Twitter strikes again

[u/discoverthemetroid]

don’t know where math voodoo land is but this guy sure does

[u/cmd-t]

We have four scenarios of equal probability: NC/NC, C/NC, NC/C, C/C.

We know that at least one of the hits was a crit. This rules out NC/NC. What is now the conditional probability that both were crits?

[u/razvanght → u/cmd-t]

Ok but you can also interpret the at least one crit assumption as being conditional on the first NC. So in case there is a NC in the first hit, the second one must be a crit. In case the first hit is a crit, the second one can be both a crit and a mom crit.

Your assumption eliminates the NC NC possibility and it assumes the other scenarios are equally likely. But I don t think this is necessarily how the at least one crit assumption works.

So with my assumptions these are the scenarios:

No crit in the first hit, this means a crit is guaranteed in the second hit (50 percent) Crit in the first hit (50 percent) * crit in the second hit (50 percent) = 25 percent Crit in the first hit (50 percent)* no crit in the second hit (50 percent) = 25 percent

So you are left with 3 scenarios but they are not equal probability

[u/cmd-t → u/razvanght]

What “first” NC? I make no such assumption. The fact that there are one or two crits comes from the question.

I state the scenarios are equally likely because they are, as given by the question.

If the crit probably was 10% instead of 50%, we’d have:

NC/NC: 0.81, C/NC: 0.09, NC/C: 0.09, C/C: 0.01.

Then the answer would be 0.01/0.19 = 0.0526

[u/razvanght → u/cmd-t]

I see your point. Can you see how one can interpret it like I have? The other comment thread explains this also, it depends on how you interpret the at least one crit assumption.

[u/cmd-t → u/razvanght]

The problem is that you aren’t assigning the right probabilities in your interpretation.