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Thread: Quick die probability question

20170603, 08:03 PM (ISO 8601)
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 Oct 2006
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 Pittsburgh, PA
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Quick die probability question
Is there an easy way to calculate my odds of rolling
a) at least three of a kind
b) at least three of a specific number
when rolling xd10? Like, if I roll 10d10, what are the odds that I'll roll three or more of the same number? What are the odds that the set will all be 3's?Hill Giant Games
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20170603, 09:49 PM (ISO 8601)
 Join Date
 Nov 2016
Re: Quick die probability question
Yes, the exact math probably escapes me but yes. Also the percentages seem dependent on how you think about things.
So on a D10 you have a 1 in 10 chance, 10% of rolling any one number.
.1 x .1 = .01% chance of rolling the any number twice in a row?
Once you roll a particular number though, you have a 10% chance of rolling it on the next roll.
To really get into the permutation though, likeliness of rolling three 4's or whatever out of ten D10's I'd have to revisit high school permutations. Maybe this will help. https://www.google.com/search?q=Perm...utf8&oe=utf8

20170604, 02:22 AM (ISO 8601)
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 Mar 2012
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 UK
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Re: Quick die probability question
There's a thread on this exact issue on the next page of the forums: link.
The best answer (with the correct forumla) is a little way down the first page, unfortunately for the "at least N successes" problem there isn't a simple formula, it's "add together the chance of exactly n successes, the chance of exactly n+1 successes, the chance of exactly n+2 successes etc. up to the number of dice in the pool".
From past experience I think the Mad Science forum is a better place for these questions (or get gets good answers faster).Last edited by Khedrac; 20170604 at 02:23 AM.

20170604, 07:48 AM (ISO 8601)
 Join Date
 Sep 2016
Re: Quick die probability question
For exactly 3 of a specific number on 3 dice with 6 sides it's
1/6*1/6*1/6=0.005=0.5%
For exactly 3 of 1 specific number turning up on 4 dice with 6 sides it's
(1/6)^3 * (1(1/6))^(43) * fact(4)/(fact(3)*fact(43))
You should be able to copy the left hand side to excel,
and then change the 6,4,3,1 to the appropriate numbers (or references) it will scale nicely. It's why I've left it looking a bit award, for a given case you could simplify it.
To find at least 3, then there's two options:
If the number of successes is high (or increases with the dice thrown) you have
(probablility of 3)+(probability of 4)
(1/6)^3*(5/6)*4 + (1/6)^4 = = 1.62%
If the number of successes is low and the number of dice is high (or changeable) then you have
1(probablility of 0)(probability of 1)(probability of 2) which is
1  (5/6)^4  (5/6)^3*(1/6)*4  (5/6)^2*(1/6)^2*6 = 1.62% (Phew!)
This won't scale quite so nicely, as you change the number of successes you need to add/remove terms from the sum. Either method will give the same answer, it's just which way requires less adding up and multiple use of the first formula.
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When you don't care what the number is (but they all need to match), you can multiply by then sides on the dice to get an approximate idea. This will NOT be exact as it double counts the cases when you win two ways on the same throw. However it is exact when the number of die needed for a success is above half of the total of die cast and pretty close near there.
Conversely if the number of die cast divided by the number of successes needed is greater than the number of sides on the die the chances are certain (so with 13 D6's your guaranteed at least one treble). And pretty close near there (with 6D6's it's (1/6)^6*fact(6)=1.5% chance of not getting a double somewhere)
So in the pairs D6 case have exact for less than 4 dice (0%,0%,16.7%,44%), exact for over 6 dices (100%), and it's a bit wobbly for 4,5,6
It would be a bit of work to do the first order correction (and then each level on) for both but that would cut down the nasty middle ground immensely. But I've probably already made too many mistakes to risk carrying on.

20170604, 08:21 AM (ISO 8601)
 Join Date
 Mar 2015
Re: Quick die probability question
Once you have the possibility of doing it for one side (xchoose3 times (1/10)^{3} (9/10)^{x3} is the one I get off the top of my head for xd10) calculating the chance of getting three of a kind on any side is a bit harder.
It is roughly 10 times of course, but not quite that high because that will probably be double counting results like [3 3 3 8 8 8] which is one result that has two different sets of 3. Mathematically I think the easiest way to calculate the result would be to have a computer list out every result and then count all the ones with 3 or more of the same. I don't know of a clean formula that will get you the same result.

