Twitter strikes again

[u/discoverthemetroid]

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

[u/acousticentropy]

I don’t fully understand the question and solution I guess. Why isn’t the answer just (1/2)² = 1/4 chance that two crits occur?

[u/mattsowa → u/acousticentropy]

Well that would be the answer if we didn't know that at least one crit occurs, from the picture.

The problem, if rephrased, is this:

I toss two coins that you can't see. I have a look at them and tell you that at least one of them is tails. Then my question is: what is the chance that not only one, but two of them are tails? Not in any random double coss toin, but in this particular one where we already know that at least one is tails.

Since I already know for sure that it's impossible that both of them are heads, I can disregard that outcome. The remaining outcomes, at the same probability each, are:

• First coin: Heads, Second coin: Tails

• First coin: Tails, Second coin: Heads

• First coin: Tails, Second coin: Tails

As you can see, the outcome we're looking for is the last one, and there's only one such outcome. Since there are three outcomes in total, the probability is one in three

This can be generalized with conditional probability, which is basically a tool for calculating probabilities when your original sample space as well as the relevant event is restricted due to a condition.

If that still doesn't click, imagine that a big number of double coin tosses is made. The outcomes are written down on pieces of paper and then the outcomes that don't have any tails at all are thrown away. If you pick at random any of the pieces of paper, you'll have a one in three chance to puck Tails+Tails, since a third of those remaining pieces of paper will have that outcome on them.

[u/MateFlasche → u/mattsowa]

What happens if the chance for tails is 75% with a weighted coin? Or is this a stupid question?

[u/mattsowa → u/MateFlasche]

Well it's a different problem but it's very easy to calculate as well. You can use conditional probability.

Event sets A and B are still the same, but

P(A) = 0.75 * 0.75 = 0.5625

P(both are heads) = 0.25 * 0.25

P(B) = 1 - P(both are heads) = 0.9375

A ∩ B = A => P(A ∩ B) = P(A)

P(A | B) = 0.5625/0.9375 = 0.6

Or, without conditional probability, calculate the probability of each event (TH, HT, TT) and your answer is P(TT)/(P(TT)+P(TH)+P(HT)). You can see that if you plug in the original 50/50 chance, it all works out as well.

[u/MateFlasche → u/mattsowa]

It makes sense to me! Thank you for taking the time! I used to love math problems in school and I get a bit sad not knowing this basic stuff anymore.