Bar dice: six-of-a-kind out of nine in one shake

[Supertuscan]

What are the odds of getting six-of-a-kind out of nine dice
in one shake?
This is the shake of the day at a tavern near me.
I know the odds of six-of-a kind-in one shake
given six dice; the three extra dice have me bumfuzzled.
I have been unable to plug the numbers into the equations
I have come across. This is probably trivial math, which adds
greatly to the irritation :)
Thank you.

 

[chiro]

Hey Supertuscan and welcome to the forums.

I'm assuming that you have nine dice and want to find the probability of getting six of the same number in a single roll.

The only piece of information I need to know is do you need to be exactly six or at least six: the answer will be very different if we make the wrong assumption.

The actual math that is used is based on combinatorics which is a fancy of finding out how many combinations and permutations you have (think in terms of how you can arrange stuff) given a set of objects.

 

[Supertuscan]

Thank you for the reply.
Exactly six of a kind in one shake.
I have been looking at the formula for this and
to me it seems ultimately non-intuitive.
Hopefully you can stick around to unpack this for me.

 

[chiro]

The answer is based on the multinomial distribution.

Here is how it works:

First I will explain the binomial because the concepts are exactly the same with the exception of having more than two probabilities.

The binomial distribution is a way of modelling independent binary outcomes. It could be yes/no, up/down, heads/tails or whatever. Usually the easiest way to introduce this distribution is to think of a Binomial(n,p) is the system of tossing a coin n times with a probability p that you get a heads, in which the distribution gives the probability of getting n heads.

So basically Binomial(n,p) has values from 0 to n corresponding to the probability of getting 0 heads to n heads given that the probability of each coin toss getting a head is p where p is between 0 and 1 inclusive (so includes both 0 and 1).

The multinomial distribution has the same idea but instead of having only two events or choices per "toss", you can have as many as you want, but all of the event probabilities have to add up to 1. If you want to model a dice roll, then you have six events.

Now for a fair coin toss each probability is 1/2. Unsurprisingly for a fair dice each probability is 1/6.

For your problem: we have probability of each event = 1/6 for a fair die. We have n = 9 since we are doing 9 rolls.

Now what we have to do is calculate a lot of different probabilities. Not only do we have to calculate the probabilities of getting a run of 6 1's, 6 2's all up to 6 6's but we have to take into account that for each of these calculations we have to consider all the possibilities within each run type (for example if we do the calculation for 6 1's we could have example 2 1's and 1 3 with none of the others or 2 3's and 1 2 and zero or the others and so on).

So basically we will need a computer to generate all the possible probability types and then calculate them and add them all up.

So yeah although the idea in terms of the multinomial distribution is intuitive, even the calculation itself is not really completely obvious and it is actually quite a pain to do by hand.

I can give you more specifics of how to calculate it if you want to (and I can help) but its up to you at this point. The main idea is that you use the multinomial formula to calculate each of the events separately and just add them all up. Here is the wiki site for the multinomial distribution:

http://en.wikipedia.org/wiki/Multinomial_distribution

Basically for this problem all of the p's are 1/6, n = 9, you have 6 p's, and the other data is what you have to fill in and generate using a computer.

The reason why the problem with six rolls is easy is because you have no room in comparison to the nine roll case. In the six roll case you just calculate the events of getting 6 1's, 6 2's, 6 3's, 6 4's, 6 5's and 6 6's and then add up the probabilities and that's it. For the 9 roll case you have to deal with the fact that you've got all these possibilities for the other 3 rolls which is why it ends up being complicated.

 

[awkward]

I think you can do this one without a computer.

The probability of rolling exactly 6 ones when rolling 9 dice, for example, is
\binom{9}{6}(1/6)^6 (5/6)^3
(using the Binomial distribution), so the probability of rolling exactly 6 of any number is just
6 \binom{9}{6}(1/6)^6 (5/6)^3

 

[chiro]

Nice work awkward :)

This is no doubt the best way to solve this problem copmutationally.

For the OP what is happening is that we are treating the events as 'not a 1' and 'a 1' and then doing that for all numbers 1 to 6.

 

[Supertuscan]

So then the odds come out to .0062 * x =1
or, 1 in 147 shakes, is this correct?

 

[awkward]

So then the odds come out to .0062 * x =1
or, 1 in 147 shakes, is this correct?

That's about right, although the odds are closer to 1 in 160 according to my calculation.

 

[Supertuscan]

Here is what I got:

(9 choose 3) x (1/6)^6 x (5/6)^3 = .001041904...

The odds of getting 6 of any particular value will be six times that, or:

.0062514288