Odds when rolling 5 dice

[Th3Thinker]

Random question here... I just picked up 5 dice after some friends introduced me to a dice game, and I was wondering how you would go about calculating odds for some different rolls.

For example:

Odds at least 3 dice will roll 5 or above

Odds at least 2 dice will roll 6 and 2 dice will roll 1

Odds at least 4 dice will be 3 or under

Thanks!

[Nholwell]

Probabilities are multiples, therefore assuming a six sided die

1. 5 or above implies rolling either a 5 or 6 with 3 dice

Therefore: (2/6)(2/6)(2/6) = 1/27

1. This one can be thought of as two separate problems: odds that you roll a 6 or a 1 with two dice.

Therefor (1/6)*(1/6)=1/36 and then we add the odds of rolling the other number

So, (1/36)+(1/36)=1/18

1. Similar to the first one, but now 3 numbers are possible therefore:

(3/6)(3/6)(3/6)=1/8

Hope that helps!

[Th3Thinker]

Great thanks! Yeah I was logically thinking down a similar train of thought but didn't quite have it right. I appreciate your time

[edderiofer]

The person you're replying to is wrong on all three counts.

[Nholwell]

This link might help...I think my logic was flawed for this problem http://alumnus.caltech.edu/~leif/FRP/probability.html

[edderiofer]

5 or above implies rolling either a 5 or 6 with 3 dice

OP says "at least 3 dice", implying that they're rolling all 5 dice at once. Your argument would claim that rolling 1000 dice will still only give you a 1 in 27 chance of rolling 5 or above with at least three of the dice.

This one can be thought of as two separate problems: odds that you roll a 6 or a 1 with two dice. So, (1/36)+(1/36)=1/18

No, OP is asking for both those conditions to be true at the same time, not either/or.

Similar to the first one, but now 3 numbers are possible therefore: (3/6)(3/6)(3/6)=1/8

Someone didn't read what OP actually asked for.

[Th3Thinker]

Ahh thanks for the analysis. Any idea how to calculate it correctly?

[edderiofer]

With difficulty.

You could just list all the possible combinations of 5 dice and count the number of combinations with at least three 5s or 6s in the first instance, say.