Can Someone Compute the Probability of Success in my New Game Mechanic?

http://anydice.com/

No, I am wrong. The dice rolls aren't sequential. I will need to think this through some more. I enjoy this sort of problem, so thank you for posting it!

I would love to help you. But calculating this proves to be very complex.

First of all, the best die is put in the first place. And no matter what you throw, you get a line of dice lined up from high to low.

The defender does exactly the same. And the 2 lines are compared.

Of course there are dice that can be discarded at a certain point. After all, the difference of the check is only 1. For example 3 versus 2 where the third die declares the winner. However, this may not be done before the throw, only afterwards. Because each die, no matter how many, gives a chance to throw a high number. (Many would fail to think of doing that correctly)

At this point, I can only conclude. It can be done.
But it takes a lot of time.
Why? Because during the calculating path, rules for calculating, "change". The calculation needs to be done in steps. (Don't get me wrong, I normally go to 10 dice, but even 3 are a nuisance right now).

Just trying to get the rules straight:
- Attacker dice lined up
- Defender dice lined up
- Only 1 die is checked.
- If a tie, next die. Until we run out of dice. (If one player has more dice, the other player has a "0")
- Still a tie? Results for the defender to win.
Correct?

With lining up, there will be doubles. Whoever calculates this, the doubles have higher chances to occur. So no scratching possibilities in this one ;).

Ok, lets start, so we get an impression of how long it would take. (And perhaps some one comes with an easier way to calculate)

First I create a 1 die versus 1 die situation.
Excel is great for this. And I simply compare 1 to 6 with 1 to 6. With 36 possibilities, an easy way to have an overview is by subtracting the defender from the attacker.
- All 0 means a tie. If it is the last die, the defender gets extra wins.
- Any lower then 0 means the defender wins.
- And any positive number is the attacker winning.

1 2 3 4 5 6
1 0 -1 -2 -3 -4 -5
2 1 0 -1 -2 -3 -4
3 2 1 0 -1 -2 -3
4 3 2 1 0 -1 -2
5 4 3 2 1 0 -1
6 5 4 3 2 1 0

More then 1 die comparison:
Attacker 15, Tie 6, Defender 15
The last die comparison:
Attacker 15, Defender 21

From this I can conclude that with 1 attacker versus 1 defender.
The win/loss is 41,7%/58,3%.

But wait, we will be needing more specific numbers if we want to use this table in the future.
When the attacker throws something specific, the chance of winning changes as well, we need to keep track of that.
Die, chance/36
1 0
2 1
3 2
4 3
5 4
6 5
And if the attacker has more die than the defender, each comparison needs to be taken in account. Then the thrown die equals the chance/36. So a 6 is 6 out of 36 that the attacker wins.

For a 2 attacker versus 1 defender. The complexity grows exponential. I need to check all combinations that the attacker has. Lining them up. Then compare them with the 6 possible outcomes of the defender.

36 possibilities, results in 21 possibilities with doubles:
High, Low, chance/36
1 1 1
2 1 2
2 2 1
3 1 2
3 2 2
3 3 1
4 1 2
4 2 2
4 3 2
4 4 1
5 1 2
5 2 2
5 3 2
5 4 2
5 5 1
6 1 2
6 2 2
6 3 2
6 4 2
6 5 2
6 6 1

As specific rule for this 2 versus 1. We have that if there is a tie, the attacker wins by default. Since the defender has only 1 die. This specific rule will change with certain combinations of the number of dice pitted out.

It just so happens:
High, chance/36
1 1
2 3
3 5
4 7
5 9
6 11

And those 6 can be compared in the table that I created at the start. In fact, the chance of getting a certain high is multiplied by the chance of winning with a specific throw. And after that, all is added up.
1 has 1 x 1 = 1
2 has 3 x 2 = 6
3 has 5 x 3 = 15
4 has 7 x 4 = 28
5 has 9 x 5 = 45
6 has 11 x 6 = 66
A total of 161
The total number of dice is 3, thus 6x6x6 = 216.
The winning chance is divided by 216.

