### Another Probability Puzzle

A man goes to the doctor, gets a prescription for a headache medicine, and dies the next day. It’s known that 60% of those who receive these prescriptions actually fill them and take the medicine. It’s also been established by investigators that if the man took the medicine, then there’s a 90% chance it killed him. What’s the probability that he took the medicine and it killed him?

The answer might depend on your auxiliary assumptions, but there is a particularly simple and natural set of auxiliary assumptions that leads to a nice clean answer. And no, that answer is not 54%. (Nor would 54% be an easy answer to defend under any reasonable assumptions I can think of.)

EDIT: I had written here For extra clarity, the phrase “the medicine killed him” should be interpreted to mean that if he hadn’t taken the medicine, he wouldn’t have died. . This seems to be confusing some readers, and I briefly posted an edit here saying to ignore it — but it actually is what I meant to say all along.

Hat tip deferred till the next posting to discourage googling.

#### 76 Responses to “Another Probability Puzzle”

1. 1 1 ed

This isn’t a math puzzle, it’s a language puzzle.

I don’t know how to translate the statement “if the man took the medicine, then there’s a 90% chance it killed him” into precise mathematical form. That is, I don’t know what this statement is meant to convey.

I’m confident that if I could figure out the precise meaning of the statement the problem would either be trivial, or perhaps incoherent.

2. 2 2 Carni

I was thinking 100%. We have to use the information that he ended up dying. We don’t expect people to regularly die the next day, so we can say that’s p = 0. The problem is that he had a headache, which must be (generally, and not just in this problem) positively correlated (at least to a small degree) with death… I’m not sure how we factor that in.

3. 3 3 Fenn

89.999%?

I assume the number of people who drop dead the day after visiting a doctor for headaches is minescule. So, from his being dead, I infer he was almost certainly in the group that fills and takes the RX.

4. 4 4 Phil

Very close to 1.

By Bayes’ Rule, we have:

We know that Pr(medicine)=0.6 and Pr(dead|medicine)=0.9. Thus Pr(no medicine) is 0.4. We don’t know Pr(dead|no medicine). So here’s where we need some auxiliary assumptions. Most people, on any given day, do not die. So call this approximately 0.

Then we have (0.9*0.6)/(0.9*0.6 + 0*0.4) = 1. But since Pr(dead|no medicine) can’t really be exactly 0, the answer has to be a hair shy of 1.

5. 5 5 n+1

Pardon my butchering of statistics terms, I have yet to take a course in probability.

I assume the full range of possibilities hinted at by the .1 chance of death from non-pill sources (given that the deceased took the pills) are completely separate from the outcomes where he died from pills, and independent of whether he took the pills… so our man would normally have died 10% of the time in the first place, so the probability that he took the pills and they caused his death is 54/64= 0.84375

Somewhat related:
If our man was scheduled to be shot at 5pm, regardless of what state he was in, (i.e. if the prior expectation of death that day was already very close to 1) wouldn’t we only be interested in situations where the pills killed him before he was shot? And wouldn’t this answer be the lame answer of .54 given above?

6. 6 6 Nick

Just under one or one depending on assumptions

7. 7 7 Mike H

I also get “very close to 100%”. Nice one.

8. 8 8 Bennett Haselton

I get 27/77.

My assumptions: If you take the medicine, there is some probability p of event P where the medicine causes a fatal condition. Whether or not you take the medicine, there is some probability q of event Q where other factors in your life cause a fatal condition. The two events are independent, and you die if either P or Q. (I said “causes a fatal condition” rather than “kills you” because two different events can both “cause a fatal condition”, but in normal semantics only one event can be said to “kill you”. The problem statement explicitly says that for an event to “kill you” means that no *other* fatal condition was present — if he hadn’t taken the medicine, then he would have lived.)

For a man starting out in the state at the beginning of the problem, he dies if Q happens, or if he takes the medicine *and* Q doesn’t happen *and* P happens, i.e. he dies with probability q + 0.6*p(1-q). The *medicine kills him* if he takes the medicine, Q doesn’t happen, and P happens, and this happens with probability 0.6*p(1-q). So, given that he’s dead, by Bayes’ Theorem, the probability that the medicine killed him, is:

0.6*p(1-q)
———-
q + 0.6*p(1-q)

Now the problem states that if the man took the medicine, the probability that the medicine killed him is 90%. Suppose it were given that the man took the medicine. The probability that event Q occurred (i.e. that he would have died even without taking the medicine) would be q. The probability that he died from the medicine (i.e. that P happened and Q did not happen) would be p(1-q). So what the 90% probability tells us is that the ratio p(1-q)/q = 9.

But, given that p(1-q)/q = 9 , you can show with some algebra that the expression above, for “the probability that the medicine killed him”, must evaluate to 27/77.

9. 9 9 Bennett Haselton

Carni, Fenn, Phil, & Nick –
You stated (or, I think in Nick’s case, assumed) that the odds of dying from non-pill causes on a given day are virtually 0.
The trouble with that is that the problem says “If the man took the medicine, then there’s a 90% chance it killed him.”
If the odds of dying from non-pill causes were 0, then you could state that if the man took the medicine, there’s a 100% chance that it killed him (because he couldn’t have died from anything else).
Therefore, since the problem states this probability at only 90%, it’s built into the problem that there is a non-negligible probability of dying from non-pill causes.

