I thought about that a while when I made my original post before concluding that it's not necessary to worry about that. I may have been mistaken, however.
Originally Posted by bashef
Let's consider the following events, where "o" is a miss, "x"is a hit, and "." is a "don't care".
To calculate the probability of any one of those happening is (1.0) for every ".", (a) for every "o", and (1-a) for every "x". Now here's the thing--if you take the probability of each of those events *given* that you end with "xo", you can divide by (a)(1-a) to remove that term. Now you have the probability that a chain will be of the appropriate length. (Note: That a *chain* will have that length. This is the probability of the length given that you *already have a chain*.)
The probability that a chain of hits will be length 1 is (a). The probability that a chain of hits will be length 2 is (a)(1-a). The probability of that it will be of length 3 is (a)(1-a)(1-a). And in general, the probability will be (a)(1-a)^(n-1).
If we sum up over all chain lengths, we see that the total is 1.0.
Now, there are two ways to take this. I believe that what my original probabilities describe is "the probability that a given hit is part of a chain of length n". And in fact, we likely would rather have "the probability that a given hit is the end of a chain of length n". Which means you are correct. Let's see what the numbers look like then:
It's interesting that there'll always be the same number of expected hit and miss chains, but it makes sense: every hit chain must be followed by a miss chain, and vice-versa.
Original Numbers (probability of a hit/miss being in a chain of length n)
1.84% 5+hit chains 5.03% 5+miss chains
2.26% 4-hit chains 4.12% 4-miss chains
5.01% 3-hit chains 7.49% 3-miss chains
11.14% 2-hit chains 13.61% 2-miss chains
24.75% 1-hit chains 24.75% 1-miss chains
New Numbers (probability of a chain being of length n)
1.01% 5+hit chains 2.26% 4+miss chains
1.24% 4-hit chains 1.85% 4-miss chains
2.76% 3-hit chains 3.37% 3-miss chains
6.13% 2-hit chains 6.13% 2-miss chains
13.61% 1-hit chains 11.14% 1-miss chains
New Numbers (expected number of chains of length n in a 600-swing fight)
6.09 5+hit chains 13.59 5+miss chains
7.44 4-hit chains 11.12 4-miss chains
16.54 3-hit chains 20.21 3-miss chains
36.75 2-hit chains 36.75 2-miss chains
81.67 1-hit chains 66.83 1-miss chains
148.5 hit chains 148.5 miss chains
So, this is the correct way to count the number of expected chains. Thanks for getting me to re-visit that.
Correct. Apologies for using poor terminology. The intent of using expected frequency instead of probability is that it becomes much easier (in my opinion) to interpret as the raw probability becomes very small. (i.e. very few people can eyeball "1% chance" vs "0.07% chance" and really understand the difference in magnitude. It's much easier to compare "happens on average once every 1m40s" vs "happens on average once every 23m48s".)
Originally Posted by bashef