The win/loss is 74,5%/25,5%.

Since there is only 1 die that is compared. If the defender has 2 and the attacker has 1 die. Then the calculated situation will revert. Instead of 161, we only have 55.

The win/loss is 25,5%/74,5%.

And there you have it, 1v1, 2v1 and 1v2.

If I continue expanding the number of dice (2 versus 2). I need to add two more steps in the calculating. And that is calculating the chance of a tie with each comparison (it differs with each number of dice). And the chance of having a certain second number/third number etc. Maybe there is logic in that last one as well.

In all fairness, I didn't claim it was right, just a roughly representative working estimate.. But you may be right, it's likely I'm still too far off to be useful ;). I think this one may be beyond me (at least in terms of the time I can devote to thinking about it...)

Not to be saying that this is easy either. I checked this one on how to calculate. (That was actually fun)
Not doing the calculations themselves (Only up to 2v1/1v2, and dice pools results up to 3)

Unfortiately making just 1 huge dice pool was not one of them. And it is the same story again as said before with the combinations/permutations. There is another problem here, and that problem is the sorting. And it is this sorting that stops calculating a big picture. You compare 2 situations, 2 sorted lines of dice. And each die in the line has its own situation.

For example (as I can recall), with 3 dice, having the first die being a 6 is 91/216. The second die being 6 was only 16/216 and the third die being 6 is of course only 1/216. For the 5 I had 61, 40 and 7 out of 216. And all these numbers are very important.

There is a way, but it takes time. I do manage to cut it into smaller problems. While I don't have the file here on my work but at home. I can say that I could do the multiplications up to 3 vs 3 now. And no one is going to be happy about that. Since it also indicates that having the 4 vs 4, will take exactly 6 times longer. Since creating the 3 dice propabilities, and sorting them out, took about 1 hour. You can guess that having 4 dice will take 6 hours.

So, while I know exactly what to do, to get the job done. I don't have the right tool for it. Excel does not support horizontal sorting, only vertical :(
If it did, I could have had all sorted propabilities up to the 10 dice already. And the next step would simply be multiplying the propabilites. (yes, that is simply) Then sum them up. And it would be done.

This is too hard to compute, so I wrote a program to simulate.