10. 10 10 Al

Approximately 93%.

11. 11 11 Bennett Haselton

n+1 — unfortunately you can’t extrapolate from “the .1 chance of death from non-pill sources (given that the deceased took the pills)” to “so our man would normally have died 10% of the time”.

Suppose Lisa and Jenny both fire pistols at Gina (why should these violent examples always involve men?). Lisa hits Gina with probability 1/3, Jenny hits Gina with probabily 1/5, and Gina dies if either or both bullets hit her. We say that “Lisa *kills* Gina” only if Lisa hits Gina and Jenny misses. With simple probability, you can show that (a) the probability of Lisa hitting Gina and Jenny missing Gina is 4/15, (b) the probability of Gina dying from either or both bullets is 7/15, (c) therefore by Bayes’ Theorem, you can say that, *given* that Gina dies, the probability that “Lisa killed Gina” is 4/7. This is quite different from the original 1/3 probability that Lisa can hit Gina with her bullet.

12. 12 12 Pyramid Head

My assumption is that even if he took the medicine, there’s a 10% chance that something else killed the man.

Therefore, by a straight Bayesian calculation, I got an answer of 93%.

Tomorrow I’ll learn where I got it wrong :)

13. 13 13 n+2

If we assume that people who are prescribed and take the pill have the same probability of dying from non-pill sources as those who are prescribed and don’t take the pill, the result is 0.84375, as n+1 said. This is true for any probability of death from non-pill sources. We do not need to assume that the chance of death from non-pill sources is 0.1.

We assume that everybody has a probability of x of dying from non-pill sources. For every person that is prescribed the pill, 0.4*x don’t take the pill and they die, 0.6*x take the pill and die from non-pill sources and nine times the latter number (9*0.6*x) die because the pill killed them (we know this because of those who take the pill and die, 90% die because the pill killed them). So, of those that die (0.4*x + 0.6*x + 9*0.6*x), 0.84375 die because they took the pill: 9*0.6*x/(0.4*x + 0.6*x + 9*0.6*x).

14. 14 14 Nick

? “if the man took the medicine there is a 90% chance it killed him”

but given we know hes dead then that doesn’t matter? I think the way it is worded is different to “the medicine has a 90% mortality rate after consumption” (which is unambiguous wording)

I suppose the other thing we dont know is how long it takes to kill the person – so depending on:

a) how long the medicine takes to kill
b) interpretation of the quoted sentence at the top

it could be from 0% to 100% I suppose.

Under my (reasonable) assumption that the medicine kills you relatively quickly and that given he took the medicine it means that it killed him, the pr(taking the medicine) = 100%

15. 15 15 n+1

@ Bennett:
Looking back, I realize I made some stupid mistakes writing that post. The ‘somewhat related’ blurb presents a trivial case (0% chance of living if he didn’t take the pills) that’s inconsistent with the conditions of the problem (and the clarification Prof. Landsburg gave.). I deserved that lecture on that point… but I’m still not convinced that the numbers in the first section are wrong, but hopefully I’ll get around to seeing that soon enough.

@ Prof Landsburg: Do you support Bennett’s reading of your problem insofar that it gave .9 as an upper bound for our answer? Or did the doctor in your example really prescribe a medication that has a 90% chance of killing his patients (if they take it)?

16. 16 16 Trevor

Let x be the probability that a person who gets the prescription would have died even if they had switched their decision on taking the pill (they would have died either way). Out of these people, .4 didn’t take the pill and .6 did, since being unaffected by the pill will not change the probability of you taking the pill.

The probability we are looking for is the probability he took the pill and wouldn’t have died if he hadn’t, which is 9*.6x over the total probability he died, since the chance the pill killed him given that he took it is .9. The total Probability of death is x + 5.4x + y, where y is the people who died and didn’t take the pill but would have lived if they had. So, the answer is 5.4x/(6.4x+y). Assuming nobody who died would have been saved by the bill,so y is zero, that is .84375 = 27/32.

y should probably be such that the probability you take the pill and die is greater than the probability you don’t take it and die, since the pill probably kills more people who would have lived than it saves people who would have died. This would give an answer of .54 if the pill saved people as frequently as it killed them. This could be the case if taking the pill only killed people “randomly”, rather than for any particular biological reason, for example your decision to take the pill or not could cause you to randomly be in the wrong place at the wrong time, or prevent this from occurring. If a certain pill acted this way i.e. had no biological way to kill you, these numbers would mean that 90% of deaths are caused by something essentially random, such as accidents that could easily be avoided.

y could be greater in the case where it is curing some disease. Then, a disproportional number of dead people would not have taken the pills, since the pills prevent them from dying, so the answer would be very low.

Since the headache drug doesn’t cure any disease, the answer should probably be close to but lower than .84 because most deaths wouldn’t be effected by random events, but some surely would.

tl;dr: Basically, the answer would be below or above .54 to the extent that the drug is more or less likely to save someone than it is to kill them. It would be .84 assuming that no one who doesn’t take it and dies would have survived if they had.

