TEN DICE Probability Questions!

What is the probability of rolling a 4 sided, 6 sided, 8 sided, 10 sided, 12 sided, and 20 sided dice all at once to get a double?

(rolling all six of these dice what is the probability of getting 2 matching die)

I'd also like to know the probability of getting 3, 4, 5, and 6 matching die.

As well as the odds of getting two triples. (example: a roll of 3 3 3 5 5 5)

Also the odds of getting a triple and a double together. (example: a roll of 3 3 3 5 5 6)

The odds of getting 2 doubles form these six dice. (example: 2 2 4 4 5 6)

The odds of getting 3 doubles from these six dice. (example: 2 2 3 3 4 4)

Finally the odds of getting 4 matching dice and a double. (example: 4 4 4 4 6 6)

I can determine dice odds with six sided dice, but i am lost when it comes to determining the probability of mixing the number of sides of dice. (using the d4-d20 together) I may be just over thinking this, but please help me if you can thank you.

It might help you resolve the fog of probability if you eliminate all of the possibilities that can't happen. You can never ever get 6 matching dice if your 20 sided dice rolls a 19.

There are 4 ways of rolling a 4-sided die.
There are 6 ways of rolling a 6-sided die.
There are 8 ways of rolling a 8-sided die.
There are 10 ways of rolling a 10-sided die.
There are 12 ways of rolling a 12-sided die.
There are 20 ways of rolling a 20-sided die.

The result of any die doesn't affect the others. So it's all independent. So there are:

4*6*8*10*12*20 permutations of results. 460800 total possible outcomes.

So work out the possibilities:

P(Pair 1) - situation where EXACTLY two dice show a "1": Each die has only a single 1 on it. So...

That is, the 4-sided die has one possible #1 value. The 6-sided die has one possible #1 value. The 8-sided die has 7 possible values that are NOT #1. The 10-sided die has 9 possible values that are NOT #1. The 12-sided die has 11 possible values that are NOT #1. The 20-sided die has 19 possible values that are NOT #1. You multiply those together: 1*1*7*9*11*19 and get 13167 possible permutations of the 4- and 6-sided dice pairing a #1 and the other dice not pairing it. (Without consideration of what they are, other than NOT #1.)

For the rest of this walkthrough, I'll leave off the 1*1* preface, along with the explanations. It's all repetitive.

If the 4-sided and 6-sided dice have a "1", the other dice cannot have a "1": There are: 7*9*11*19 (=13167) different permutations of this happening.
If the 4-sided and 8-sided dice have a "1", the other dice cannot have a "1": There are: 5*9*11*19 (=9405) different permutations of this happening.
If the 4-sided and 10-sided dice have a "1", the other dice cannot have a "1": There are: 5*7*11*19 (=7315) different permutations of this happening.
If the 4-sided and 12-sided dice have a "1", the other dice cannot have a "1": There are: 5*7*9*19 (=5985) different permutations of this happening.
If the 4-sided and 20-sided dice have a "1", the other dice cannot have a "1": There are: 5*7*9*11 (=3465) different permutations of this happening.

If the 6-sided and 8-sided dice have a "1", the other dice cannot have a "1": There are: 3*9*11*19 (=5643) different permutations of this happening.
If the 6-sided and 10-sided dice have a "1", the other dice cannot have a "1": There are: 3*7*11*19 (=4389) different permutations of this happening.
If the 6-sided and 12-sided dice have a "1", the other dice cannot have a "1": There are: 3*7*9*19 (=3591) different permutations of this happening.
If the 6-sided and 20-sided dice have a "1", the other dice cannot have a "1": There are: 3*7*9*11 (=2079) different permutations of this happening.

If the 8-sided and 10-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*11*19 (=3135) different permutations of this happening.
If the 8-sided and 12-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*9*19 (=2565) different permutations of this happening.
If the 8-sided and 20-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*9*11 (=1485) different permutations of this happening.

If the 10-sided and 12-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*7*19 (=1995) different permutations of this happening.
If the 10-sided and 20-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*7*11 (=1155) different permutations of this happening.

If the 12-sided and 20-sided dice have a "1", the other dice cannot have a "1": There are: 3*5*7*9 (=945) different permutations of this happening.

For a total of... 66319 permutations matching the condition (Pair of 1s) out of 460800 possible permutations (~14.4%). This includes situations where, say, the dice read: 1,1,2,2,8,18 - a condition of "two pair." If you want the exclusive percentage of *just* a single pair, you'll need to figure out the probabilities of each combined event and subtract them out of the larger percentage for just 1 pair.

P(Pair 2, 3 or 4) will be the same as P(Pair 1)

[Side note - interesting that about 9/16ths of all rolls will have a pair of 1-4] This would make for an interesting D&D Event that you wanted to have happen slightly more than half the time.

