Probability 3D6 check

I wouldn't mind a thread on all sorts of probability (card/die/etc...) ... A friend asked me last night if I could figure out a probability for him in the following scenario:

You roll 9d6, and choose the highest three numbers and add them together... What's the most likely sum?

I know how to do simple dice probability (which is why he approached me with it) but the 'choose the top three' was what made it difficult to work out...

In the end, I wrote a Perl script to roll 9d6, sort them into an array and sum the last three elements of the array... and had it run for 100,000 iterations... Doing that, I came up with:

3 occurred 0 times
4 occurred 1 times
5 occurred 0 times
6 occurred 0 times
7 occurred 28 times
8 occurred 81 times
9 occurred 232 times
10 occurred 625 times
11 occurred 1439 times
12 occurred 3052 times
13 occurred 5925 times
14 occurred 10258 times
15 occurred 15920 times
16 occurred 22478 times
17 occurred 21958 times
18 occurred 18003 times

Which isn't 'exactly' what I expected... close, but not exactly... but I've run it several times and tested it with smaller problems to verify the logic... (ie, a 3d6, choose top 2 problem)

What I'd love to know is how to calculate this with math... instead of resorting to programming (though I did enjoy the programming aspect).

TrekNoid

Instead of using the random method generation, like you did, simply generate all the possible rolls. There should be 6^9 possibilities, which mean : 10077696 permutations.

For each permutation, totalise the 3 highest dices and increment the counter result array ( 3 to 18 ). You should now have the number of permutation that generate each result.

Now for each result, take (nb_permutation / 10077696) * 100 to determine the probability percentage to roll this result. Cumulate the % together if you want to determine the "roll higher than X" or "roll lower than X".

Much more easier to evaluate than my Xd6, Y dice must be >= Z.

I started with that approach, but short of leaving the program running overnight, 10,077,696 iterations isn't an easy task to get done...

I may go that route to be certain, but it would still be cool to know a more mathematic way to do it...

Thanks,

TrekNoid

Still, in perl script, it might be a bit slower than in C++. I'll try in C++ ... if have have some time.

Okay... you shamed me into it :)

Actually, it turns out I had a bug in the program... which is why it was never returning... fixed it, and re-ran it, and came up with the following:

For 9 dice, choosing the top 3 rolled, here's all the possibilities:

3 occurred 1 times
4 occurred 9 times
5 occurred 45 times
6 occurred 547 times
7 occurred 2340 times
8 occurred 6948 times
9 occurred 23806 times
10 occurred 63612 times
11 occurred 140049 times
12 occurred 314245 times
13 occurred 604863 times
14 occurred 1037223 times
15 occurred 1607641 times
16 occurred 2257245 times
17 occurred 2222676 times
18 occurred 1796446 times

So, 16 is the most likely roll in this scenraio...

For all you completists, here's the 9 through 3 most likely rolls:

9d6(top3): 16
8d6(top3): 16
7d6(top3): 16
6d6(top3): 15
5d6(top3): 14
4d6(top3): 13
3d6(top3): 10/11 equally likely

Thanks, Larienna :)

TrekNoid

Damn, your are faster than me! (`_`)

I made a C++ programm that generated the probabilities from 3 to 10 dices. Here is the table of result.

http://www.bgdf.com/files/My_Uploads/Larienna/k3d6onxd.pdf

On my 333 Mhz, the process start to slow down for these probabilities.

7 Dice - 1 Sec
8 Dice - 6 Sec
9 Dice - 40 Sec
10 Dice - 4 Min 40 Sec

Then again, It is not the most optimised code. For each permutation I had to make a backup copy of the 10 dice array and I hade to sort the whole array each time. I have now realizes that I could make it by only sorting the 3 first number to save time.

I made a C++ programm that generated the probabilities from 3 to 10 dices. Here is the table of result.

http://www.bgdf.com/files/My_Uploads/Larienna/k3d6onxd.pdf

Well, I had an unfair advantage... I already had written the die rolling and 'choose top three' routines before you heard of it :)

Nice chart though...

Of course, now my friend came back and asked me to settle some other dice-related question he was debating with another friend...

Starting to think I need to add all this to my web page!

Thanks for nudging me into doing it... it was actually kinda fun to figure out.

TrekNoid

Re: a more purely mathematical approach:

I'm not sure about 9 dice, choosing the 3 highest, but here is some thinking that might get people's thinking going:

If you have D dice and simply choose the single highest die out of those, then the distribution has the form

P(X) = k*X^(D-1)

where k is a scaling constant that makes sure all probabilities together sum to 1, X is the potential result whose probability you're interested in, and D is the number of dice you are selecting the single highest from.

So... can anyone extend this?

The way I coded it is that the probability for a TN is subtracted from 1 and further TN prob are substrated from this result. The rest is the probability for rolling this TN or higher. So if one TN is round up, it might remove 0.01% more to the total changing the prob slightly for the next TN. But in general it is almost the same probabilities.