A Bayesian Riddle

A murder has been committed. The suspects are:

• Bob, a male smoker.
• Carol, a female smoker.
• Ted, another male smoker.
• Alice, a female non-smoker.

You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.

Now your crack investigative team, in which you have total confidence, reports that, on the basis of new evidence, they’ve determined that the culprit is definitely female.

Who’s the most likely culprit, and with what probability?

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80 Responses to “A Bayesian Riddle”

1. 1 1 Roger

You don’t say that the info is independent, so I say that the culprit is Alice, with a probability that could be anything from 33% to 100%.

It’s Alice (female non-smoker) with 2/3?

I’m a bit unsure about what the 2:1 probability of the perp being a smoker means.
Is it like this:

The victim was NOT tortured with a burning cigarette, and we know smoking perps do that quite often, so this speaks against the perpetrator being a smoker, but since most of our suspects are smokers, the perp is probably still a smoker?

3. 3 3 Bennett Haselton

Assuming a priori odds of 25% for each suspect, I also get 2/3 probability for Alice. I have a feeling we’re being set up, where tomorrow’s post will be a slight variation on the same puzzle, which leads to a radically different answer that will blow our minds. Or maybe we’re just really smart today.

4. 4 4 Xorph

Same here, from the first evidence i get Alice at 1/3 and the three others at 2/9 each ; and from the second evidence, Alice is at 2/3 and Carol at 1/3.
The two evidences look independant (it says new evidence), but obviously the two events (being female and being a smoker) are not.

5. 5 5 Xorph

Maybe Steven is going to use this to show that it is a grave mistake to discard evidence, even inconclusive, in a trial, because a beam of independant evidences can point to a culprit with a conclusive probability ?

6. 6 6 Partial Spectator

You have calculated the odds of 2 to 1 before knowing that the murderer was female. Therefore, before the new evidence appeared, the odds were 2/9 for Bob, Carol and Ted each (making 6/9 combined), and 3/9 for Alice. In other words, Alice was 1,5 times more likely suspect than Carol. After taking the new evidence into account, its Alice 60%, Carol 40%.

7. 7 7 Steve Reilly

I have Alice at 60% and Carol at 40%. But as a good Bayesian, I see others disagree and I update my belief that I’m correct downward accordingly.

8. 8 8 Frank McGahon

I have Alice at 60% and Carol at 40% too. Starting out at 1/4 each. Then updating for 2:1 smoker:nonsmoker Bob, Carol, Ted at 2/9 and Alice at 1/3 (or 3/9), then updating for 1:0 Female:Male means Alice is 3 to 2 more likely than Carol. 60% to 40%.

I suppose the riddle is because the result depends on when you find out that the killer is female at 100% probability. If you start at 1/4 each and then find out that the killer is female first, you get Carol and Alice at 1/2 and then when you apply the smoker:nonsmoker probability you get Carol at 2/3 to Alice at 1/3 so it makes Carol the more likely at 66.666% compared to Alice at 33.333%

The problem presumably arises because of the 100% probability that the killer is female. It’s a bit of a bait and switch. You start with 4 suspects but are told at some stage that in fact there are only two.

9. 9 9 Xorph

@Partial Spectator and Steve Reilly
You’re correct, it’s 60% and 40% anot not 2/3 and 1/3.

10. 10 10 David Johnson

I tried to woork this out using Bayes’ theorem but I was having so much trouble defining my prior distribution and information that I abandoned it. I’m not convinced this is a Bayesian question.

First thing to note is that if the smoker/non-smoker calculation relies on the murderer being a female or male with equal probability (which I think we are tempted to assume) then we don’t have enough information. Our evidence is not independent.

However if we take the smoker/non-smoker probability to be true and unaffected by the new information I think we need the following train of thought.

Pr(Bob or Carol or Ted) = 2/3
Pr(Alice) = 1/3

Note that I’m not making any further assumptions about the probability for Bob, Carol or Ted individually.

Now when it turns out that the culprit is female we remove the males from the probabilities.

Pr(Carol) = 2/3
Pr(Alice) = 1/3

So Carol is twice as likely to be the culprit if we assume that all the statements made about the evidence remain true.

I came to this conclusion by replacing the suspects with balls in a dark bag, which bizarrely made it much easier to visualise.

11. 11 11 loveactuary

It is 60% likely that the culprit is Alice, and 40% likely that it is Carol.