20170605, 08:42 PM (ISO 8601)
 Join Date
 Nov 2010
Re: Quick die probability question
No easy solutions, but some shortcuts. Experience from dealing with a certain Pathfinder feat helps me here...
Your first roll does not matter, call it a.
You have 9/10 chance that your second is different from your first, so you have a b, otherwise, you have a a.
With the third roll, you have 9 * 8 / 100 that all three are different (a b c), 1 / 100 that all three are the same (a a a), and otherwise you have two of one type and one of the other (a a b) (27 / 100).
So with three dice, you have a 1% chance of three of a kind, with no differentiation, so 0.1% of any specific three.
When you roll a fourth die, it only matters in the 99% of cases where you don't have a triple. 27% of these you have a pair, and 72% you have three singles. The fourth die can only get you three of a kind if you started with a double, so you get another 2.7% chance of a triple (3.7% total). Otherwise, you can have the following:
2 pairs (2.7%)
1 pair and 2 singles (43.2%) (21.6% from a pair and a single, 21.6% from three singles)
4 singles (50.4%) (7/10 chance if you had three singles)
With a fifth die, you're looking at the 96.3% of cases where you didn't get a triple yet. You get one 2/10 of the time if you start from 2 pairs, 1/10 of the time if you start from one pair  a total of 4.86% chance, so your global chance with five dice is 8.56%. Otherwise...
2 pairs and a single (10.8%, from (2 pairs) and (1 pair and 2 singles))
1 pair and three singles (50.4%, from (1 pair and 2 singles) and (4 singles))
5 singles (30.24%, from 4 singles)
With six dice, you start out with 91.44% of cases where there are no triples among the first five. You have another 7.2% chance from the existing pairs, for a global total of 15.76%.
3 pairs (1.08%) (only from 2 pairs and a single)
2 pairs and 2 singles (22.68%)
1 pair and 4 singles (45.36%)
6 singles (15.12%)
And you can continue the trend until 10 dice, beyond which you must have pairs (since they are d10s). At 20 dice, you can only fail to have triples with 10 pairs, which is rather rare (20! / (2^10 * 10^20), so 0.0024%), and at 21 or more dice your odds are 100%.
Basically, using a table, you can establish the number of pairs and their odds for N dice, and easily go on to the odds for N+1 dice. It gets more intricate if you are looking for quadruplets, as you have to keep track of the triples and pairs, rather than just the pairs.
You're also not looking at the odds of having more than one triple, but the approach used is indifferent to the values of a triple used, so your odds of getting a specific triple are 10% of the overall number of triples (which is more than the chance if you have 6+ dice, as 2 or more triples are possible). If it matters, you can set up a parallel table keeping track of the remaining pairs and singles after you've got a triple (or more). This assumes that you're after triples, so a sextet is treated as two triples...

20170606, 09:35 AM (ISO 8601)
 Join Date
 Dec 2015
Re: Quick die probability question
I spent a lot of time fiddling around with http://anydice.com/ over the weekend trying to wrap my head around a roll and keep system.
The expression for the probability of getting three 1's in five d10's is:
output [count 1 in 5d10] >= 3
The probability for a general three of a kind is ten times the probability of three 1's.

20170606, 10:10 AM (ISO 8601)
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 Oct 2014
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Re: Quick die probability question
With five dice, yes, with ten not entirely, which is what makes the original question hard to compute.
Because we're only interested in three of a kind out of ten dice it's possible to get two or even three three of a kinds in a single roll. Those rolls still only count as one succes, but they are counted in the odds for both or all three of the numbers you got three of a kinds for. So the total odds of throwing at least one three of a kind are slightly lower.
But for five dice it's an exact answer, and I didn't know that function of anydice, so that's cool.The Hindsight Awards, results: See the best movies of 1999!

20170607, 03:20 AM (ISO 8601)
 Join Date
 Nov 2010
Re: Quick die probability question
Fine. If you're looking for two triples among six dice, then:
At three dice, you get three different values 72% of the time, which does not allow two triplets at six dice.
You get a pair and a single 27% of the time, which is possible
You get a triple 1% of the time, which is also possible.
At four dice, if you started with a pair and a single, you get:
 A triple and a single 2.7% of the time
 Two pairs 2.7% of the time
 A pair and two singles 21.6% of the time, which does not allow two triples at six dice.
If you started with a triple, you get:
 A quadruple 0.1% of the time
 A triple and a single 0.9% of the time (total is 3.6%)
At five dice, if you started with a quadruple, you get a quintuple 0.01% of the time
If you started with a triple and a single, you get a triple and a pair 0.36% of the time
If you started with two pairs, you get a triple and a pair 0.54% of the time (total 0.9%)
At six dice, you get a sextuple 0.001% of the time.
You get two triples 0.09% of the time.
Otherwise, you have one or fewer triples.
So with six dice, your odds are:
No triples: 84.24%
One triple: 15.669%
Two triples: 0.09%
Six of a kind: 0.001%
Given that you have a triple (or more), the odds of getting any specific value are the same.
One triple: 1.5669% chance of the triple being for the number you want
Two triples: there are 45 number pairs, of which 9 have the number you want (combined with each of the other possible 9 values), so you have your target triple 1/5 times  a chance of 0.018%
Six of a kind: 0.0001% chance of the six dice coming up with the number you want.
So the odds of getting a specified triplet with six dice are 1.585%, slightly more than a tenth of your chance of getting any triplet (15.76%).
Which matches the correction above, and means my conclusion was not correct. The numbers for the totals are accurate, but the possibility of having multiple pairs means that some double counting occurs.