, if either one fails to work, you can turn immediately to the other.