I have to capture the (8) -- and I want to maximize my chances of being able to capture the (20) on a subsequent turn -- but which die should I use? ONE die choice is obviously wrong, I didn't have to do the math to know that -- but the other two seemed pretty close. I did do the math and discovered about a 4% difference between them, maybe that's too close to tell by general principles without crunching the numbers -- but one is a bit above 50% and the other a bit below.
5 comments:
This is quite interesting. First I thought it was just another example of the classic "Highest Lower Number" (https://beatpeopleup.cheapass.com/strategy/the-highest-lowest-number/) but the added possibility of a skill attack makes it a bit more tricky.
It's in part a version of "highest lower number", but (A) there are 3 dice, not just 2, so 2nd and 3rd highest numbers both matter (B) you have to weight the probability of the outcome with the product of the probabilities AD will succeed on two consective rolls of his 20-sided die. [I used the probability he would win, rather than the probability I would win, to keep the formula simpler].
It's easy enough to see that if you reroll the 12-sided die, the lowest two dice might be 1&6, 2&6, 3&6, 4&6, 5&6, 6&6 each with probability 1/12, and 6&7 with probability 1/2. and AD's probability of winning in each case is 13/20*19/20, 13/20*18/20, .... 13/20*14/20 for the first six, and 12/20*14/20 for the last case.
He has a 47.8% chance
with the 17-sided die, the computation is similar, but the lower two dice have values 1&5, 2&5..., 5&5 with probability 1/17 each and 5&6 with probability 12/17, giving him a 51.9% chance
So the 8-sided die is the one you said was obviously wrong?
Because that actually seems a bit better than the 17-sider to me:
The lowest two dice would be 1&5, 2&5, 3&5, 4&5, 5&5, 6&5 with probability 1/8 each and 7/5 with probability 1/4.
So we'd have probabilities of 13/20*19/20 etc. for the first cases and 12/20*15/20 in the second case.
This should give a total of 51.47%, so slightly better for you than rerolling the 17-sider, right?
I thought the obviously wrong one was the 17-sider because it wasted a lot of probability on rolls which were high but useless, because they'd be captured anyway.
hmm. You're right that the significantly higher probabilities of 1-5 are partially offset by the chance of rolling a 7 or 8, though I think there are errors in your computation, I'm getting that AD has a 54.5% chance, i.e. it is the worst, but it's closer than I thought.
[I get the probabilities of 1/8*14/20 * 19/20 for the 1&5 case,
1/8*14/20*18/20 for the 2&5 case... up to 1/8*14/20*15/20 for the 5&5 case, then 1/8*13/20*15/20 for the 5&6 case, and 1/4*12/20*15/20 for the 5&7 case.]
It doesn't matter so much that many rolls of the 17 are "high but useless" what matters is that so much more of the probability is in the best case above the threshold of the other two numbers. But it seemed clear to me before doing any math that rerolling the 8 had to be worse than rerolling the 12 -- the 8 BOTH begins higher, so you are giving up a larger pre-existing number AND is more likely to roll low numbers than the 12 is.
Oh right, of course. My calculations were based on the (12) showing a 6, which is obviously untrue. :)
And your argument makes a lot of sense.
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