17. 17 17 Harold

Without looking at all the above posts, here are my thoughts. My very first instinct was 54%. After more thought, it is better to say that before we knew he dies, the chances of the drug killing him is 54%. That is, if we were asked: what is the chances of a man who has just been given such a prescription being killed by the medicine? then the answer is 54%. This means that he has a 46% chance of not being killed by the medicine. In this case, we know that he dies, so we must factor in the prior probability of his dying the next day. The probability of dying on any particular day is very small. The chances of dying not of the medicine is therefore very small x 0.46 = very, very small. Therefore the chances of the medicvine having killed him is very close to 1.

I will now look at the other posts and decide whether to post this!

Yes, having looked above, I can see my error. I read “90 % chance the drug killed him” as “the drug has 90% mortality rate”. Clearly the investigators have found that there was a 10% chance of him dying anyway.

18. 18 18 wellplacedadjective

no med => no death
==
death => med

prob(med|death) = 1 [exactly 1]

19. 19 19 Michael

Like Harold, I interpreted “90% change the drug killed him” as the drug having a 90% mortality rate.

I wonder how much of, what I had considered, my limited aptitude for mathematics, is really just a problem with the ambiguity of language.

20. 20 20 math_geek

This is the answer I’m pretty sure.

Probability of death without medicine is P.
Probability of death with medicine is 10P, as 90% of deaths with medicine are medicine caused.

Critical assumption is that non medicine caused deaths are independent of taking medicine (possibly false if the medicine can prevent death also, but it’s headache medicine and I’ve never died of a headache).

Thus probability of death is .6 * 10P + .4P = 6.4P

Let M be the event that the person took the medicine and D be the event that he died, then by Bayes rule
P(D)P(M|D)=P(M)|P(D|M)
6.4P * P(M|D) = .6 * 10P
P(M|D) = 6P/6.4P

and since we know he died, the probability he took the medicine is P(M|D) and the probability that that death is medicine caused is 90%. Thus the answer is 5.4/6.4 = 84.375% I’m not convinced this is a nice clean answer, but maybe I made different assumptions.

21. 21 21 Pat

If he hadn’t taken the medicine, he wouldn’t have died.

He died.

So the probability he took the medicine died from it is 100%

I agree with Wellplacedadjective. This seems too easy (I have to be missing something) and I would be hesitant to turn this in for a grade!

22. 22 22 Roger Schlafly

If a headache is severe enough to see a physician (instead of just using over-the-counter painkillers), then there is a significant chance that it is an undiagnosed brain aneurysm, and it could very well kill next day without treatment. Somehow that probability needs to be estimated to do the Bayesian analysis.

23. 23 23 Dave

Here’s an interesting and probable set of assumptions that throws a wrinkle in the analysis we’ve seen so far:

“60% of those who receive these prescriptions actually fill them and take the medicine” – assumption: this rate is so low because it is well known that there is a good chance that the medicine is lethal.

“It’s also been established by investigators that if the man took the medicine, then there’s a 90% chance it killed him” – assumption: this isn’t determined after the fact, but it’s well understood before hand.

final assumption: the man with the headache is rational. He’s not going to take medicine that will almost certainly kill him unless he’s very likely to die anyway.

Given these assumptions (all of which seem reasonable to me), the probability that the medicine killed the man is less than 50%. Since knowing what he knows, he otherwise would have foregone.

24. 24 24 Scott F

Just as a sidenote: You used the wrong symbol for medicine. The caduceus was Hermes staff who is mostly associated with trade and battle. The symbol you’re looking for is the Rod of Asclepius. Although in a world where doctors prescribe pills that kill with 90% chance when taken, I suppose the caduceus is probably more apt.

25. 25 25 Super-Fly

@Pat et al,

I think the point is that he could have taken the medicine, lived, but he got hit by a bus, for example.

One thing I’d also like to assume is that the medicine only alleviated his headache, in other words, not taking it wouldn’t increase his chance of dying.

26. 26 26 Neil

The statement “if he hadn’t taken the medicine he wouldn’t have died” implies a natural mortality risk of zero, so since he’s dead the chance he took it is 100%.

27. 27 27 Henry

First impression without doing any work: the purpose of this puzzle will be to illustrate the base rate fallacy.

28. 28 28 Henry

First impression without doing any work: the purpose of this puzzle will be to illustrate the base rate fallacy.

29. 29 29 Jeff Semel

I agree with Pat and wellplacedadjective. The statement “If he hadn’t taken the medicine, he wouldn’t have died” is equivalent to “If he died, he took the medicine.” In formal logic, it’s a contrapositive. Perhaps Steve wants us to think in terms of indirect proof, a staple of mathematical reasoning.

By the time I saw this post, Steve had apparently added this clue (“for extra clarity”), taken it away because it confused people, and then put it back when he decided it was clear after all. Of course, Steve did not actually say, “No, I was wrong. It is clear!” Less humorously, he just said “it actually is what I meant to say all along.” I’ve heard a math joke with a plot like that, except the punchline is, “No, I was wrong. It is obvious!”

