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"Question for the math guys!" Topic


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otherone04 Dec 2020 8:07 a.m. PST

Hi.
I was wondering what the percentage chance of rolling at least two "1s" on three dice?

Thanks!

Rob

Stryderg04 Dec 2020 8:44 a.m. PST

.074 %

There are 16 combinations that result in either 2 or 3 1's being rolled out of 216 possible combinations.

I brute forced it: went into Excel, wrote out the dice combinations:
1 1 1
1 1 2
1 1 3
etc

Then looked for lines where there are 2 or 3 1's.
I'm sure are more elegant ways to figure it out, but that's my answer. What did I win?

Swampster04 Dec 2020 9:27 a.m. PST

That'd be 7.4% :)
16 combos is right though.
1,1,1.
There are three places where another number can be and 5 numbers other than 1, so 15 other combos with two ones.

Personal logo Parzival Supporting Member of TMP04 Dec 2020 9:33 a.m. PST

There are 6X6X6 or 216 possible combinations of 3 d6. A combination of 1, 1, 1 is therefore only 1 of the possible specific combinations (that it is 1, 1, 1 is only relevant in that there is no ordinal importance to the result, thus making it only 1 possible result).

Therefore the odds are simply 1/216, or approx 0.46%.

One can also calculate the odds by multiplying the odds of each separate die result, which for each is of course 1 in 6, or 16.66… percent.
16.66… cubed (multiplied by itself 3 times) is approx 0.46%.

EDIT: Correctedó forgot to account for the percentage! *face palm*

MajorB04 Dec 2020 10:09 a.m. PST

A combination of 1, 1, 1 is therefore only 1 of the possible specific combinations (that it is 1, 1, 1 is only relevant in that there is no ordinal importance to the result, thus making it only 1 possible result).

Except the OP asked for at least 2 1s on any of the 3 dice, not 3 1s.

pnguyenho04 Dec 2020 1:01 p.m. PST

The chance of rolling a 1 on a D6 is 1/6.
The chance of not rolling a 1 (ie 2-6) on a D6 is 5/6.

There is only one possible combination of all three 1s, so 1/6 x 1/6 x 1/6 = 1/216.

If you label the dice as A, B, and C, in order to roll two 1s only, then the combinations would be:
A1, B1, C2-6 so 1/6 x 1/6 x 5/6 = 5/216
A1, B2-6, C1 so 1/6 x 5/6 x 1/6 = 5/216
A2-6, B1, C1 so 5/6 x 1/6 x 1/6 = 5/216

So chance of rolling at least two 1s is 1/216 + 3 x 5/216 = 16/216 (7.4%).

Stryderg04 Dec 2020 1:26 p.m. PST

That'd be 7.4% :)

{mumble / grumble} stupid percent sign {mumble / grumble}

Thanks for the catch. :)

Personal logo Parzival Supporting Member of TMP04 Dec 2020 8:40 p.m. PST

Ah, missed that detail of 2 out of 3… mea culpa!

Rudysnelson04 Dec 2020 9:29 p.m. PST

1,1,1.
1,1,2. 1,1,3. 1,1,4. 1,1,5. 1,1,6
1,2,1. 1, 3,1. 1,4,1. 1,5,1. 1,6,1
2,1,1. 3,1,1. 4,1,1. 5,1,1. 6,1,1

Those are the combinations for your parameters. Looks like 16 valid rolls.

Dice probability makes a great science fair project. My son took his in middle school and won at the regional competition a few years ago.

otherone07 Dec 2020 3:03 p.m. PST

You guys are the best! Thanks!

Rob

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