P(Pair 5 or 6) can't involve the first die and will resolve independently of what it rolls. The answer will have one less "paragraph," since you'll only need to iterate through combinations of 6-, 8-, 10-, 12- and 20-sided dice, with the result of the 4-sided dice being irrelevant (4*... will precede every set of combinations, since any value of the 4-sided dice is valid for these)

P(Pair 8-10) can't involve the first 2 dice and will resolve independently of what they roll. Each "paragraph" of the answer will start with: 4*6*...

Etc. Etc.

I have found out my percentage for a double appearing and my percentage for a triple appearing is there any way i can use this information to determine how often a double and triple appear together? such as (5 5 5 4 4 2)

Post Reserved

What does a four sided die look like?

Quote: Ayecarumba

What does a four sided die look like?

A tetrahedron.

Quote: Ayecarumba

What does a four sided die look like?

Probably shaped like a tetrahedron.

Edit: Ooops! Reading several threads at once on different tabs and slow to post my reply to this one.

I cheated and wrote a program to loop through the 460800 combinations:
six of a kind: 4
five of a kind and a single: 224
2 three of a kinds: 240
four of a kind and a pair: 388
3 pairs: 2488
four of a kind and 2 singles: 4608
three of a kind, pair, and a single 13184
three of a kind and 3 singles: 44544
2 pair and 2 singles: 76784
6 singles: 100800
pair and 4 singles: 217536

Of course I'm known for bainos and/or typeos.

Quote: thecesspit

A tetrahedron.

The funny thing about d4s is that some of them have the base of the die determine the number rolled and others have the pointy tip do the same.
Examples:

miplet, i did some calcualtions

these are the percentages i have found from doing permutations.

chance for any double = 84.2664% (your number was half of this percentage and 84% seems to happen when rolling dice)

chance of any triple = 12.6233% (your number was near 10% which is very close)

chance for any 4 matching dice = 1.0867% (this was spot on with your numbers)

chance for any 5 matching dice = 0.0486% (this was spot on with your numbers)

chance for any 6 matching dice = 0.0000868% (is only off by a 0) (i had 1/460800 multiplied by 4 for this result, yours may be more accurate)

i believe all of my calculations are correct

of course i have no way to determine the odds of those percentages i have not determined yet but your result may help me find the math.

your program may allow dice to vary based on a positive or negative luck, but even with this to have only 217536 pairs with 4 singles is very unlikely with 460800 rolls.

keep in mind all 4,6,8,10,12,20 dice are used and rolled all at once. my math may be wrong though, but i doubt it, i rolled these six dice hundreds of times and never had 5 or 6 match but the others matched with similar odds. I have to admit though the odds of rolling four of a kind with 2 singles is spot on. thank you for your help, may something was slightly off with how the program was setup, obviously with no knowledge of the program i have no idea.

I looked carefully at my program for typos, and didn't find any. I even did a separate program and got the same results. Let's look carefully at any trips:
They can be made with 2 three of a kinds like 1 1 1 1 2 2 2 or 1 1 1 5 5 5. This occurs 240 ways. I think you are counting these twice.
They can be made with a 3 of a kind, and 3 singles like 2 2 2 3 4 12 This occurs 44544 ways.
They can be made with a 3 of a kind, 1 single, and a pair like 2 2 2 8 8 12 This occurs 13184 ways.
This brings us to 57968 ways or 12.579861 %. You are coming up with 58168 ways which is 200 too many.
I'm heading to bed now, but feel free to ask any questions.

i believe you are right miplet, i was counting them twice thank you for pointing that out. yeah doing these numbers out by using permutations is very time consuming and can lead to many many mistakes. Thank you so much for helping me figure this out it saves me alot of time from doing these on paper.

final results thanks to miplet and his program
number of rolls needed to obtain the result on average is rounded.

six of a kind: 4/460800 = 0.000868% (115,200 rolls)
five of a kind and a single: 224/460800 = 0.0486% (2,057 rolls)
2 three of a kind: 240/460800 = 0.0486% (1,920 rolls)
four of a kind and a pair: 388/460800 = 0.0842% (1,188 rolls)
3 pairs: 2488/460800 = 0.54% (185 rolls)
four of a kind and 2 singles: 4608/460800 = 1% (100 rolls)
three of a kind, pair, and a single 13184/460800 = 2.861% (35 rolls)
three of a kind and 3 singles: 44544/460800 = 9.667% (10 rolls)
2 pair and 2 singles: 76784/460800 = 16.663% (6 rolls)
6 singles: 100800/460800 = 21.875% (5 rolls)
pair and 4 singles: 217536/460800 = 47.208% (2 rolls)