30. 30 30 Steve Landsburg

Pat, wellplacedadjective, Jeff Semel, etc:

The man is dead. There is a 90% chance the medicine killed him; that is, there is a 90% chance that if he hadn’t taken the medicine, he wouldn’t have died. In your arguments about contrapostives, at least some of you seem to be ignoring the “90% chance” part.

31. 31 31 Phil Birnbaum

The key is the phrasing: “If the man took the medicine, there’s a 90% chance it killed him.”

That’s not the same as “The medicine is 90% fatal.” It’s also not the same as, “If the man took the medicine, there was a 90% chance that he would die from it.”

Remember, *investigators determined* that if he took the medicine, there’s a 90% chance it killed him.

So it’s a trick question. The investigators are Bayesian. The 90% is already the posterior for the case where he took the medicine. So the answer is a shade less than 90%.

32. 32 32 BW

Let M denote the event that he took the medicine and let K denote the event that the medicine killed him.

The question is “what is the probability that he took the medicine and it killed him?” We are therefore looking for P(M & K) not P(M/K). P(M & K) = P(K/M)*P(M)

We see in the edit that “if he hadn’t taken the medicine, he wouldn’t have died”. This implies that if he is dead he took the medicine. We know he is dead so P(M) = 1. This does not necessarily imply that the medicine directly killed him. The probability that if he took the medicine it killed him, P(K/M), is 90%.

So the probability that he took the medicine AND it killed him is 1*0.9 = 90%.

This of course changes if the statement “if he hadn’t taken the medicine, he wouldn’t have died” somehow does not imply that the probability he took the medicine is 1.

33. 33 33 Super-Fly

The guy is dead. There are three possibilities.
1. He took the medicine and it killed him
2. He didn’t take the medicine, but died anyway
3. He took the medicine, it didn’t kill him, but he died from something else.

We’re looking for the prob of statement 1.

Here’s my current guess:

(.6)(.9) – (.4)*(Prob. Something else killed him) – (.6)(.1)(Prob something else killed him)

34. 34 34 Super-Fly

Let me redo my current guess. I think it should be

1-.4*(prob something else killed him)-.06(prob something else killed him)

The statement “P(He took the medicine and it killed him)” is the same as saying 1-P(He is dead and the medicine didn’t kill him) (I’m mangling the Bayesian wording to put it in terms of our problem

35. 35 35 Phil Birnbaum

OK, just in case it’s not a trick question, and “If the man took the medicine, there’s a 90% chance it killed him” really means “In the absence of other causes of death, the medicine kills the man 90% of the time,” here’s an answer. It depends on the chance of the man dying otherwise.

If he has zero chance of dying from other causes, the answer is 100%.

If he has only a normal mortality rate from other causes, the answer is very slightly less than 100%.

Otherwise, it depends what his mortality rate is. For instance, suppose that the doctor prescribed him that medicine because he had a terminal brain tumor that would kill him with 50% probability today, and 50% probability tomorrow. The medicine had a 10% chance of curing him tomorrow, and a 90% chance of killing him today.

Then, it gets more complicated.

The chance he died of the drug is the chance he filled the prescription and took the drug, times the chance he didn’t die of the tumor, times the chance that the drug killed him and not cured him. That’s 60% times 50% times 90%, which is 27%. (This assumes that if the tumor kills the man on the first day, it always kills him before the drug does.)

The chance he died of the tumor today and didn’t take the drug is 50% times 40%, which is 20%.

The chance he died of the tumor today and did take the drug is the chance that he filled the prescription and took the drug, times the chance that the tumor killed him (again, we’re assuming that if the tumor kills the man on the first day, it does so before the drug does). That’s 60% times 50%, which is 30%.

So:

p = .27: takes drug, dies of drug.
p = .2: doesn’t take drug, dies of tumor.
p = .3: takes drug, dies of tumor.
p ~= 0: dies of other causes.

That’s .77 total. So the chance he died of the drug is .27/.77, which is about .35.

Of course, that depended on my very specific assumptions about the tumor.

36. 36 36 math_geek

Phil,

The Problem with your answer seems to be that given that he dies and took the drug, he either dies to the drug or dies to the tumor. The probability that he dies of the drug is .27/(.27+.3) which is not equal to .9. We don’t know what the man’s chance of dying otherwise is, but we do know its one tenth of the chance of him dying if he takes the drug.

37. 37 37 Economists Do It With Models

My probability is a little rusty, but here goes: We observe that the dude is dead. If the probability (given that he took the drug and is dead) that the drug itself killed him is 90 percent, then by definition the probability that something else killed him is 10 percent, since he is 100% dead. (I think this is why Steve was so careful with his wording regarding the “he wouldn’t have died if he hadn’t taken the drug” point, since without this we can’t conclude that “drug” and “other causes” are mutually exclusive.) Therefore, the drug is 9 times as risky for those who take it as everything else put together is, since 9 times as many of the bad outcomes are associated with the drug than with everything else.

