### Update

This morning’s probability puzzle, as originally posted, contained a remark at the end saying For extra clarity … “the medicine killed him” should be interpreted to mean that if he hadn’t taken the medicine, he wouldn’t have died.

I’ve realized that for some readers that wording might be subtracting more clarity than it’s adding. It is, however, correct as originally stated.

#### 8 Responses to “Update”

1. 1 1 math_geek

That’s where all those 1′s and 90%s came from in the comments to the problem.

2. 2 2 Bennett Haselton

The assumptions I used in my attempt to solve it, were:
- If you take the medicine, then “a fatal condition arises” with probability p; call that event P.
- Whether or not you take the medicine, some other fatal condition arises with probability q; call that event Q.
- The two events are independent.
- If ONE fatal condition X arises, and no OTHER fatal condition arises, then you die and we say that “X killed you”. (So “the medicine killed the man” means that P happened and Q did not.)
- If two or more fatal conditions arise, then you still die, but it is undefined which of the two events killed you.

3. 3 3 Steve Landsburg

Bennett: Yes. “The medicine killed the man” means that P happened and Q did not. This is EXACTLY right.

Everyone else: I hope this helps to clarify!!!

4. 4 4 Jonathan Campbell

Bennett: I believe that P and Q do not need to be independent.

However, Q does need to be independent of whether the medicine was taken.

If the latter is true, i get 27/32. (I see that you got a different answer, although the first few statements you make I agree with. If I have time I will try to see why you are getting a different answer.)

5. 5 5 Bnenett Haselton

I would think that if Q is independent of whether the medicine was taken, then that would force P and Q to be independent as well.

Because if P and Q are not independent, then if you know that the man took the medicine, that changes the probability of P (from zero to some nonzero value), and P and Q are not independent therefore that also changes the probability of Q, and hence knowing that the man took the medicine, changes the probability of Q, and they’re not independent.

Now regarding your 27/32 answer, how easy would be to type your work into a web form? :) Or at least, would you be able to post the equation(s) that you started with, and I’ll see if I can reproduce the algebra on my own.

6. 6 6 Mike H

I want to change my answer!

Consider three events, M = he took the medecine, K = it killed him, D = he died.

Let
a = P(M & K & D)
b = P(M & K & ~D)
c = P(M & ~K & D)
d = P(M & ~K & ~D)
e = P(~M & K & D)
f = P(~M & K & ~D)
g = P(~M & ~K & D)
h = P(~M & ~K & ~D)

Clearly, K => M & D, so b=e=f=0.

The investigators, assuming they know he died, but don’t know if he took the medecine, calculate that a=9c

it is reasonable to assume that the medecine does some good if it doesn’t kill you, so that g > 0.4/0.6*c

The probability that he took it and it killed him, given that he died, is 9c / (10c+g) < 9c / (10.667 c) = 27/32.

The probability is at most 27/32, and this assumes the medecine has no effect. If the medecine approaches 100% effectivity, c approaches zero, and the probability that he took the medecine and it killed him approaches zero. There exists some value of c that gives an answer 54%, but I'm not going to work it out.

7. 7 7 Glen

In the previous thread, I said I thought the answer was approximately 35%. But I rechecked my math and found an error. Now I think the answer is approximately 84%. Here is my work:

Let p = probability that a person will die of some other cause
Let q = probability that a person who has taken the medicine will die from it

These are independent probabilities, so a person who has taken the medicine could conceivably “die of both.” Given Steve’s clarifications in response to earlier questions, we will count this outcome as *not* definitively dying from the medicine.

There are six possible outcomes and associated probabilities:

(1) Man doesn’t take medicine, dies from something else: 0.4p
(2) Man doesn’t take medicine, lives: 0.4(1-p)
(3) Man takes medicine, dies from both medicine and something else: 0.6pq
(4) Man takes medicine, dies from something else only: 0.6p(1-q)
(5) Man takes medicine, dies from medicine only: 0.6q(1-p)
(6) Man takes medicine, lives: 0.6(1-p)(1-q)

What we already know, based on the investigators’ conclusion, is that the probability that he died solely from the medicine, assuming that he took the medicine, is 0.9. We assume the investigators were good Bayesians. Then mathematically, they have told us:

0.9 = (5) / [(3) + (4) + (5)]

That is, if we consider all outcomes in which he took the medicine and died, in 0.9 of those cases he from the medicine and only the medicine. Subbing in, we get:

0.9 = 0.6q(1-p) / [0.6pq + 0.6p(1-q) + 0.6q(1-p)]

With some algebra, this reduces to:

(q – pq) = 9p (**)

We’ll use this conclusion (**) later.

Using Bayes rule, what we want to know is the value of

(5) / [(3) + (4) + (5) + (1)]

That is, of all the cases in which the man died, we want to know the fraction in which he both took the medicine and died solely from the medicine. Subbing in:

0.6q(1-p) / [0.6pq + 0.6p(1-q) + 0.6q(1-p) + 0.4p]

With some algebra, this becomes:

0.6(q – pq) / [p + 0.6(q - pq)]

Subbing in (**) from above, this becomes:

0.6(9p) / [p + 0.6(9p)]

= 5.4p / 6.4p = 5.4/6.4

= 0.84375, or approximately 84%

So I conclude the probability that the man both took the medicine and died solely from the medicine is about 84%.

8. 8 8 John S

Correction. The penultimate paragraph should read:
Thus, if, contrary to how it was advertised, the question had allowed for the “medicine” to actually behave like a medicine, only the relationships of A to B and of A+B+C to D+E would have been known, your “assumption” would not have been justified, and the question would not have been capable of a solution.