Offense: 1 Defense: 1 Offense Wins 41%
Offense: 1 Defense: 2 Offense Wins 25%
Offense: 1 Defense: 3 Offense Wins 17%
Offense: 1 Defense: 4 Offense Wins 12%
Offense: 1 Defense: 5 Offense Wins 9%
Offense: 1 Defense: 6 Offense Wins 7%
Offense: 1 Defense: 7 Offense Wins 5%
Offense: 1 Defense: 8 Offense Wins 4%
Offense: 1 Defense: 9 Offense Wins 3%
Offense: 2 Defense: 1 Offense Wins 74%
Offense: 2 Defense: 2 Offense Wins 47%
Offense: 2 Defense: 3 Offense Wins 34%
Offense: 2 Defense: 4 Offense Wins 25%
Offense: 2 Defense: 5 Offense Wins 19%
Offense: 2 Defense: 6 Offense Wins 15%
Offense: 2 Defense: 7 Offense Wins 12%
Offense: 2 Defense: 8 Offense Wins 9%
Offense: 2 Defense: 9 Offense Wins 8%
Offense: 3 Defense: 1 Offense Wins 82%
Offense: 3 Defense: 2 Offense Wins 65%
Offense: 3 Defense: 3 Offense Wins 48%
Offense: 3 Defense: 4 Offense Wins 38%
Offense: 3 Defense: 5 Offense Wins 30%
Offense: 3 Defense: 6 Offense Wins 24%
Offense: 3 Defense: 7 Offense Wins 19%
Offense: 3 Defense: 8 Offense Wins 16%
Offense: 3 Defense: 9 Offense Wins 13%
Offense: 4 Defense: 1 Offense Wins 87%
Offense: 4 Defense: 2 Offense Wins 74%
Offense: 4 Defense: 3 Offense Wins 62%
Offense: 4 Defense: 4 Offense Wins 49%
Offense: 4 Defense: 5 Offense Wins 40%
Offense: 4 Defense: 6 Offense Wins 33%
Offense: 4 Defense: 7 Offense Wins 27%
Offense: 4 Defense: 8 Offense Wins 22%
Offense: 4 Defense: 9 Offense Wins 18%
Offense: 5 Defense: 1 Offense Wins 90%
Offense: 5 Defense: 2 Offense Wins 80%
Offense: 5 Defense: 3 Offense Wins 69%
Offense: 5 Defense: 4 Offense Wins 59%
Offense: 5 Defense: 5 Offense Wins 49%
Offense: 5 Defense: 6 Offense Wins 41%
Offense: 5 Defense: 7 Offense Wins 35%
Offense: 5 Defense: 8 Offense Wins 29%
Offense: 5 Defense: 9 Offense Wins 25%
Offense: 6 Defense: 1 Offense Wins 92%
Offense: 6 Defense: 2 Offense Wins 84%
Offense: 6 Defense: 3 Offense Wins 75%
Offense: 6 Defense: 4 Offense Wins 66%
Offense: 6 Defense: 5 Offense Wins 58%
Offense: 6 Defense: 6 Offense Wins 49%
Offense: 6 Defense: 7 Offense Wins 42%
Offense: 6 Defense: 8 Offense Wins 36%
Offense: 6 Defense: 9 Offense Wins 31%
Offense: 7 Defense: 1 Offense Wins 94%
Offense: 7 Defense: 2 Offense Wins 87%
Offense: 7 Defense: 3 Offense Wins 80%
Offense: 7 Defense: 4 Offense Wins 72%
Offense: 7 Defense: 5 Offense Wins 64%
Offense: 7 Defense: 6 Offense Wins 57%
Offense: 7 Defense: 7 Offense Wins 49%
Offense: 7 Defense: 8 Offense Wins 43%
Offense: 7 Defense: 9 Offense Wins 37%
Offense: 8 Defense: 1 Offense Wins 95%
Offense: 8 Defense: 2 Offense Wins 89%
Offense: 8 Defense: 3 Offense Wins 83%
Offense: 8 Defense: 4 Offense Wins 77%
Offense: 8 Defense: 5 Offense Wins 70%
Offense: 8 Defense: 6 Offense Wins 63%
Offense: 8 Defense: 7 Offense Wins 56%
Offense: 8 Defense: 8 Offense Wins 49%
Offense: 8 Defense: 9 Offense Wins 43%
Offense: 9 Defense: 1 Offense Wins 96%
Offense: 9 Defense: 2 Offense Wins 91%
Offense: 9 Defense: 3 Offense Wins 86%
Offense: 9 Defense: 4 Offense Wins 81%
Offense: 9 Defense: 5 Offense Wins 75%
Offense: 9 Defense: 6 Offense Wins 68%
Offense: 9 Defense: 7 Offense Wins 62%
Offense: 9 Defense: 8 Offense Wins 56%
Offense: 9 Defense: 9 Offense Wins 49%