In order to get to an actual numerical answer, you need to make some assumption about what this baseline risk is. For a given baseline risk rate, let’s call it “other”, the probability of taking the medicine and having said medicine kill you is equal to the probability of taking the medicine times the probability of dying from the medicine given that you took the medicine. This is just 0.6*(9*other).

This makes intuitive sense, since if the baseline risk of death is quite low then the probability of taking the drug and dying should be very low as well even if most of the deaths are attributable to the drug. On the other hand, if the baseline mortality rate is high and most of the deaths are still attributable to the drug, then the drug itself must be quite dangerous.

38. 38 38 Thomas Bayes

Steve,

Which, if any, of these statements are true?

(Each probability is motivated by the parenthetic quote.)

1. The probability that the medicine killed him GIVEN that he was given the prescription AND he took it AND he died is equal to 0.9:

Pr[killed by medicine | took, prescribed, died] = 0.9

(“established by investigators that if the man took the medicine, then there’s a 90% chance it killed him”)

2. The probability that the medicine killed him GIVEN that he was given the prescription AND died is equal to 0.9 (without knowledge of whether or not he took the medicine):

Pr[killed by medicine | prescription, died] = 0.9

(“The man is dead. There is a 90% chance the medicine killed him”)

3. The probability that he would be alive GIVEN that he was given a prescription AND did not take it is equal to 0.9:

Pr[not dead | prescription, didn't take medicine] = 0.9
or
Pr[dead | prescription, didn't take medicine] = 0.1

(“there is a 90% chance that if he hadn’t taken the medicine, he wouldn’t have died”)

I think it should be 1., but I’m not certain.

Thanks.

39. 39 39 Economists Do It With Models

Actually, I made an error in my earlier post. The overall probability should be 0.9375*9*other, since we need to take all probabilities given that we observe that the person is dead. The probability that the guy took the medicine given that he is dead is the probability that he took the medicine and is dead divided by the probability that he is dead (regardless of whether he took the medicine). This is just:

(0.6*10*other)/(0.6*10*other + 0.4*other)

The others cancel out and you are left with 15/16, or 0.9375. Depending on your interpretation of the phrase “it killed him,” you would either do what I did above or just calculate 0.9375*0.9, which makes more sense now that I think about it. (This equals 0.84375.)

Basically what you are trying to do is calculate Pr((med and kill)/dead). This expands to Pr(med and kill and dead)/Pr(dead), which is (0.6*9*other)/(0.6*10*other + 0.4*other), and this gives the 0.84375 figure. All you have to assume is that the probability of getting hit by a truck is the same regardless of whether you took the medicine.

40. 40 40 wellplacedadjective

27/32

nM = doesn’t take meds
M = takes meds
nD = doesn’t die
DM = dies from meds
DO = dies from other causes

want: Pr{M,DM|DM or DO}

assume:
Pr{nM,DM} = 0 [so you can't die from meds you don't take.]
Pr{DO|M} = Pr{DO|nM} [dying from other stuff is unrelated to meds.]

so that
Pr{M,DM|DM or DO)
= Pr{M,DM}/(Pr{M,DM}+Pr{M,DO}+Pr{nM,DM}+Pr{nM,DO})
= .6/(.6+.6/9+0+.4/9)
= 27/32

41. 41 41 Phil Birnbaum

math_geek,

My answer was based on a specific interpretation of the 90% figure, which I specified in the first paragraph of my answer. Your interpretation of the 90% is explicitly different from mine.

Not that yours is wrong, it’s just not what I used in that particular comment.

42. 42 42 Phil Birnbaum

math_geek,

BTW, I do agree that your interpretation is better. It’s the one I assumed in my 12:36pm post.

43. 43 43 Steve Landsburg

Thomas Bayes: 2 is the correct assumption. Nobody knows whether he took the medicine. The investigators, who know it was prescribed but don’t know whether it was taken (and who know he is dead) have concluded that there is a .9 chance that the medicine killed him. To say that the medicine killed him is to say that he took it, and if he hadn’t taken it he wouldn’t have died.

44. 44 44 Neverfox

“if he hadn’t taken the medicine, he wouldn’t have died” means P(D | ~M) = 0% or, by modus tollens, P(M | D) = 100%. And since P(D) is 100%, P(M) = 100%

P(DM | M) = P(DM | 100%) = P(DM)

P(M & DM) = P(M) * P(DM) = P(M) * P(DM | M) = 100% * 90% = 90%

45. 45 45 math_geek

Phil,

I understand what you mean now. I just didn’t understand your interpretation of the 90% figure the first time I read it. I agree the alternative interpretation is better, as the 90% number was in reference to someone already dead.

46. 46 46 Neverfox

“All you have to assume is that the probability of getting hit by a truck is the same regardless of whether you took the medicine.”

But that’s precisely what Steve says we shouldn’t assume, if I read him strictly. He said (and confirmed again), “if he hadn’t taken the medicine, he wouldn’t have died.” I realize that it’s a really strange assumption because, of course, any real human being can die from all kinds of causes at any time and, in this universe, people are apparently immortal as long as they don’t take this particular medicine. So I guess we’re supposed to assume that in 10% of the cases someone might react to his taking the drug and kill him in response (or perhaps the drug has indirect effects like making someone commit suicide or wander into the road).

