Prince Valiant mechanic help?

[Sparky]

I'd like to explore this as an opposed roll mechanic, but I need a formula for figuring out the percentages (if it can be had.)

I hand wrote a result tree for 6 dice but I'd like to do all the way up to 11.
It's not only tedious to count out all the permutations (1 vs 5 dice, 7 vs 3, etc) but I'd like to avoid the errors that WILL occur with this method.

For my purposes, I only care what percentage of the time that x dice beats y dice, where x and y are every number including and between 1 and 10. All that is required to win is a single 'odd' greater than the opposer.

I hope that makes sense. Can anybody with some math-fu help me out?

Chris

[Ian Charvill]

My quick-and-dirty way of working stuff like this out is to assume that equal amounts of coins cancel.  So the chance of victory would be the chance of getting a head in the excess coins.

This may well be the response from the department of I just pulled a statistic out of my hat.

[Sparky]

Ian,

That's a very good idea.  I very well may run with that to simplify things, but I'd still like to see how the odds stack up. I have to see if (say) 6 vs 3 has nearly the same odds of winning as 4 vs 1. I suppose I could also use a common divisor and consider the 3 vs 1 option as well.

(I still want the lower side to roll at least one randomizer.)

I'm mystified that no one else here has already done the statistical analysis for this mechanic, given its source. But that just might be me reading too much into the reviews I've read. Does the book have a chart in it?

Thanks.

Chris

[Eero Tuovinen]

The Prince Valiant mechanic? It's die pools against each other, counting successes with a 50% chance of success? No criticals or anything like that? Shouldn't be too difficult... Here they are, assuming I didn't make any calculation errors. Did the permutations in my head, so better check my reliability for today against your calculations.

Code Select Expand


Probabilities for the left column to win, in percentage:
   0  1  2  3  4  5  6  7  8  9 10
0  00 00 00 00 00 00 00 00 00 00 00
1  50 25 13 06 03 02 01 00 00 00 00
2  75 50 31 19 11 07 03 02 01 00 00
3  88 69 50 34 23 14 09 12 03 02 01
4  94 81 66 50 36 29 17 11 07 04 03
5  97 89 77 64 50 38 27 20 15 09 06
6  98 94 86 75 62 50 39 29 23 16 11
7  99 96 91 77 73 61 50 40 30 23 16
8  00 98 95 89 81 71 61 50 40 31 23
9  00 99 97 93 87 79 69 60 50 41 32
10 00 00 98 96 91 85 77 69 60 50 41


I've calculated some numbers with internal rounding, so there might be a percent of error there in some numbers due to it (didn't have a passable calculator). Otherwise the numbers are accurate to the percent assuming I didn't make any stocastical fumbles.

The easiest way to calculate this kind of system is to calculate the odds for the ties in each case and notice that for m against n, the probability for m winning is (m-1 wins n)+(m-1 ties n)/2. You can see this by noting that if m-1 ties n, there is a 50% change of another coin deciding it for m. With the recursive function in place you just have to have some baselines, and the easiest is the case of 1 against 0: that's clearly 50% win, 50% tie. After that it's easy to calculate down the column by the recursive function, and sideways by remembering that if m wins n a% of time and ties b% of time, then n wins m 100-a-b % of time.

You can reverse-engineer the probability of the tie (should you need it) between n and m from 100-(n wins m)-(m wins n).

[Sparky]

Alright! This is just what I wanted. Thank you very much, Eero!

Chris