ideal percentages

[1afay3tt3]

can anyone tell me how to figure out the ideal percentages, without actually counting each outcome? 1v1 was simple, but the rest just seem impossible. also, I don't need to know about the 1v2 rolls.

[Genghis Khan CA]

Here are the basic odds for each roll... bear in mind these are the number of dice you are rolling with, so if you have 4 on a territory you are rolling with 3, if you have 3 you are rolling with 2 etc...

Code: Select all

3v2	37.17%	   33.58%     29.26%
3v1	65.97%	              34.03%
2v2	22.76%	   32.41%     44.83%
2v1	57.87%	              42.13%
1v2	25.46%	              74.54%
1v1	41.67%	              58.33%

First column represents a win for the attacker, second column a split between attacker and defender (1 each) and the third column a win to the defender.

[Bean_]

can anyone tell me how to figure out the ideal percentages, without actually counting each outcome? 1v1 was simple, but the rest just seem impossible. also, I don't need to know about the 1v2 rolls.

As far as I know, you do have to count each outcome. It's probably best to count the failing outcomes instead of the successful ones.

So, for 2v1, if D rolls a 1, A fails only on 1,1 (1); if D rolls a 2, A fails on 1,1; 1,2; 2,1; and 2,2 (4); if D rolls a 3, A fails on 1,1; 1,2; 1,3; 2,1; 2,2; 2,3; 3,1; 3,2; 3,3 (9). You'll notice these are squares, so if D rolls a 4, A fails on 16, if D rolls a 5, A fails on 25, and if D rolls a 6, A fails on 36.

In total, A fails on 1+4+9+16+25+36, or 91 ways out of 6^3, or 216 possible combinations. A succeeds (216-91)/216 of the time, or about 57.87%.

For 3v1, same thing. If D rolls a 1, A fails only on 1,1,1; if D rolls a 2, A fails in 8 ways (1,1,1; 1,1,2; 1,2,1; 1,2,2; 2,1,1; 2,1,2; 2,2,1; 2,2,2). This progression is by cubes, so A has 1+8+27+64+125+216, or 441 ways to fail out of 6^4. A succeeds (1296-441)/1296 of the time, or about 65.97%.

2v2 and 3v2 I will have to work out later.

[1afay3tt3]

thanks for showing me. counting the failures is a lot easier.

[Bean_]

2v2 is considerably more difficult. The tables for attacker wins both, and defender wins both, are as follows:

Attacker wins both:

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A	1	2	 3	 4	 5	  6
1	0	0	 0	 0	 0	  0
2	0	1	 3	 5	 7	  9
3	0	3	 4	 8	12	 16
4	0	5	 8	 9	15	 21
5	0	7	12	15	16	 24
6	0	9	16	21	24	 25

Total = 295/1296

Defender wins both:

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D	 1 	2	 3	 4	 5	 6
1	 1	 3	 5	 7	 9	11
2	 3	 4	 8	12	16	20
3	 5	 8	 9	15	21	27
4	 7	12	15	16	24	32
5	 9	16	21	24	25	35
6	11	20	27	32	35	36

Total = 581/1296

Split = 6^4 - 295 - 581 = 420/1296

For the "defender wins both" table, assume the first roll is D1 and the second roll is D2, the number of successes is 2*D1*D2-(lesser of D1,D2)^2. For example, for D1=6 and D2=5, successes = 2*6*5 - 5^2, or 35.

For the "attacker wins both" table, the calculation is essentially the same, but you have to subtract 1 from each roll because defender wins ties. Thus, the number of successes is 2*(A1-1)*(A2-1)-((lesser of A1,A2)-1)^2.

The interpretation is this: suppose D1 = 4 and D2 = 2. The formula indicates that successes = 2*4*2 - 2*2 = 12. It wins where the highest die is no more than 4, and the second highest die is no more than 2. Thus: attacker rolls of 1,1; 2,1; 3,1; 4,1; 1,2; 2,2; 3,2; 4,2 (which is 8 ), and the same rolls flipped, i.e., 1,1; 1,2; 1,3; 1,4; 2,1; 2,2; 2,3; 2,4 (which is another 8 ). The number of duplicates counted will be the square of the lesser number (here, 1,1; 1,2; 2,1; 2,2 were counted twice because flipping them did not make a difference). So these 4 are subtracted from the 16, making 12. (By flip, I mean interchange D1 and D2.)