If instead Steve said, “if he hadn’t taken the medicine, he wouldn’t have died from the medicine” that would just be a trivial statement and hardly worth pointing out explicitly. What is strange is that Steve later says, “Nobody knows whether he took the medicine,” but in this world of near-immortals, the fact he is dead at all should have been all they needed to come to this conclusion.

47. 47 47 ice9

It seems to me that the interesting part of this whole story is how the investigators came up with such an odd sounding conclusion. Either the investigators are being wise-asses in presenting their findings, they are not very good investigators, or there is something unusual in the circumstances of the man’s death that the investigators are keeping to themselves. I would ask the investigators for more information. And probably investigate the investigators.

48. 48 48 Super-Fly

Neverfox:

If he takes the medicine, it has a 90% chance of killing him.

If he does not take the medicine, he does not die *because of the medicine*

If he does not take the medicine, he can still die from something else. (If he couldn’t, why would he ever take the medicine?)

49. 49 49 Phil Birnbaum

If he took the medicine, there’s a 90% chance it killed him. If he didn’t, the would have been an X% chance of him dying. As X approaches zero, the probability that he took the medicine approaches 100%. So, as X approaches zero, the probability that he took the medicine and it killed him approaches 90%.

That’s the answer to the problem.

What about the other 10%? Since the guy would still be dead, the death must have something to do with the act of taking the medicine, since if he didn’t take the medicine, he wouldn’t be dead (as the “edit” tells us).

So, since we presume the investigators to be rational, something like this must have happened: he took he medicine, and, at the same time, he slashed his hand on the medicine bottle while closing it afterwards. The investigators aren’t sure which actually killed him: the medicine, or the bleeding. But they’re 90% sure it was the medicine.

50. 50 50 Harold

1)” It’s also been established by investigators that if the man took the medicine there is a 90% chance it killed him”

2) “The investigators, who know it was prescribed but don’t know whether it was taken (and who know he is dead) have concluded that there is a .9 chance that the medicine killed him”

These do not seem to be equivalent statements. Surely 1) says that there is a 90% chance it killed him IF he took the medicine. Therefore there is a 10% chance that something else did even if he took the medicine. 2) says there is a 90% chance that he took the medicine AND it killed him.

It seems to me that he had a serious condition with a high likelihhod of dying. He also had a strong likelihood of being killed by the medicine. Possibly a rare allergic reaction, as it seems unlikely such a deadly medicine would be prescribed much if it killed so may people.

51. 51 51 Neil

Another day wasted. The drug is 100% lethal, but the probability the guy took the drug and was killed by it is 84.375% as others have already shown. But that “if he hadn’t taken the drug he wouldn’t have died” stuff is confusing because there is a 10% chance of dying even if you don’t take the drug.

52. 52 52 Phil Birnbaum

Here’s another try:

Suppose there’s a disease that always kills within a day. If you hve the disease and take the medicine, it can’t harm you, and there’s a (1-X)% chance it cures you. If you DON’T have the disease and take the medicine, it always kills you.

This fulfils the important condition that if the medicine kills you, you wouldn’t be dead if you didn’t take it.

Suppose Z% of patients who are prescribed the medicine actually have the disease. Then, four cases that lead to death:

No disease, no medicine, die: 0%, by the conditions of the problem.
No disease, medicine, die of medicine: 60% times 100% times (1-Z).
Disease, no medicine, die of tumor: 40% times 100% times Z.
Disease, medicine, die of tumor: 60% times X times Z.

We know that if he took the medicine, there’s a 90% chance that it was caused by the medicine. So

0.9 = (.6(1-Z)) / (.6(1-Z) + .6XZ) [equation 1]

What’s the chance that he took the drug and died from it? That’s represented by this expression:

.6(1-Z)) / (.6(1-Z) + .4Z + .6XZ) [expression 2]

Is there a way to sub equation 1 into expression 2 and get a more simplified answer? I can’t see one.

But if you make an assumption about Z — maybe that it’s 0.5 or something — then it probably simplifies pretty easily.

53. 53 53 Snarky

>”But that “if he hadn’t taken the drug he wouldn’t have died” stuff is confusing because there is a 10% chance of dying even if you don’t take the drug.”

I can disprove Fermat’s Last Theorem in one line. I have to say, though, the “n>2″ in the statement of the theorem is confusing because my disproof assumes n=2.

54. 54 54 Thomas Bayes

Original question:
“What’s the probability that he took the medicine and it killed him?”

Recent comment:
“The investigators, who know it was prescribed but don’t know whether it was taken (and who know he is dead) have concluded that there is a .9 chance that the medicine killed him.”

I feel like I must be missing something . . .

Pr[took medicine, killed by medicine | prescribed, dead]
=
Pr[killed by medicine | prescribed, dead]
=
0.9

If the investigators say there is a 90% chance that the medicine killed him, and, if I believe the investigators, I’d have to say there is a 90% chance that the medicine killed him. (He can’t be killed by the medicine if he doesn’t take it, so the original question doesn’t need to include the part about him taking the medicine.)