After the first conditional statement we learned that there is a (2/3) chance that the culprit is (Bob or Carol or Ted), and a (1/3) chance that it is Alice. Bob, Carol, and Ted’s individual chances are (2/9) each with no other information.

With the knowledge that the culprit is female, this narrows it down to Carol and Alice. Alice we know is 50% more likely to be the culprit than Carol (1/3 vs. 2/9), so with the knowledge that the killer is certainly female, Alice is 60% likely to be the culprit, and Carol 40% likely.

12. 12 12 Steve Reilly

What I don’t understand is this: I got the 60/40 using Partial Spectator’s method.

But imagine the investigative team had worked a bit faster, and my evidence in favor of a smoker had taken a bit longer to work out. Then We’d start by eliminating the men and giving Carol a 2/3 chance of being the murder and Alice a 1/3 chance (as other people seem answered.)

So my own response is based on the order of independent information, and I don’t see why the order should be relevant.

13. 13 13 Harold

Both answers of 60% and 2/3 appear reasonable, but one must be wrong. If we had received the information in a different order, would we get a different result? If we knew first that the culprit was female, we would have equal probablilities of Carol and Alice, and can forget Bob and Ted. If we then learn that there is a 2:1 chance it was a smoker, we have 2/3 chance it is Carol.

How can the “red herring” of the males alter this result? If we introduce an elephant as a possible culprit, then rule it out, surely it can’t alter the odds?

14. 14 14 ThomasBayes

For those of you who have an answer, what do you get (or use) for the following probabilities?

A. Pr[Carol]
B. Pr[Evidence | Carol]
C. Pr[Evidence | Alice]

I assume that everyone is using Pr[Alice] = 1/3 (she is the only smoker), and I assume that everyone is using Pr[Evidence | Bob or Ted] = 0 (you have total confidence in the investigation team). Here, Evidence = “investigation team says the culprit is definitely female.”

15. 15 15 Ken B

I’m curious. How many here catch the allusion?

16. 16 16 Steve Reilly

I did, but I don’t how much of the puzzle refers to it.

17. 17 17 Lucas

Before reading this thread, I had Carol at 2/3, based on the fact that this is the chance the killer smokes and she’s the only female smoker.

Breaking down, I went like this:

Before anything: each suspect has a 1/4 chance of being the killer

After the “smoking gun” (sorry): Bob, Carol and Ted with a (2/3)x(1/3) = 2/9 chance each, Alice with a (1/3)x(1/1) = 1/3 chance

After the men are out: Carol with a (2/3)x(1/1) = 2/3 chance and Alice with a (1/3)x(1/1) = 1/3 chance

But now I’m not sure. I would be a crappy detective.

@Harold,

your reasoning is incorrect. The order is tied to probability we learn of.
When Steve L. tells us the probability is 2:1 he is referring to the suspect population at hand (4 smokers, 1 non-smoker).
If you switch around the order the evidence appears, the suspect population changes (1 smoker, 1 non-smoker), and so does the probability that the killer is a smoker. It is now 40% smoker, 60% non-smoker.
I believe my initial assessment of Alice 2/3 was wrong.

19. 19 19 Ken B

@Steve Reilly: Well the puzzle is about combinations, with some on top and others on the bottom ….

20. 20 20 JonathanKariv

I’m pretty sure that this is underspecifed. That is to say that it depends on how dependent the evidence you have is on the evidence your team has.

For example:
World 1: Everyone starts thinking all 4 are equally likely:
You determine that Bob is innocent (this fits the you think 2/3 it’s a smoker).
Your team determines that Bob and Ted are innocent (fits with them thinking 100% chance of female)
In this case Alice and Carol are both 50-50 to be the culprit;
World 2: Again everyone starts out thinking all 4 are equally likely.
You determine carol is innocent (again fits with 2/3 chance)
Team determines bob and ted innocent (again fits with 100% chance female)
In this world we know alice did it.

@Ken B: I think I know what steve is alluding to but no idea how.

21. 21 21 Mike D

FWIW, if the odds are 2-1 that it was a smoker, then that tells me the odds of it being a smoker are only 33.3%. That puts Alice at 66.7%, or 1-2 on a non-smoker.

(I take this from parimutuel betting – if I put a bet on a 2-1 horse, I’m taking a 33% chance to win 3x my bet).

