VoiceOfUnreason
Deity
- Joined
- Dec 5, 2005
- Messages
- 3,663
Edit: Oops, I get it now; you'd look for the odds that one of your first two axemen win, but that the other two fail. So, you would have to look at the odds that:
Since your first two axemen will be fighting different archers, you'd look at the odds of winning, times two. One is the inverse of the other. Ok, got that far. What about the odds of the third axeman? Man, that seems like a pain to figure out, am I right? Then again, you must've already done that to make this calculator
- both of the first two axemen win
- both of the first two axemen lose
- the third axemen loses
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Close - you've got the right idea in your head, but your vocabulary is a little bit off.
If I want the odds that the first two axemen win, then I take the probability that one axe kills an archer X(1) and square it - so the answer is X(1) * X(1). You should see that relationship in the first two rows of every table.
The odds that a single axeman loses is the complement of the odds that he wins. !X(1), if you like, which is 1 - X(1). So if I want the odds that both of the first two axes lose, I can take advantage of the fact that we know both axes fight a healthy archer, and the probability is
!X(1) * !X(1)
= ( 1 - X(1) ) * (1 - X(1))
= ...
Now, if the first two axes die, can we predict the odds that the third will face? Yes - given the information in the spreadsheet; you just have to figure out how likely it is that the stronger of the surviving archers has 100, 73, 66 etc hit points remaining. Tedious, but not difficult.
Oddly enough, it's NOT something I actually solved along the way though. Given that the case I was interested in was "kill ALL the archers", I didn't have to look at cases where I didn't have enough axes. That allows me to take a short cut.
Essentially, because I'm going for a kill in each case, I can pretend that I'm fighting the archers in order. What are the odds of killing 2 with 3: P(2,3)?
A kills #1, B kills #2, C naps. That's X(1) * X(1)
A kills #1, B dies, C kills #2. That's X(1) * X(2)
A dies, B kills #2, C kills #1. Thats ? * X(1) * ?? - but multiplication commutes, so we can write that as X(1) * ? * ??, and it's clear that we have X(1) * X(2) again. To make it clearer that this isn't the same case as above, I order the terms to match the archers. X(2) * X(1)
So you see, I can get the answer to all of the successful outcomes by computing X(n) -- it turns out that X(8) was as high as I needed to go for these cases -- and then jumbling things around.
In other words, I can calculate the odds that two axes out of four survive against two archers without needing the result that one axe survives out of three against two archers.