thread: 2006-04-13 : T equals Zero
On 2006-04-19, Ghoul wrote:
Yeah... the probabilities aren't really hard to do with only two dice, of course. They do get trickier with more, naturally.
For d6 + d12, the distribution is this...
1 1 in 72 1.39%
2 3 in 72 4.17%
3 5 in 72 6.94%
4 7 in 72 9.72%
5 9 in 72 12.50%
6 11 in 72 15.28%
7 6 in 72 8.33%
8 6 in 72 8.33%
9 6 in 72 8.33%
10 6 in 72 8.33%
11 6 in 72 8.33%
12 6 in 72 8.33%
And for d8+d10, it's this...
1 3 in 80 3.75%
2 5 in 80 6.25%
3 7 in 80 8.75%
4 9 in 80 11.25%
5 11 in 80 13.75%
6 13 in 80 16.25%
7 8 in 80 10.00%
8 8 in 80 10.00%
9 8 in 80 10.00%
10 8 in 80 10.00%
Convolute those and you get the following...
P[d6+d12 loses] = 35.677%
P[ tie ] = 9.549%
P[d6+d12 wins] = 54.774%
This makes G go "On Ties"
Note that I didn't resolve those ties, but most of them would resolve against the d6+d12 side, since the lower of d8+d10 will usually be higher than the lower of d6+d12. Maybe one in 8 of them would remain ties. And, of course, this is a roll of 2 dice, not a roll of a d12 and an advantage d6, which would be added to the high roll.
This makes G go "Ooops!"
There I go... Said this was "not really hard", but I went ahead and did the d8+d10 wrong anyway.
1 1 in 80 1.25%
2 3 in 80 3.75%
3 5 in 80 6.25%
4 7 in 80 8.75%
5 9 in 80 11.25%
6 11 in 80 13.75%
7 13 in 80 16.25%
8 15 in 80 18.75%
9 8 in 80 10.00%
10 8 in 80 10.00%
That re-works out to P[win] = 41.997%, P[tied higher die] = 9.549% and P[lose]= 48.455%, which is MUCH closer, and more in synch with the results from the average.
If you break up the 9.549% ties, they end up 3.472% wins, 1.771% still tied, and 4.306% loses.