Here's the code, which I haven't figured out how to make formatted correctly.

~~~~~~~~~~~~~~~~~~~~~
package net.zag;

import java.util.Arrays;
import java.util.Random;

public class DiceSim {

private static final int TRIALS = 1000000;
static Random rnd = new Random(System.currentTimeMillis());

static double calc(int offCount, int defCount) {
int offenseWinCount = 0;
for (int trial = 0; trial < TRIALS; trial++) {
if (offenseWins(offCount, defCount)) {
offenseWinCount++;
}
}

return ((double) offenseWinCount) / (double) TRIALS;
}

static boolean offenseWins(int offCount, int defCount) {
int[] offArray = getRolls(offCount);
int[] defArray = getRolls(defCount);

int shorter = offCount > defCount ? defCount : offCount;
for (int i = 0; i < shorter; i++) {
// Because Arrays.sort returns in ascending order, lower roll wins.
if (offArray[i] < defArray[i]) {
return true;
} else if (defArray[i] < offArray[i]) {
return false;
}
}
return offCount > defCount;
}

private static int[] getRolls(int count) {
int[] result = new int[count];
for (int i = 0; i < count; i++) {
result[i] = rnd.nextInt(6);
}
Arrays.sort(result);
return result;
}

public static void main(String[] args) {
for (int offCount = 1; offCount < 10; offCount++) {
for (int defCount = 1; defCount < 10; defCount++) {
double result = calc(offCount, defCount);
int resultPercent = (int) (result * 100);
System.out.println("Offense: " + offCount + " Defense: "
+ defCount + " Offense Wins " + resultPercent + "%");
}
}
}
}

~~~~~~~~~~~~~~~~~~~~~

Thanks for sharing this fun problem! Here's how I'd suggest thinking through it. And I have run a simulation to verify these results, but would appreciate any 3rd party confirmation.

Start by thinking about the chances of rolling each possible value 1-6 as your highest die given different numbers of starting dice. For one die this is easy, each die value has the same chance of being your highest roll: 1/6 = 16.6% chance each.

If you start writing out the possibilities when rolling 2, 3, and 4 d6s, you are likely to notice a pattern develop. Specifically, your chance of rolling a high value of X across N dice is: (X^N - (X-1)^N)/6^N. The way to understand this formula is that X^N is the number of ways you can have N dice contain values that are less than or equal to X. But we need to ensure that we have at least one value of X. So we subtract all of the arrangements of dice that do not include an X, and there are (X-1)^N of those. The final piece of this formula is simply dividing by the number of possible ways of rolling N dice which is 6^N.

You can use this formula to build a table with your chance of rolling a high value of 1-6 for as many different numbers of dice as you'd like. From this table you can derive another slightly different table with your chances of rolling a high die that is less than every possible die face with different numbers of dice. For instance, your chance of rolling a high die that is less than 5 on 3D6 can be found by summing the the chances of rolling high die values of 1, 2, 3, and 4 on 3D6.

If you're still with me, we now have two tables. One essentially lets us look up our chance of rolling each value with any number of dice. And the other essentially lets us look up our chances of losing (or rolling a high die smaller than each value) with any number of dice. We can combine these two table into a single table that gives us our chances of strictly winning with our highest die (no ties) between any two competing numbers of dice. For each possible value 1-6, we multiply the our chance of rolling that value with the chances that our opponent will roll less than that number. Then we add up all six of those products to find our chance of strictly beating an opponent (without ties). Here's what the table looks like in my spreadsheet:

W\L 1D6-- 2D6-- 3D6-- 4D6-- 5D6-- 6D6--
1D6 0.417 0.255 0.174 0.126 0.095 0.073
2D6 0.579 0.390 0.281 0.211 0.162 0.128
3D6 0.660 0.472 0.352 0.271 0.212 0.169
4D6 0.707 0.526 0.403 0.315 0.250 0.201
5D6 0.738 0.564 0.440 0.348 0.279 0.226
6D6 0.760 0.591 0.468 0.374 0.302 0.246

From this table we can look up our chances of strictly beating or losing to another number of dice without any ties. However we can find our chance of ties by subtracting our chances of winning and losing from 1. For instance our chances to tie when rolling 4D6 against an opponents 2D6 is 1 - 0.526 - 0.211 = 0.263. And if we want to find our chances of winning with ties we can add our chances of winning without ties to our chances of winning with ties: 0.526 + 0.263 * (chance of win when high dice are tied). Our chances of winning when the high dice are tied, are basically the odds of winning with 3D6 against 1D6 (subtract one tied die from each side), which we can also look up in the table. Also notice that a tie here will result in the player with more dice winning, since the other player runs out of dice after their second die ties. According to the table, 3D6 beats 1D6 with a chance of 0.660, loses to it 0.174, and therefore ties 1-0.660-0.174=0.166. So we can plug these chances of winning after the tie into our previous formula above to get our final answer for 4D6 beating 2D6: 0.526 + 0.263 * (0.660 + 0.166) = 0.743.