Before the “killer is female” intel, the smokers have an equal 11.1% of being culprit and Alice 66.7%, or 6x. Removing the two men at 22.2%, then distributing their odds proportionally between the two remaining women gives me 85-86% Alice, 14-15% Carol.

My interpretation relies on how you view the odds statement then how you allocate remaining risk. Since my calculations are unique to the conversation thus far, I allow I may have misinterpreted.

22. 22 22 ECSL

it’ll depend on whether the new evidence (100% female) overturns the prior 2-to-1 odds of a smoker being the culprit.

If so, Advo is right (along with most of the comments). If not, Harold is right with Carol and 2/3 as this is the only solution consistent with both pieces of evidence.

23. 23 23 Ted Levy

It’s unclear who did it. Which one is the butler?

24. 24 24 Ron

This is the Monty Hall problem in disguise. Alice is door number 1.
A 3:1 bias for a smoker would give each the same odds (1/4 each).
The 2:1 means Alice is 1/3 likely to be the killer, while each
smoker has a 2/9 chance. Eliminate two of the smokers (doors), and
the probabilities collapse into the remaining smoker (door). This
leaves Alice at 1/3 odds and Carol at 2/3 odds.

This is the same result as if one detective team first reported that
it was definitely a woman, then the other team independently reported
that 2:1 it was a smoker.

25. 25 25 Martin-2

Alice with 60% probability.

” the odds are 2-to-1 that the culprit is a smoker”

So P(Alice) = 1/3 = 3/9
P(Bob) = P(Carol) = P(Ted) = 2/9

“the culprit is definitely female”

Instead of writing out the theorem I’ll just shave off Bob and Ted’s probability. P(Alice) + P(Carol) = 5/9 so I’ll scale by 9/5 so they add to 1.
P(Alice|E) = 3/9 * 9/5 = 3/5 = 60%

26. 26 26 Dave Smith

I understand how everyone is getting Alice at .6, but agree with #10 David Johnson. This is not a situation where you would Bayes’ Thm.

27. 27 27 DMS

The problem is underspecified. The implication is that given no other data, the odds are 2:1 that the murderer is a smoker, which then places the individual odds at 22.2% for Bob, Ted, and Carol, and 33.3% for Alice. Then, given the above, there is evidence the murderer is known to be a female, and therefore the odds are (33.3) / (33.3 + 22.2) = 60% that it is Alice (or 40% Carol).

However, if one starts with the prior knowledge that the murderer is female, then the individual odds are 50% for Carol and Alice. But nowhere does the problem specify what the odds are that the murderer is a smoker, given the data that the murderer is female. The prior evidence of 2:1 odds that the murderer is a smoker, are odds GIVEN NO OTHER DATA (or at least that is the implication).

One could take a stab at resolving as follows:

Scenario 1: Given evidence that the murderer is a smoker with odds of 2:1, then it is known with certainty that the murderer is a female, and then Alice is 60% the likely culprit, while Carol is 40% the likely culprit.

Scenario 2: Given no evidence relating to smokers (i.e. given no other data), it is known with certainty that the murderer is a female, and that is all that can be known. Then Alice is 50% the likely culprit, while Carol is 50% the likely culprit.

Resolution: Assume the likelihood of each scenario is equal. For example, every time I look at smoker likelihood evidence I also find the female certainty evidence AND every time I don’t look at smoker likelihood evidence I find the female certainty evidence. Reasonably I can look at smoker likelihood evidence as often as I don’t look at smoker likelihood evidence, and vice versa. Thus the Bayesian kluge solution is to weight the scenarios equally, implying a 55% likelihood for Alice and a 45% likelihood for Carol as the culprit.

28. 28 28 Yancey Ward

Monty, put the cuffs on Carol.

@ECSL:
If the new evidence DOESN’T overturn the 2:1 likelihood, then it would follow that it provides no new information.

30. 30 30 Roger

All of these supposed solutions make unwarranted assumptions. As David #10 showed, the probability for Alice can be as low as 1/3 with one set of assumptions.

OTOH, assume that you have eliminated Carol and found the others equally likely. That gives the stated 2:1 odds for smokers. Then you get the female info, and conclude that the culprit must be Alice, 100%.

31. 31 31 Alex

Clarifications that would help me understand this puzzle and thread:

1. Does “odds are 2-to-1 that X” mean X has 2/3 probability?
2. Are the two conclusions (smoker and female) independent of knowledge about the suspect pool? (For instance: a cigarette with red lipstick was found at the scene.)

