This week the question is: Did xkcd Get its Math Right?

The question is, very simply: Is this comic mathematically correct?

Highlight to reveal (possibly incorrect) solution:

10 arrows in all. 5 of them poisoned. I choose 2. What is the probability 0 of them are poisened? For the first arrow I have to choose, among the 10, one of the 5 good ones. For the second arrow I choose one of the 4 among the 9. I could have chosen the same 2 arrows in the opposite order, so to avoid double counting, divide by 2. My probability is 5/10 * 4/9 * 1/2 = 1/9, appr. 0.1111.

Simply looking at 3d6, it’s easy to find a table with the probabilities, like this one. Let me replicate some of that table and add a little extra information.

3d6 sum	p	d4	p
12	27/216	n/a	0
13	21/126	4	1/4
14	15/216	3+	2/4
15	10/216	2+	3/4
16	6/216	1+	1
17	3/216	1+	1
18	1/216	1+	1

Going through the columns: Given a sum of 3d6, what is the probability of this sum, what should the d4 roll to hit a sum of 16+, and what’s the probability of that? To get the complete probability, multiply columns 2 and 4, and add these products together. (21/4 + 15*2/4 + 10*3/4 + 6 + 3 + 1) / 216 = 30.25/216, appr. 0.1400.

The 2 probabilities aren’t quite the same.