I'm sure that this tie breaking calculation could be worked into a new table of its own, but I don't currently see a non-tedious way of building this into my current spreadsheet. And the strictly beating with your highest die table that I posted above seems more generally useful, so I'll stop here for now. I hope you find this interesting and understandable, and would love to hear if it raises any questions. My enjoyment of solving and attempting to explain this kind of problem has lead me to start a blog on BGG, for anyone interested in reading more: http://boardgamegeek.com/blog/2948/probability-in-games

Here's the corrected table.

Offense: 1 Defense: 1 Offense Wins 42%
Offense: 1 Defense: 2 Offense Wins 25%
Offense: 1 Defense: 3 Offense Wins 17%
Offense: 1 Defense: 4 Offense Wins 13%
Offense: 1 Defense: 5 Offense Wins 9%
Offense: 1 Defense: 6 Offense Wins 7%
Offense: 1 Defense: 7 Offense Wins 6%
Offense: 1 Defense: 8 Offense Wins 5%
Offense: 1 Defense: 9 Offense Wins 4%
Offense: 2 Defense: 1 Offense Wins 75%
Offense: 2 Defense: 2 Offense Wins 47%
Offense: 2 Defense: 3 Offense Wins 34%
Offense: 2 Defense: 4 Offense Wins 26%
Offense: 2 Defense: 5 Offense Wins 20%
Offense: 2 Defense: 6 Offense Wins 15%
Offense: 2 Defense: 7 Offense Wins 12%
Offense: 2 Defense: 8 Offense Wins 10%
Offense: 2 Defense: 9 Offense Wins 8%
Offense: 3 Defense: 1 Offense Wins 83%
Offense: 3 Defense: 2 Offense Wins 66%
Offense: 3 Defense: 3 Offense Wins 49%
Offense: 3 Defense: 4 Offense Wins 38%
Offense: 3 Defense: 5 Offense Wins 30%
Offense: 3 Defense: 6 Offense Wins 24%
Offense: 3 Defense: 7 Offense Wins 20%
Offense: 3 Defense: 8 Offense Wins 16%
Offense: 3 Defense: 9 Offense Wins 13%
Offense: 4 Defense: 1 Offense Wins 87%
Offense: 4 Defense: 2 Offense Wins 74%
Offense: 4 Defense: 3 Offense Wins 62%
Offense: 4 Defense: 4 Offense Wins 49%
Offense: 4 Defense: 5 Offense Wins 40%
Offense: 4 Defense: 6 Offense Wins 33%
Offense: 4 Defense: 7 Offense Wins 27%
Offense: 4 Defense: 8 Offense Wins 23%
Offense: 4 Defense: 9 Offense Wins 19%
Offense: 5 Defense: 1 Offense Wins 91%
Offense: 5 Defense: 2 Offense Wins 80%
Offense: 5 Defense: 3 Offense Wins 70%
Offense: 5 Defense: 4 Offense Wins 60%
Offense: 5 Defense: 5 Offense Wins 50%
Offense: 5 Defense: 6 Offense Wins 42%
Offense: 5 Defense: 7 Offense Wins 35%
Offense: 5 Defense: 8 Offense Wins 29%
Offense: 5 Defense: 9 Offense Wins 25%
Offense: 6 Defense: 1 Offense Wins 93%
Offense: 6 Defense: 2 Offense Wins 84%
Offense: 6 Defense: 3 Offense Wins 76%
Offense: 6 Defense: 4 Offense Wins 67%
Offense: 6 Defense: 5 Offense Wins 58%
Offense: 6 Defense: 6 Offense Wins 50%
Offense: 6 Defense: 7 Offense Wins 43%
Offense: 6 Defense: 8 Offense Wins 37%
Offense: 6 Defense: 9 Offense Wins 31%
Offense: 7 Defense: 1 Offense Wins 94%
Offense: 7 Defense: 2 Offense Wins 88%
Offense: 7 Defense: 3 Offense Wins 80%
Offense: 7 Defense: 4 Offense Wins 73%
Offense: 7 Defense: 5 Offense Wins 65%
Offense: 7 Defense: 6 Offense Wins 57%
Offense: 7 Defense: 7 Offense Wins 50%
Offense: 7 Defense: 8 Offense Wins 43%
Offense: 7 Defense: 9 Offense Wins 38%
Offense: 8 Defense: 1 Offense Wins 95%
Offense: 8 Defense: 2 Offense Wins 90%
Offense: 8 Defense: 3 Offense Wins 84%
Offense: 8 Defense: 4 Offense Wins 77%
Offense: 8 Defense: 5 Offense Wins 70%
Offense: 8 Defense: 6 Offense Wins 63%
Offense: 8 Defense: 7 Offense Wins 57%
Offense: 8 Defense: 8 Offense Wins 50%
Offense: 8 Defense: 9 Offense Wins 44%
Offense: 9 Defense: 1 Offense Wins 96%
Offense: 9 Defense: 2 Offense Wins 92%
Offense: 9 Defense: 3 Offense Wins 87%
Offense: 9 Defense: 4 Offense Wins 81%
Offense: 9 Defense: 5 Offense Wins 75%
Offense: 9 Defense: 6 Offense Wins 69%
Offense: 9 Defense: 7 Offense Wins 62%
Offense: 9 Defense: 8 Offense Wins 56%
Offense: 9 Defense: 9 Offense Wins 50%

