Rolling Dice

If you roll 5 (six sided) dice, is it just as likely you roll 3 of a kind as you do a pair or am I under thinking this a bit?

Never mind, That is definitely not correct.

I think I figured it out...

the number of all possible cases is 6^5

The number of favorable cases is

C(5,3)=10, (the number of ways you can select 3 of 5 dice, distinctly)

C(5,3)*6=60, (each one of the ten ways above has 6 choices. e.g.1-6)

C(5,3)*6*5*5 = 1500, (each one of the 60 choices has a remaining 5 choices each for the two remaining selections)

P(3 of a kind) = 1500/6^5.

Before, when I thought P(1 pair)=P(3 of a kind) was because I failed to factor in the remaining two numbers(thus leaving out the factor of 125 and 25 respectively). And since C(5,3)=C(5,2) thought they were equal...in case anyone was wondering what all this fuss was about.

The chances of a three of a kind and a two of a kind are equal, but only on dice with just two different sides, such as coins.

Going through all the possibilities I got

five-of-a-kind: 6
four-of-a-kind: 150
full house: 450
three-of-a-kind: 1200
pair: 5400
two pairs: 1800
straight: 240
no combination, gap straight: 480

..which is not right as it should add up to 7,776 but doesn't. Probably a permutation where there should be combination somewhere, or vice versa.

Originally posted by talzamir
The chances of a three of a kind and a two of a kind are equal, but only on dice with just two different sides, such as coins.

Going through all the possibilities I got

five-of-a-kind: 6
four-of-a-kind: 150
full house: 450
three-of-a-kind: 1200
pair: 5400
two pairs: 1800
straight: 240
no combination, gap straight: 480

..which is not right as ...[text shortened]... but doesn't. Probably a permutation where there should be combination somewhere, or vice versa.

Full house: C(5,3)*6*5*1 = 300
pair: C(5,2)*6*5*4*3 = 3600
Straight: 6*5*4*3*2 = 720

which when removing that "gap straight" adds to 7776

So.. fixed. Seems we posted at the same time. =)

five-of-a-kind: 6
four-of-a-kind: 150
full house: 300
three-of-a-kind: 1200
two pairs: 1800
one pair: 3600
straight: 240
no combination, gap straight: 480

Originally posted by talzamir
So.. fixed. Seems we posted at the same time. =)

five-of-a-kind: 6
four-of-a-kind: 150
full house: 300
three-of-a-kind: 1200
two pairs: 1800
one pair: 3600
straight: 240
no combination, gap straight: 480

Can you walk me through the two pair calculation? I'm struggling to come up with it.

I can't get away from 3600, the number for 1 pair which is obviously wrong, so what am I double counting?

C(5,2)*6*C(3,2)*5*4=3600

are aces wild?

I went by yazzy rules. So a straigth is 1-2-3-4-5 or 2-3-4-5-6, but not 4-5-6-1-2. And for aces or deuces wild.. *grins* we'd get a whole different set of probabilities.

Joe wrote: Can you walk me through the two pair calculation? I'm struggling to come up with it.


Actually.. I got annoyed at the mistakes I did in the first calculation, so I did it the easy way, and wrote a bit of code that went over the 7,776 combinations and checked what they are.

One pair is indeed in 3,600 combinations. In the first go I did it like this.

Pick the number for the pair; six possibilities. They can be in (5,2) = 10 different places in the sequence.

The other numbers that go into the gaps are one of five, one of four, and one of three possibilities.

6 x 10 x 5 x 4 x 3 = 3,600.

Originally posted by talzamir
Joe wrote: Can you walk me through the two pair calculation? I'm struggling to come up with it.


Actually.. I got annoyed at the mistakes I did in the first calculation, so I did it the easy way, and wrote a bit of code that went over the 7,776 combinations and checked what they are.

One pair is indeed in 3,600 combinations. In the first go I did it ...[text shortened]... ps are one of five, one of four, and one of three possibilities.

6 x 10 x 5 x 4 x 3 = 3,600.

Ok, I was asking how you calculated "two pair's" number of calculations, but don't worry about it. I did some research and found why I was twice as high, because in each one of the combinations the order of the pairs can be switched (e.g. 1,1,2,2,5 has a compliment combination where the order of the pairs are switched i.e. 2,2,1,1,5), I missed that entirely.

Its more complicate with aces wild, as in perudo.

Here is a puzzle - in perudo you throw your dice and score for the largest number that are all the same, except with aces wild, eg, 1,1,3,5,5,6 would be four fives.

I threw N dice which, with aces wild were just enough dice to make it more likely I would throw three of a kind than a pair. How many dice did I throw?

Note three of the kind has to be the largest set to count, the above throw is four fives, and is NOT a pair OR three of a kind.