Math help: doubles, triples, quadruples on 2-3-4d10

[overgeeked]

I'm wondering about the likelihood of doubles, triples, and quadruples on 2d10, 3d10, and 4d10.

I know it's 10% chance of doubles on 2d10. There's 1/100 for each number to come up, and there's 10 possible doubles so 10/100.

On 3d10, it's 30/1000 getting doubles, either the first two dice (10/1000), second and third dice (10/1000), or the first and third dice (10/1000).

On 3d10, it's 10/1000 of getting triples.

On 4d10 is where my brain starts hurting.

The quadruples is easy, 10/10,000.

My hunch is that each pairing of dice (1,2; 1,3; 1,4; 2,3; 2,4; 3,4) would have the same 10/10,000 of yielding doubles. There are six possible pairings, so 60/10,000 for doubles.

Likewise, each set of 3 dice (1,2,3; 1,2,4; 1,3,4; 2,3,4) would have 10/10,000 chance of yielding triples. There are four possible pairings, so 40/10,000 for triples.

Any help on doubles and triples on 4d10?

 

[Umbran]

So, here's a question - do you care if the double is also a triple or quadruple?

 

[overgeeked]

So, here's a question - do you care if the double is also a triple or quadruple?

Yes. It’s exclusive. Finding matches and setting them aside. Bigger taking precedence. So quadruples first, triples, then doubles.

 

[Kobold Stew]

I think it goes this way.
We calculate the liklihood of not getting doubles.

* with 2d10, the answer is .9 or 90%, which means there's 10% you do.

* with 3d10, the answer is .9x.8=.72. That means there is a 28% likelihood that you will get doubles, 280 of the 1000 possibilities. From these you are excluding the ten triples, so that gives us 270 possibilities of the 1000 rolls, or 27%.

*With 4d10, the doubles become .9x.8x.7=.504 that you don't, so .496 that you do, less ten quadruples and thirty triples that are not quadruples -- so 49.2%?

 

[TwoSix]

Quadruple = 10/10000 = 1/1000.
Triple = 4*9*10/10000 = 360/10000 = 36/1000.

Single = 10/10 * 9/10 * 8/10 * 7/10 = 5040/10000 = 504/1000.

Doubles must be whatever is left, so 1000 - (504+36+1) = 459/1000.

 

[tomBitonti]

Another problem .. with four numbers, double doubles are possible (2244), (1177), etc.
(Obscure note: This problem of double doubles feels related to interesting things that happen with rotations in 4 dimensions, where two independent rotation axis are possible.)
TomB

 

[TwoSix]

Triples in a quadruple.

10 numbers that can be triples. (1,1,1; 2,2,2; etc.)

Nine combinations where the last number does not make a quadruple. (1,1,1,2; 1,1,1,3; etc.)

4 mirrors of those combinations. (1,1,1,2; 1,1,2,1; 1,2,1,1; 2,1,1,1)

So 10*9*4, all over 10^4. 360/10000.

 

[TwoSix]

Another problem .. with four numbers, double doubles are possible (2244), (1177), etc.
(Obscure note: This problem of double doubles feels related to interesting things that happen with rotations in 4 dimensions, where two independent rotation axis are possible.)
TomB

Double-doubles in a quadruple.

45 combinations (1,1,2,2 to 9,9,0,0) x 6 permutations = 270/10000 = 27/1000.

Doubles, but no double-doubles.

10 doubles x 6 mirrors x 9 values for a 3rd digits x 8 values for 4th digit = 4320/10000 = 432/1000.

Add the 2 and we get 459/1000.

 

[J.Quondam]

"Double, double, toil and trouble..!"

 

[overgeeked]

So, putting that all together.

Chance of doubles on 2d10: 10/100. 10%.

Chance of doubles on 3d10: 270/1000. 27%.

Chance of only one set of doubles on 4d10: 4590/10,000. 43.7%.

Chance of two sets of non-matching doubles on 4d10: 270/10,000. 2.7%.

Chance of triples on 3d10: 10/1000. 1%.

Chance of triples on 4d10: 360/10,000. 3.6%.

Chance of quadruples on 4d10: 10/10,000. 0.1%.

Those right?

And thanks everyone.