I still feel like I’m missing something about this question.

55. 55 55 Steve Landsburg

THomas Bayes: Sorry; I misspoke in the comment (though it’s right in the original blog post). The investigators have determined that *if* he took the medicine, *then* there’s a 90% chance that it killed him, not that there’s a 90% chance unconditionally.

56. 56 56 Neil

The investigators say there is a 90% the drug killed him *if* he took it, meaning that of any ten dead people who took the drug, 9 were killed by it and 1 was not killed by it (because he was going to die anyway.) I take this to imply there is a 10% chance of dying apart from the drug.

Say 100 people are presribed, then 60 would take the drug. Of these, 6 would die anyway, so 54 are at risk of being killed by the drug. But all at risk must have been killed by the drug if the investigators say 9 out of 10 dead drug-takers were killed by it. Thus 54 people were killed by the drug (it is 100% fatal), and 10 people are dead from other reasons (6 who took the drug and 4 who didn’t). Thus the probability that any observed dead guy was killed by the drug is 54/64.

I cannot reconcile this logic, though, with Steve’s comment “To say that the medicine killed him is to say that he took it, and if he hadn’t taken it he wouldn’t have died.” That statement implies the only possible cause of death is the drug and is inconsistent with the investigator’s statement that 1 out of 10 dead drug-takers die for other reasons.

57. 57 57 Hein

We know he died the next day, so we need to think about what are the odds something else besides the medication killed him.

Odds that a person in such a situation would fill Rx/Take medication AND subsequently die: 54%

Odds of this person dying due to some other reason: Approximately 151,399 / 6,602,224,175 (a figure stolen from the web, I will assume correct)

.54 vs 0.000023

Four orders of magnitude difference would mean that a person in this situation would have taken the medicine and had it kill him be 99+%

58. 58 58 Neil

I think all is well if Steve means “the medicine killed him is to say that he took it, and if he hadn’t taken it he wouldn’t have died AS A RESULT OF TAKING THE MEDICINE.”

(Sorry for the caps, I do not know how to format italics for this forum.)

59. 59 59 Anshuman

Prof. Landsburg,

It seems that a major source of confusion in these problems appears to be the wording. OK, so here’s my take and please so let me know if I am right.

P(taking medicine) = 0.6

P(death through natural causes) = 0.5 (my simplifying assumption)
The above probability is independent of medicine taking.

P(killed by medicine | having taken medicine) = a (to be determined)
P(killed by medicine | having taken medicine AND died) = 0.9

Now,
P(not killed by medicine | having taken medicine) = 1-a
P(not killed by medicine but died of natural causes | having taken medicine) = (1-a)/2

So,
P(dead | having taken medicine) = a + (1-a)/2

and

P(killed by medicine | having taken medicine AND died)

= P(killed by medicine and dead | having taken medicine)/P(dead | having taken medicine)

= a/(a + (1-a)/2)
=0.9

Solving for a, we get a = 0.81

If I change my assumption of probability of death from natural causes to ‘b’ rather than 1/2 then that changes my answer to

a = b/(.1+b)

So there.
Thanks,
Anshuman

60. 60 60 Mark R.

Let P(man dies on any given day) = x.

Then the relevant numbers are
P(man takes medicine and dies) = .6*.9 =.54
P(man doesn’t take medicine and dies) = .4*x

So, P(man died from medicine, given that the man died) = .54 / (.54 + .4*x)

Let’s say that x = 1/10,000. Then the approximate answer is 99.993%, i.e. close to certain that the medicine killed him.

61. 61 61 Mark R.

Oops, misread the question.

62. 62 62 Mark R.

I posted an incorrect solution earlier based on my misreading of the question. Feel free to delete it. New attempt:

Let P(the man dies on a particular day) = x.

Then

P(man took no medicine and died) = .4x
P(man took medicine and died from other causes) = .6x
P(man took medicine and it killed him) = .6*9x

And P(medicine killed the man, given that he died) = .6*9x/(.4x + .6x + .6*9x)

= 84.375%

63. 63 63 Neverfox

@Super-Fly

“If he does not take the medicine, he can still die from something else.”

Perhaps in the distant future but not the next day. This was explicitly stated. How else do you interpret “if he hadn’t taken the medicine, he wouldn’t have died”? I take that to mean that if he does not take the medicine, he cannot die from something else. P(DO | ~M) = 0%

“If he couldn’t, why would he ever take the medicine?”

Hey, I didn’t write the problem. As I spent a whole comment discussing, yes, it’s very far-fetched but it’s what Steve wrote. And not only did he write it, he went out of his way to clarify that he meant to say exactly that. That said, Steve is under no obligation to make the problem realistic since he’s not trying to make a point about reality. It’s just a puzzle. If he wants to have the probability of pink dragons flying out of the man’s ass be 50%, he’s entitled to do so.

64. 64 64 Thomas Bayes

Here are some definitions:

Pr[medicine kills him | takes medicine] = P1

Pr[something else kills him] = P2

I assume that both P1 and P2 are small probabilities.

