Dice Probability Help

[Aqua]

Hai~
My mind is pretty frazzled right now, and I'm having trouble thinking about this...

Could someone tell me and explain the following if I am wrong?

Given 5d6, what is the probability that you roll any kind of triple, quadruple, or quintuple?
The order in which they are rolled does not matter, as long as a triple, quadruple, or quintuple can be formed.
So... 1-1-1-2-3 is the same as 2-3-1-1-1.

What I got was 630 out of 7776 possibilities or 8.102%.  Is this correct?

[ShadowPierce]

Yay! I'm correcting the great Aqua!

Sorry, but no... That's a permutation where a specific order of numbers are counted as one combination... If you have a scientific calculator nearby, just use the "nCr" function or combination to get a combination of 5 numbers out of 6 where repetition is allowed and order doesn't matter...

I'm too lazy to compute it myself so go ahead and do it yourself... Sorry...

~EDIT:
Had free time after all... There were:

126 triples minus the possibility of getting a similar number for the 4th and/or 5th number which is 42 = 84
36 quadruples minus the possibility of getting a similar number for the 5th number which is 6 = 30
6 quintuples

In total, there are 120 repetitions of triples and above out of 6C5 or 252 combinations which equals to 47.619047619047619...%

Glad I helped...

[winkio]

That doesn't account for the extra dice.

6 quintuples
(1s, 2s, 3s, 4s, 5s, 6s)
6 quadruples * 5 positions the extra dice can be * 5 numbers the extra dice can be = 150 quadruples
(XOOOO OXOOO OOXOO OOOXO OOOOX)             (every number but the quadruple = 6 - 1 = 5)
6 triples * 10 positions the extra dice can be * 25 numbers the extra dice can be = 1500 triples
(XOOOX OXOOX OOXOX OOOXX                     (every number but the quadruple on two dice = (6 - 1) * (6 - 1) = 25
XOOXO OXOXO OOXXO
XOXOO OXXOO
XXOOO)

Total probability
(6 + 150 + 1500) / 6^5 = 21.3%

[ShadowPierce]

I don't play DND, but 5d6 means 5 6-sided dice right? If so, then I still stand by my answer...

She also said that positions don't matter...

[winkio]

If you are rolling dice, then each of the unique rolls has a different chance of appearing.  For example, 55555 has a 1/6^5 chance of occurring, while 55551 has a 5/6^5 chance of occuring, because it is the same as 55515 and 55155 and 51555 and 15555.  Your method does not include this information, and thus is incorrect.

[Blizzard]

I agree with winkio. You have to consider all 5 possible permutations of a combination such as 55551.
Also, I think you made a mistake, winkio. Isn't it supposed to be 5 * (1/6)⁵ and not (5/6)⁵? (Except if this was actually what you meant to write since there are no parenthesis and it kinda confused me.)

[winkio]

yes, that is what I meant: 5/(6^5).  I know parenthesis are clearer, but exponents are first in order of operations.

[ShadowPierce]

I still don't see it, but you're the physicist/physics expert here so I guess I'll leave the math up to you... No further explanation needed for me...

[Aqua]

Oooh I see where I went wrong.
I multiplied the number of combinations that worked (21) by the number of numbers (6) by the number of possible positions (5).
Thanks winkio and Blizz.  (And SP for trying.)

Now that I know this chance for realz... time to find the chance for other combinations...
I'll try to do it myself, but might need help later XD

(Other combinations I need for 5d6:
Step: 3 consecutive numbers:  123, 234, 345, or 456
Jump: 135 or 246
I'll leave that there if you guys /really/ wanted to help me~ <3)



Edit:
Step: 14.19753086%
Jump: 7.098765432%
Lol that's what I got by making quick assumptions...

[Blizzard]

I'm a bit rusty in combinatorics, but I'll give it a shot.

Step:

- the easy ones

2 favorable base quintuples (12345 and 23456) with each having 5! permutations = 2 * 5! = 2 * 120 = 240

- strict combinations of only 4 consecutive numbers

2 favorable quadruples (1234, 3456) with each having 4! permutations = 2 * 4! = 2 * 24 = 48
48 quadruples * 5 positions the extra dice can be * 1 number the extra dice can be = 48 * 5 = 240
(if there are 1234, then the only number allowed is 6 because it does not continue the sequence as 5 would since that would be a quintuple which we already counted)

- strict combinations of only 3 consecutive numbers

2 favorable triples (123, 456) with each having 3! permutations = 2 * 3! = 2 * 6 = 12
12 triples * 10 positions the extra dice can be * 4 numbers the extra dice can be = 12 * 10 * 4 = 480

2 favorable triples (234, 345) with each having 3! permutations = 2 * 3! = 2 * 6 = 12
12 triples * 10 positions the extra dice can be * 1 number the extra dice can be = 12 * 10 * 1 = 120
(if there are 234, then the only number allowed is 6 because it does not continue the sequence as 1 or 5 would)

- combinations of 4 numbers where the 5th one is a repetition

3 favorable quadruples (1234, 2345, 3456) with each having 3! permutations = 3 * 3! = 3 * 6 = 18
18 quadruples * 10 positions the repeating dice can be * 4 numbers the extra dice can be = 18 * 10 * 4 = 720

Now I'm not entirely sure if I am going in the right direction with this and if there is a simpler method. There are 3 more combination group that have to be counted.

- combinations of 3 numbers with one repetition
- combinations of 3 numbers with two repetitions
- combinations of 3 numbers with two repetitions of the same number

I wrote a small program to count them manually and the result should be 3480 (~45%). I have counted 1800 so far so the remaining 1680 should be in the remaining 3 groups. If I'm right that is.

[Aqua]

Lol... When I saw that this had a new post.
I was like... "WHO IS HK-47 AND WHY IS HIS FIRST POST IN THIS!?!"
Then I clicked and was like... Oh... it's Blizzy.

Then I read the post, and was like.
WAT.

I understand your reasoning and see why it's more complicated than what I did.  Just trying to think to myself if 45% seems right.
Lol I suck at this :x

Thankies.

[Blizzard]

As I said, I ran it through a script and ~44.75% is definitely right.
I've tried reverse counting by counting all combinations with 3 and then removing duplicates, but that would have been even more complicated. I'm not sure if the remaining 3 groups can be counted together somehow (I think they can't).