Message text Let's assume that there are 'n' chits in the game, and 'm' Red chits are extracted in a damage "roll". Moreover, if I remember correctly, a chit cannot be extracted twice in the same damage calculation.
The probability of any given chit to be extracted is
pc(m) = 1 - (n-m)/n = m/n
since (n-m)/n is the probability that a chit is _not_ extracted in a damage calculation. The probability for either the PK or PE to be extracted (that is, the one-shot kill) is
pKE(m) = 2*pc(m) - [pc(m)]^2 = m*(2n-m)/n^2.
Also, assuming that I have a good memory, BM has 36 chits. Removing one of the two red 'pk's, the number of red chits should be equal to 36-1 = 35.
This analysis holds for a single damage roll. It gets (a lot) more complicated if you want to calculate what is the probability of an aircraft to be downed by one-shot kills. I don't even know whether or not it is possible, since the outcomes of a fire roll, the presence of blue chits, the sturdiness of aircrafts: the sturdier the aircrafts, the easier they are to suffer one-shot kills.
--- Message edited by Calsir |
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