32. 32 32 nobody.really

A MURDER?!? Oh my.

I pray for the poor victim’s soul, and that he or she might recover. If only I knew someone to pray to….

33. 33 33 Martin-2

Darn, I thought this problem looked easy. Here’s a narrative I constructed that will hopefully do away with this order-of-evidence nonsense.

I examine the crime scene with two instruments: A Female Detector and a Smoke Detector. Both instruments are completely accurate in their measurements. However, the Smoke Detector rolls a concealed 3-sided die and lies when it lands on 1.

Now I suppose I can just do a bunch of case studies like a sucker.

34. 34 34 Martin-2

Oh and Steve, are you aware that TBQ is rot13 for GOD?

35. 35 35 Ken B

@Roger 29:
Roger and a few others are right of course, we really need to know if the new data is independent. I believe that is Thomas Bayes’s point too. HOWEVER this concern hides a nicety. To a died-in-the-wool Bayesian you can still apply the rule just assuming maximum entropy (in the relevant distribution). That was done in #6. It’s still new info for your Bayesian grist mill. A Bayesian will prefer to know if its independent of course, but that’s just missing info, you go with what you know.

So if you are happy answering 60% you are a true Bayesian; if you have qualms you might be a frequentist. [I would need repeated trials to tell :) ]

36. 36 36 Ken B

@33: Ubyl fuvg.

37. 37 37 JonathanKariv

@nobody.really: Might be what you where refferrencing but SL does want the sainthood nomination secured..

38. 38 38 Ben

I think the answer is 1/3 for Alice and the key mistake the 60% people are making is that P(B), P(C), and P(T) are not 2/9 after the first evidence; they are 2/7.

Let’s say I want to construct a hat from which I can draw slips of paper with people’s names on them, and I will draw names in proportion to how likely it is for that person to be guilty. I start by putting a slip with Alice’s name on it in the hat. Because smokers are twice as likely to be the culprit, I put 2 slips with Bob’s name in. And then 2 slips with Carol’s name, and 2 slips with Ted’s name. Now there are 7 slips in the hat and smokers have twice as much exposure to being pulled out of the hat as non-smokers (2:1 odds).

Starting from this hat, if we then find out that the culprit is female, we simply remove all of Bob and Ted’s paper slips. That leaves us with 2 Carol slips and 1 Alice slip — Alice is the culprit with 1/3 probability.

Reversing the order that evidence appears, we put 1 slip for Alice, 0 slips for Bob and Ted, and 1 slip for Carol because the killer is female. Then, we add another slip for Carol because the odds for smokers are 2:1. Same result — Alice @ 1/3, Carol @ 2/3.

I think the confusion is in the wording of the smoker odds. It’s unclear whether 2:1 has been determined based on the fact that 3 of the suspects are smokers, or whether that fact stands alone from the composition of the suspect pool. My answer assumes the latter, I think the 60% answer assumes the former.

39. 39 39 JohnW

I think it would clarify things if it were explained exactly how “you’ve concluded that the odds are 2-to-1 that the culprit is a smoker”.

40. 40 40 nobody.really

It’s a simple case of Schrodinger’s Defendant: the guilty party is 60% Alice, 33% Carol, and 7% Phillip Morris.

Plus, it’s a fair cop that society’s to blame.

41. 41 41 Ken B

Since the boxcar threads are closed I’ll drop this here. How to build a rocket from a match http://www.popsci.com/diy/article/2013-06/blast-turn-matches-tiny-rockets

42. 42 42 Sol

Ben has it right, I think. In the real world, the police would never report that non-smoker Alice was the most likely suspect by saying the odds were 2-to-1 the killer was a smoker. That’s why the result that she is most likely suspect feels very weird.

43. 43 43 Ken B

Jonathan Kariv 20 “I think I know what steve is alluding to but no idea how.”

I have just noticed that whenever Steve needs 4 people in his example he always uses Bob and Carol and Ted and Alice. It’s come up several times; he’s quite promiscuous with this. I don’t claim he has any deeper point, I just find these things interesting.

I too have no deeper point here. I’m not trying to queer his sainthood!

