How to calculate the odds in a multihold multitoss game

Ok, I have been twiddling some more bits.

I have determined that there are 6^6 possible final outcomes. I have further relearned some math to figure out that there are 6! / 4! possible winning combinations. I thought I had the tiger by the tail as the odds seemed pretty easy at (6!/4!) / (6^6 - (6!/4!)) ... That comes out to a very small percentage 0.0643417836%

I assume that is the odds of doing it all in one throw.

Further I determined that there are 32 different possible outcomes (regardless of the actual dice numbers) By this I mean, Throw 6 Keep 6 = 1 way. Throw 6 keep 5, throw 1 keep 1. That is 2. And so on all the way to Throw 6 keep 1, throw 5 keep 1 and so on down to throw 1 keep 1.

For what it is worth, if you ever throw all numbers that do not include a "keepable" 2,4,6 you can't win a perfect 24 score.

So in the best winning throw you get all perfect in 1 roll. In the worst winning throw you will make 6 rolls totaling 21 dice. At one point I thought this might even be a 21 choose 6 solution, but that actually comes out to more than 6^6 so I am not sure it is correct. I am not sure how to compensate for the permutations in 21 choose 6 where I know there will be duplicate values.

An interesting paytable would be to pay on 21-24. I think even money on 21 and a bit more on the others with, perhaps some bonuses on the lesser number of throws. For example 24 in 1 throw at .06% could be a good candidate for a jackpot of some sort :) But I digress.

When I looked at the brute force site I think it was flawed in the "go for broke" strategy, because I think taking 6's over 4's and 2's early would be a better strategy, since taking 4's or 2's reduce your odds of a safe throw. But again I digress.

When trying to calculate the game, since there are 32 ways to get to the end. Do I have to do all those calculations as well? My 7% calc came from doing t6 h1, t5 h1, t4 h1... but assumes that you keep 6's first then the 4 and 2. This gives you the best possible chances that I can see numerically. I would assume that if I do it in reverse, of holding the 4 and 2 first, I immediately reduce the odds of a safe roll and subsequent ones as well. I guess I could calculate all 30 permutations. Many of which would be the same and some just a bit different. Or do I have to calculate the chances of getting more than one keepable die and each of the 32 possible ways to finish. This would require lots of calculations (30 * 32) of them to be exact. Am I at least thinking the right direction? It has been years since game theory and even more since statistics and algebra :(

Here is a generic answer. To implement it requires some programming.

1. Determine every way to roll 6 dice as well as the likelihood of each outcome happening. This isn't quite as bad as it sounds. Although there are 6^6 outcomes, we don't care about the order. This reduces the number of unique rolls to 462. Here is a sample of the list. We don't care about the order, just the number of each dice. The weight is the number of ways that this can be thrown. For instance, in the second row, the 5 can be in any one of the 6 positions. Research combinatorials for this.

Qty 1s	Qty 2s	Qty 3s	Qty 4s	Qty 5s	Qty 6s	Weight
0	0	0	0	0	6	1
0	0	0	0	1	5	6
0	0	0	0	2	4	15
0	0	0	0	3	3	20
0	0	0	0	4	2	15
0	0	0	0	5	1	6
0	0	0	0	6	0	1
0	0	0	1	0	5	6
0	0	0	1	1	4	30
0	0	0	1	2	3	60
0	0	0	1	3	2	60
0	0	0	1	4	1	30
0	0	0	1	5	0	6
0	0	0	2	0	4	15

2. Loop through every way of holding the dice. For the first roll, you may hold everything, reroll 1, 2, 3, 4, or 5 of the die. There are 63 total ways to hold/re-roll the first time.

3. For each way to discard, loop through all the ways to re-roll the dice. There are 252 ways to throw 5 dice, 126 ways to throw 4 dice, 56 ways to throw 3 dice, 21 ways to throw 2 dice, and 6 ways to throw 1 die. For each of these, calculate the best possible pay. This means recursively searching how to discard the remaining dice. The recursion ends when you get down to discarding 0 dice, and you just calculate your score. At each step, choose the best possible average pay.

If this is implemented to use a paytable, it would also be easy to calculate the "go for broke" by setting the pay for a perfect score to 1 and the pay for everything else to a 0.

I know this is an overly simplistic description, but I think it is the general approach of any strategy based gambling game.

I see where you are getting 462 .. that is C(n+(s-1),s-1) where n = number of dice and s = number of sides on the dice. 11 choose 5.
I think I see the other 'weight' values as a divided factoral such as 6!/5! for 1 specific number with 5 duplicated numbers then 6!/4!/2! for the 15 then 6!/3!/3! comes out to 20 and I will assume on down the table.

Am I following you well? Now, what are you doing with that weight?

I am not sure I follow your counting of ways to throw and reroll dice. For example, if you have 3 dice you can only roll 6 different ways. I am not sure where you are getting 56. I must be missing something...

R K
3 3
3 1
3 2
2 2
2 1
1 1

This thread got really quiet.

Quote: Absinthe

This thread got really quiet.

Still working on it. I'm trying to learn c++, so it may take awhile.

Am waiting for a call back from Albert Einstein.

I can help with the C++ if you like. Though I have been focusing on C# lately.