With exactly the same amount of dice for the attacker and the defender. Each die comparison has a "41,7-41,7%" chance. And if it is a tie, it goes to the next die comparison. However, the last one has that offset to 41,7% as a win ratio, just like the 1 vs 1.

The chance that there is a tie is 16,7%. Actually 1/6th to be precise. So with this knowledge I can calculate the arms race results.

1 vs 1, 41,6667%; becomes 42%, you have 42
2 vs 2, 48,6111%; becomes 49%, you have 47
3 vs 3, 49,7685%; becomes 50%, you have 49
4 vs 4, 49,9614%; becomes 50%, you have 49
5 vs 5, 49,9936%; becomes 50%, you have 50

Or I am doing something wrong. Especially with 2 vs 2 dice.
Or you have some more rounding problems to fix.

@SugarPillStudios.

Your numbers are still different then mine. What am I doing wrong?

Edit:
Did not think about having extra ties for the second and more dice.
I estimated 2 vs 2. I did not yet calculate it exactly in excel. Will commence shortly.

Edit 2:
Silly me, I already had this knowledge. Yet I forgot.
The chance on having a certain number with 2 dice:

Outcome first second
1 1 11
2 3 9
3 5 7
4 7 5
5 9 3
6 11 1

It is obvious that not only the second die has a higher chance on a tie. The first one as well. So the system with more dice tend to go to a tie much much faster than any one anticipated.

First die
win 505/1296=38,966%
Tie 286/1296=22,0679%, go to second die
Los 505/1296=38,966%

Second die
win 505/1296=38,966%
Los 791/1296=61,034%

The 2 combined
win1 505/1296=38,966%
win2 144430/1679616=8,599%
los1 505/1296=38,966%
los2 226226/1679616=13,469%

win 47,565%, which is not 48,611% as I said before.
Los 52,435%

Nor is it the 47% of Zag24.
Nor is it the 48,2% in your latest table.
Los 505/1296=38,966%

You guys, someone has to do this right. Or I loose my mind. :D

I must admit that I was not able to follow your tables labeled: outcome, first, second, etc. or how you calculated those numbers. Having said that, it does look like your final calculations were very similar to mine... which produced close but still incorrect results. The difference between your results and mine appear to be the second die win percentage. You used 38.966% and I used 41.666%, but I now believe that both of these values are incorrect. One way to think about the incorrectness of these values is to think about the difference between having the two highest dice tied at a value of six versus tied at a value of one. Tied sixes give the attacker the best possible chance of winning, versus tied ones give the attacker no chance of winning. On top of this: when a tie occurs, you are much more likely to tie with sixes than with ones.