I anticipate 3v2 follows essentially the same principle, except that for the attacker's dice we are into 3 dimensions instead of 2.

[bob3603]

All of you are wrong, these are obviously the ideal percentages (atleast in my eyes).

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3v2	100.00%	   0.00%      0.00%
3v1	100.00%	              0.00%
2v2	100.00%	   0.00%      0.00%
2v1	100.00%	              0.00%
1v2	100.00%	              0.00%
1v1	100.00%	              0.00%

[Bean_]

3v2 is done as follows:

For the defender wins both scenario, you have D1 and D2. Say Dl(D1,D2) is the lesser of the two and Dg(D1,D2) is the greater of the two. The number of situations in which both will win, given D1 and D2, is 3*Dl*Dl*Dg - 2*Dl^3. (This is the 3-dimensional equivalent of the formula in the 2v2 analysis -- you rotate a solid block among 3 axes and then subtract the small cube twice.)

You get the following results:

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D	 1	 2	  3	  4	  5	  6	
1	 1	 4	  7	 10	 13	 16	
2	 4	 8	 20	 32	 44	 56	
3	 7	20	 27	 54	 81	108	
4	10	32	 54	 64	112	160	
5	13	44	 81	112	125	200	
6	16	56	108	160	200	216

Totaling 2275. 2275 divided by 6^5 =~ 29.26%.

(Notice that this also = 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + 6^4. Strange how that works.)

For the attacker wins both scenario, you have A1, A2 and A3. Say Ag(A1,A2,A3) is the greatest, and Al(A1,A2,A3) is the second highest. You can then use the 2v2 computation as if Ag and Al were the dice that were rolled.

If you do this, you get a total of 2890, and 2890 divided by 6^5 =~ 37.17%.

As before, subtract 2890 and 2275 from 6^5 to get the split probability combinations, which is 2611 or ~ 33.58%.

[Bean_]

A further note on 2v2. The formula 2*Dl*Dg - Dl^2 is also the same is Dg^2 - (Dg - Dl)^2. You can reduce the formula for defender wins both to:

1^2 + 3*2^2 + 5*3^2 + 7*4^2 + 9*5^2 + 11*6^2 -
10*1^2 - 8*2^2 - 6*3^2 - 4*4^2 - 2*5^2, which works out to 581.

Similarly, you can reduce the formula for attacker wins both to:

1^2 + 3*2^2 +5*3^2 + 7*4^2 + 9*5^2 -
8*1^2 - 6*2^2 - 4*3^2 - 2*4^2, which works out to 295.

May be able to do something similar with 3v2, but you get the idea.

[1afay3tt3]

Bean_ thank you for all your help and explanations.

[Bean_]

You're welcome. It's an interesting question.

To close the loop, the 3v2 calculation on attacker's dice can also be reduced to a pattern of squares. The general approach is to take the 216 attacker possibilities, and create a "solid" using the 3 dice rolls, calculate the number of cases in which that "solid" wins, and rotate the solid (some -- triplets like 111, 222 -- cannot be rotated; others -- duplets like 112, 225, etc -- can be rotated 3 times; and the rest can be rotated 6 times).

The ending number, 2890, is:

1 x [1^2 + 2^2 + 3^2 + 4^2 + 5^2] +

3 x [1 x 1^2 + 2 x 2^2 + 3 x 3^2 + 4 x 4^2 + 5 x 5^2] +

(3+6) x [1 x 2^2 + 2 x 3^2 + 3 x 4^2 + 4 x 5^2
- 4 x 1^2 - 3 x 2^2 - 2 x 3^2 - 1 x 4^2] +

6 x [1 x 3^2 + 3 x 4^2 + 6 x 5^2
- 6 x 1^2 - 3 x 2^2 - 1 x 3^2]

(sorry, tried to get it to line up but it is too long! )

Bottom line, it can be done without counting each individual set of rolls, but it is not completely trivial.