[stats] Whats the odds of three sixes in yd6?

[Morrus]

Yep, it's one of those dice stats questions! Bonus XP for help from statisticians!

Say I need three sixes or more to succeed at a task. A difficult one, albeit.

What is my chance of success if I am rolling 5d6?

What is it when I roll 6d6, 7d6, and so so on?

 

[Muad'dib Pendragon]

AnyDice has your answer: output [count 6 in 3d6]

 

[Nagol]

Simplest way to calculate it is to take 1 - (not rolling at least three sixes) using the binomial theorem.

Chance of rolling 0 sixes = (5 / 6) ** number of dice * (number of dice choose zero which simplifies to 1)
Chance of rolling one 6 is (1/6) * (5/6) ** (number of dice - 1) * (number of dice choose 1 which simplifies to number of dice)
chance of rolling two 6s is (1/6)**2 * (5/6)**(number of dice - 2) * (number of dice choose 2) which simplifies to n * (n-1) / 2

 

[Oldtimer]

Here is a nice graph of it:

 

[Morrus]

Very helpful. Thank you guys!

 

[Henry]

That program is excellent training for Liar's Dice. Thank you for the info!

 

[tomBitonti]

Is that chart correct? At 18 dice, it seems that the odds should not be above 50 %.

Thx!
TomB

 

[Morrus]

This is what I got.

 

[Nagol]

Is that chart correct? At 18 dice, it seems that the odds should not be above 50 %.

Thx!
TomB

The chance of getting at least 3 sixes on 18 dice is about 59.7%

P = 1 - [ power(5/6,18) + power(5/6,17) * (1/6) * 18 + power( 5/6, 16) * power(1/6,2) * 18 * 17 / 2 ]
~= 1 - [ 0.037 + 0.135 + 0.23 ]
~= 0.598 (with rounding errors; it's closer to 0.597)

 

[tomBitonti]

Ah, ok, I see where I went wrong. With 18 rolls, the expected number of 6's is 3. But that doesn't mean 100% chance of success, since a lot of rolls have more than 3 sixes. I turned that 100% into 50% and thought the chance should be less than 50%. (Easier to think about with n d2 looking for n/2 2's.)

Thx
TomB