To account for this, we'll start with your first die win probability of 38.966%, but we'll break the term for a single tie (22.0179*38.966) into 6 different terms for the six different values of ties that might occur: each term will include a (Chance of Tie Occurring) * (Chance of Winning).

If you followed the logic in my original post above, I proposed a formula for calculating the odds of rolling a high value of X when rolling N dice: (X^N - (X-1)^N) / 6^N. We can square this formula to get the chances of both players rolling a high X with N = 2. This gives us the following chances to tie with each possible high value.

Chance of Tie for each High Die Value
6: 9.336%
5: 6.250%
4: 3.780%
3: 1.929%
2: 0.694%
1: 0.077%

Next we can work on our chance of winning with each of these high die values. As I did in my first post above, I'll work through this by first calculating the chances that our second die ends up rolling each possible value. Then I'll work out the chance that each of these rolls beats our opponents' possible rolls. Then we can aggregate these values into a single chance of winning for each possible high die value.

Be careful about re-using these calculations from before, as they will be slightly different here. For instance, if we know that our high die is tied at six, it is less likely that our second die is also a six (9.090%) than any other value (18.181%). Here's the table of our chances at rolling each second die value for every possible value of higher tied dice.

L\H 6-HIGH 5-HIGH 4-HIGH 3-HIGH 2-HIGH 1-HIGH
:1: 18.18% 22.22% 28.57% 40.00% 66.66% 100.0%
:2: 18.18% 22.22% 28.57% 40.00% 33.33%
:3: 18.18% 22.22% 28.57% 20.00%
:4: 18.18% 22.22% 14.28%
:5: 18.18% 11.11%
:6: 09.09%

Just like I did in my first post above, I'll use this table to create a new table that tells us the chances of our opponent rolling less than each of these values. For instance, the chance of rolling less than six is the sum of the chances of rolling a five + rolling a four + rolling a three ... from the previous table.

W\L 6-HIGH 5-HIGH 4-HIGH 3-HIGH 2-HIGH 1-HIGH
:1: ----0% ----0% ----0% ----0% ----0% ----0%
:2: 18.18% 22.22% 28.57% 40.00% 66.66%
:3: 36.36% 44.44% 57.14% 80.00%
:4: 54.54% 66.66% 85.71%
:5: 72.72% 88.88%
:6: 90.90%

Now we can finally calculate the chance of winning with each possible high tied value. We do this by multiplying the two tables above, and then adding up each column of products.

W\L 6-HIGH 5-HIGH 4-HIGH 3-HIGH 2-HIGH 1-HIGH
::: 41.32% 39.50% 36.73% 32.00% 22.22% ----0%

And these are the last numbers that we needed to fill in our formula from above: 38.966% + Σ(6->1) (Chance of Tie Occurring) * (Chance of Win). And here's that equation filled with the numbers from above:

38.966% + (9.33% * 41.32%) + (6.25% * 39.50%) + (3.78% * 36.73%) + (1.92% * 32%) + (0.69% * 22.22%) + (0.07% * 0) = 47.45%

This 47.45% appears to match what I've found in simulation and does not appear to conflict with zag24's simulation data, so I suspect that it is correct. It would certainly be nice to find an easier way of accounting for these different scenarios of ties. Although I suspect that our previous numbers (without considering ties) should be more than sufficient for working in the context of the game described. I know there's a lot here, but hope you find in helpful. Best, Gary.