Expert: Paul Klarreich - 12/20/2006

Question
First, thanks.

Equations like (---) or (----)^4 are like alien language to me. Originally, I just used a smaller problem, roll 3 dice six times and drop the lowest (to find out how many could roll 12) and wrote out all possible combinations by hand.

I then counted how many times each number came up:

12 came up 18x
11  "   "   3x
10  "   "  18x
9   "   "   6x
8   "   "   18x
7   "   "   9x
6   "   "   15x  
5   "   "   6x
4   "   "   9x
3   "   "   3x
2   "   "   3x


The number 12 came up eighteen times, out of 108 possibilities. So I divided 108/18 and concluded that the chances of getting at least one 12 in the six rolls is 1 in 6. If I didn't err, it seems the chances of getting a 10 or 8 is the same as the chances of getting a 12.

Anyway, figuring out four dice (instead of three) and 18s (instead of 12s) is a little bit more daunting. Not only that, but figuring the chances of two 18s (out of six rolls), three 18s (out of six rolls), and so on. That's why I asked your help.

And if my listed results are good, since the 10s and 8s have the same chances as the 12, then similarly for the real problem (four dice rolled six times, drop lowest), a few numbers might have the same chance of being rolled as 18s would (maybe 14s and 12s?)

Again thanks, but I really need a more straightfoward solution. Or just the answers :)
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The text above is a follow-up to ...

-----Question-----
Hi, we play a game where 4 six-sided dice are rolled. You add the three highest dice (dropping the fourth lowest die) and write down their total sum. You repeat this process five more times, so in the end you have six numbers written.

The highest sum possible is 18.

So the question is "what's the probability (%) chances of rolling one 18? Also, if you would be kind to list the probability for two 18s, three 18s, 4 18s, 5 18s, and six 18s? Because we drop one die each time, it's difficult to figure out the probabilities.

Thank you.
-----Answer-----
Questioner:  boozerker
Category:  Advanced Math
 
Subject:  Probability of rolling high with "biased" dice

Question:  Hi, we play a game where 4 six-sided dice are rolled. You add the three highest dice (dropping the fourth lowest die) and write down their total sum. You repeat this process five more times, so in the end you have six numbers written.

The highest sum possible is 18.

So the question is "what's the probability (%) chances of rolling one 18?

Also, if you would be kind to list the probability for two 18s, three 18s, 4 18s, 5 18s, and six 18s? Because we drop one die each time, it's difficult to figure out the probabilities.

Thank you.
...........................................
Hi, Boozerker,

Difficult, it's true, but maybe not impossible.

You roll four dice.  There are 6^4 possible outcomes, assuming that we can distinguish the dice.  That means, for the purpose of this discussion, that:

6-5-4-3  is different from 4-5-6-3

Now how many rolls yield a total of 18?  Assuming the FIRST three are 6's, there are 5 ways:

6-6-6-x, where x = 1..5.  That is 5 rolls.

But the x could be one of the first three, such as:

6-6-x-6,  6-x-6-6,  x-6-6-6.  That's 15 more, giving 20.

BUT there is also 6-6-6-6.  I think we also consider that ONE roll to be an '18'.
So there are a total of 21 rolls that yield an 18-count.

So on a given roll, we have:
         21     21
p(18-count) = ---- = ---- (which I won't reduce
         6^4    1296

Suppose we now denote  p18  as that probability, and q18 as its complement, that is:

p18 = probability that a single roll yields 18
q18 = ...................................... less than 18
And, of course, q18 = 1 - p18, giving
     1275
q18 = ----
     1296
Now the problem for multiple rolls reduces to this analogous situation:

You toss an unfair coin. (Unfair means the two sides don't have equal probability of coming up.)  One side of this coin says '18', the other side says 'sorry-no-dice'.  

The probability that '18' comes up is p18, and the probability that 'sorry' comes up is q18.

Now in this analogous game, you basically have your standard binomial expansion: (I will stop writing p18 and q18 now, and just use p and q for those numbers.)

(p + q)^6 = binomial expansion, which you work out from standard stuff.

Now here's the clue:  The binomial expansion that has, say  p^2 q^4, in it, along with a binomial coefficient of  C(6,2), is the probability of getting exactly 2 '18's and 4 'sorrys'.  Evaluate each of these 7 terms, keeping in mind that

p = 21/1296,  q = 1275/1296.

For example, the probability that you get exactly 2 '18's will be:
         6!
p(two 18's) =  ---- p^2 q^4
         2!4!

         6 5    21      1275
p(two 18's) =  ---- (----)^2 (----)^4
         2 1   1296     1296

(I'll leave the computation to you.)

and you have your 7 probabilities. (7, because there is also the case of no 18's)

Yes, it's going to be a bunch of arithmetic, but as I said : Difficult, but not impossible.


Answer
Hi, Boozerker,

I think there's not much else I can do for you on this.  The calculation for '12 on two out of three dice' is similar, though simpler, so I'm going to leave that to you -- the outlines are the same.

After computing the probability, you can THEN compare it with your empirical results.

Good luck. (which is what anyone needs when they play with dice, I suppose.)