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"3 d6 Probabilties (for geniuses)" Topic


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490 hits since 2 Jan 2017
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alexjones Inactive Member02 Jan 2017 10:22 a.m. PST

I am struggling with the calculations of probability for rolling 3 6 sided die and getting the following results:

i) 6 on any of the 3 die rolled.
ii) 5 or 6 on any of the 3 die rolled
iii) 4 ,5 or 6 on any of the 3 die rolled.

Can anyone help please?

gbowen02 Jan 2017 10:34 a.m. PST

There are 3 dice of 6 faces, 6*6*6 possible results = 216.
A unique result such as 6, 6, 6 would be 1/216
For one face on 6 the chance is of a 6 on 1 die and any other result on the other 2.
1/6 * 5/6 * 5/6 = 25/216
This can happen with either of the 3 dice so add up the 3 scores.
75/216 = 35%

JMcCarroll02 Jan 2017 10:41 a.m. PST

No genius but…
i) 3/6 or 50% of 1 success.
ii) 6/6 or 100% of 1 success.
iii) 9/6 or 150% of 1 success.
hope this helps!

alexjones Inactive Member02 Jan 2017 11:01 a.m. PST

to re phrase to make it a bit clearer

i) 6 on at least one of the 3 die rolled. (includes rolling 2 sixes or 3 sixes)

ii) 5 or 6 on at least one of the 3 die rolled (includes rolling 2 sixes or 3 sixes or 2 fives or 3 fives)

iii) 4 ,5 or 6 on at least one of the 3 die rolled.(includes rolling 2 sixes or 3 sixes or 2 fives or 3 fives or 2 fours or 3 fours).

MajorB02 Jan 2017 11:15 a.m. PST

For one face on 6 the chance is of a 6 on 1 die and any other result on the other 2.
1/6 * 5/6 * 5/6 = 25/216
This can happen with either of the 3 dice so add up the 3 scores.
75/216 = 35%

But you also need to add:
The possibility of 6 on 2 dice and any other score on the other:
1/6 * 1/6 * 5/6 = 5/216
And this can also happen in 3 ways so:
15/216
Then there is also the possibility of 6 on on all 3 dice:
1/6 * 1/6 * 1/6 = 1/216
75 + 15 + 1 = 91
91/216 = 42%

A 5 or 6 on 1 die and any other result on the other 2.
2/6 * 4/6 * 4/6 = 32/216
This can happen with any of the 3 dice so add up the 3 scores.
96/216
Then there is the possibility of 5 or 6 on 2 dice and any other score on the other:
2/6 * 2/6 * 4/6 = 16/216
And this can also happen in 3 ways so:
48/216
Then there is also the possibility of 5 or 6 on on all 3 dice:
2/6 * 2/6 * 2/6 = 8/216
96 + 48 + 8 = 152
152/216 = 70%

A 4, 5 or 6 on 1 die and any other result on the other 2.
3/6 * 3/6 * 3/6 = 27/216
This can happen with any of the 3 dice so add up the 3 scores.
81/216
Then there is the possibility of 4, 5 or 6 on 2 dice and any other score on the other:
3/6 * 3/6 * 3/6 = 27/216
And this can also happen in 3 ways so:
81/216
Then there is also the possibility of 4, 5 or 6 on on all 3 dice:
3/6 * 3/6 * 3/6 = 27/216
81 + 81 + 27 = 189
189/216 = 87%

martin goddard Sponsoring Member of TMP02 Jan 2017 11:30 a.m. PST

Good chat. Very interesting
may I chip in with some thoughts.

It may be simpler to calculate the probability of total failure . Take this away from 1 and what is left is the probability of success!

Chance of not gettinga 6 is 5/6. Chance of this happening on 3 dice is 5/6 x 5/6 x 5/6= 125/216

So if that is prob of failure. all othet outcomes woudl give 1 or more successes!

So prob is 91/216 of success.

A 5 or 6 would be 4/6x 4/6 x 4/6= 64/216 of failure. Thus 152/216 for success.


So in closing , find the chance it will not happen and take it way from 1.

DesertScrb02 Jan 2017 11:38 a.m. PST

Play around on this website: anydice.com

alexjones Inactive Member02 Jan 2017 12:22 p.m. PST

thanks for the solutions and especially how you calculated it! very useful.

wakenney02 Jan 2017 12:43 p.m. PST

There are 216 possible combinations of results. Out of those, 91 of the sets will contain a 6 on at least one die. Meaning that the probability of rolling one or more 6s is 91/216 or 42.13%.

For 5&6, P = 70.37%

For 4,5&6, P = 87.5%

Basically, confirmation of everyone else's answers.

Personal logo Shaun Travers Supporting Member of TMP02 Jan 2017 1:45 p.m. PST

I am with Martin. I find it easier to do the chance of it not occurring, and then subtract that from the total possible results.

(Phil Dutre)03 Jan 2017 2:34 a.m. PST

If you're not afraid of a little math: look up binomial distributions. Those describe the number of x successes, on y tries, each with a z probability for success.

But with the low numbers of dice you need, it's indeed possible to compute it exhaustively as shown above.

Personal logo John Treadaway Supporting Member of TMP16 Jan 2017 7:50 a.m. PST

If it helps, I found wakenney's answer the easiest to understand…. wink

Other maths too much for my tiny brain…

John T

Edit: oh and thankfully I have more than one set of percentile dice!

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