KM = killed by medicine
TM = took medicine

Here is my answer:

Pr[KM, TM | D]
~=
1/(1 + (5/3)*(P2/P1))

If P2 is much smaller than P1, then Pr[KM, TM | D] ~= 1
If P2 is much larger than P1, then Pr[KM, TM | D] ~= 0
If P2 and P1 are approximately equal, then

Pr[KM, TM | D] ~= 0.375.

Or, then again, I could have made a mistake somewhere.

65. 65 65 Steve Landsburg

Neverfox:

This was explicitly stated. How else do you interpret “if he hadn’t taken the medicine, he wouldn’t have died”? I take that to mean that if he does not take the medicine, he cannot die from something else.

You have entirely misread this.

“If he hadn’t taken the medicine, he wouldn’t have died” was not an assertion. It was a clarification of what is meant by the statement “The medicine killed him”.

The problem asks for the probability that he took the medicine and it killed him. Asking for this probability is not an assertion that he took the medicine, and it’s not an assertion that the medicine killed him.

66. 66 66 Steve Landsburg

Neil:

I think all is well if Steve means “the medicine killed him is to say that he took it, and if he hadn’t taken it he wouldn’t have died AS A RESULT OF TAKING THE MEDICINE.”

No, this is not at all what I meant.

I meant that the phrase “the medicine killed him” is to be interpreted as “he took the medicine, and if he hadn’t taken the medicine, he wouldn’t have died for any reason”.

Suppose the man takes a fatal dose of medicine and simultaneously has a fatal heart attack. Did the medicine kill him? You could answer this either way. I am resolving this ambiguity by saying that in this case, he would have died anyway, so we are counting this as a case where the medicind did not kill him.

67. 67 67 Neverfox

@Super-Fly, my reply to you is in moderation (I suppose because I used a mildly naughty word). Hopefully, it will show up soon.

To add to my take on this, 54% would be the answer if we weren’t aware that he had died and all we knew was that he was prescribed the medicine. In other words, P(DM | Rx) = 54%. But we do know that he died. Because of that the fact that P(M | Rx) = 60% is a red herring. It’s not needed. The strange, albeit clear, statement “if he hadn’t taken the medicine, he wouldn’t have died” causes P(M) to collapse to 100%. This is why the fact that P(DM | M) = 90% is already the answer we seek, i.e. it is equal to P(M & DM).

In case my notation isn’t clear:
M = takes medicine
DM = dead from medicine
Rx = prescribed medicine

Assuming I’m right about the strange assumption of immortality if you don’t take the medicine, is my logic valid? And what about the assumption? Can anyone say why I’m reading it incorrectly, i.e. can you tell me why it’s not sound as well? Steve didn’t correct me but that might not mean anything.

68. 68 68 Neverfox

I meant to add that Neil’s last two comments echo my own thoughts, both about the 54/64 logic and about how it can’t be reconciled with Steve’s statement taken literally (but could be reconciled with an amended statement). I do admit that the probability that Steve meant the amended version is > than the probability that he didn’t.

So again I say the answer is 90% (literal) and 84.375% (amended).

69. 69 69 Neil

@Steve: “Suppose the man takes a fatal dose of medicine and simultaneously has a fatal heart attack. Did the medicine kill him? You could answer this either way. I am resolving this ambiguity by saying that in this case, he would have died anyway, so we are counting this as a case where the medicind did not kill him”.

No problem with that. But can he have a fatal heart attack if he does not take the medicine? If so, he can die even though he did not take the medicine.

70. 70 70 Thomas Bayes

Pr[KM,TM|D]
=
Pr[KM|TM,D]*Pr[TM|D]
=
(0.9)*Pr[D|TM]*Pr[TM]/Pr[D]
=
(0.9)*(0.6)*Pr[D|TM]/Pr[D]
——————————–
Then I think the rest goes like this . . .

Pr[KM|TM,D]
=
Pr[D|KM,TM]*Pr[KM|TM]/Pr[D|TM]
=
Pr[KM|TM]/Pr[D|TM]
=
0.9

and, for my model, I have

Pr[KM|TM] = P1,

and

Pr[D] = P2 + 0.6*P1*(1-P2) ~= P2 + 0.6*P1,

so that

Pr[D|TM] = P1/0.9,

Pr[D|TM]/P[D]
~=
(P1/0.9)/(P2 + 0.6*P1),

and

Pr[KM, TM | D]
~=
1/(1 + (5/3)(P2/P1))

71. 71 71 Neil

I would add that if one has a fatal heart attack only if one takes the drug, then logic requires us to attribute that death to the drug.

72. 72 72 Glen

I believe the answer is approximately 35%. Others have gotten this answer as well, with reasoning not identical to mine but probably equivalent. I eagerly await Steve’s confirmation. :)

73. 73 73 Math Problems

Mathematical tips and tricks given sites are more useful to the math knowledge seeker, Now a day’s children and Peoples are seeking the easy way of moths learning. Like this mathematical web site is having so many math learning shortcuts and tricks.

74. 74 74 Floccina

Is it 93%?

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