44. 44 44 Glen Whitman

Ben: “Let’s say I want to construct a hat from which I can draw slips of paper with people’s names on them, and I will draw names in proportion to how likely it is for that person to be guilty. I start by putting a slip with Alice’s name on it in the hat. Because smokers are twice as likely to be the culprit, I put 2 slips with Bob’s name in. And then 2 slips with Carol’s name, and 2 slips with Ted’s name. Now there are 7 slips in the hat and smokers have twice as much exposure to being pulled out of the hat as non-smokers (2:1 odds).”

This seems like a very odd way to interpret the 2-1 odds. If there are 7 slips of paper in the hat (2 each for Bob, Carol, and Ted, and 1 for Carol), then there is a 6/7 chance that I will pull a smoker from the hat. That doesn’t sound like 2-1 odds for the culprit being a smoker; it sounds like 6-1 odds.

The right way to put slips into a hat would be 2 each for Bob, Carol, and, and 3 for Alice. That results in 2/3 of the slips being smokers, and thus 2-1 odds of the culprit being a smoker. That’s before learning the culprit is definitely female. Upon learning that, you’d eliminate Bob and Ted’s slips, leaving 2 for Carol and 3 for Alice.

Mark me down with the “60% for Alice” team. I’ll also agree with Steve Reilly (#7): “But as a good Bayesian, I see others disagree and I update my belief that I’m correct downward accordingly.”

45. 45 45 Glen Whitman

Typo correction: The right way to put slips into a hat would be 2 each for Bob, Carol, and *Ted*, and 3 for Alice.

46. 46 46 Dave

From the first round of investigation:
P(B) = 1/6 = 16.6667%
P(C) = 1/6 = 16.6667%
P(T) = 1/6 = 16.6667%
P(A) = 3/6 = 50.0%

remove B and T

P(A) = 50/(50+16.66667) = 75%

47. 47 47 Harold

Bob and Alice are nearly always used for quantum mechanics and cryptography scenarios. I had thought convention filled in the next characters based on initial letters c, d etc. So we might expect Chas and Dave, (particularly if you are a fan of Cockney musical humour) or Carol. Ted comes out of left field.

Back to the problem, I had assumed you knew the 2/3 chance of being a smoker independantly of the other information.

48. 48 48 Harold

There was a 1969 film Bob and Carol, Ted and Alice.

49. 49 49 Ken B

@Harold: Almost. The title was Bob & Carol & Ted & Alice, and the lack of a comma separating the couples is significant! It was the first mainstream movie to have a plot centering on group sex.

50. 50 50 Martin-2

JohnW and others: I think the natural assumption for where you get 2-1 odds the murderer is a smoker is you consulted an oracle on the matter who lies 1/3 of the time.

51. 51 51 Dave

Sorry, late the party…

My first inclination was to do the default Bayes Rule calculation and come up with 60% for Alice.

But after thinking it through, I think it depends on the basis for this statement:

“Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.”

Are the odds “2-to-1 that the culprit is a smoker” because you’ve added up the probabilities of each suspect using all the available evidence (which did not include anything to do with smoking) and assigned these probabilities arbitrarily using the categories of smokers vs non-smokers (resulting in the initial results of Alice having a 1/3 probability and the others having 2/9 each)? In that case, when the new evidence arrives, the answer is Alice at 60%.

But if the statement “the odds are 2-to-1 that the culprit is a smoker” is based on evidence related ONLY TO BEING A SMOKER, then the answer is Carol at 2/3, because you can then eliminate the men and reallocate the smoker probabilities to the remaining two women.

“…after carefully examining the evidence” is not specific to the type of evidence. If the initial evidence that influenced the initial odds distribution includes something pertaining to smoking specifically as well as other evidence, then Alice’s final p is between 1/3 and 60%.

52. 52 52 Eliezer

I only read up to comment 25 until I posted this, so if I’m repeating someone you’ll know why.

Once the investigation learns that the culprit was definitely female, we erase Bob and Ted from the suspect pool. The fact that they were smokers has nothing to do with the case anymore. We’re only looking at Carol and Alice now. We only have one piece of evidence now that there’s a 2 to 1 chance that the killer is a non-smoker. Therefore P(Alice)= 2/3.

53. 53 53 Ben

@Glen Whitman, I totally agree with you if we understand the 2:1 odds statement as including information about the composition of suspects. I instead interpret it to mean something like “For any two given suspects I have in an interrogation room, if one of them smokes and the other one doesn’t, then the smoker is twice as likely to be the culprit” (which is independent of the composition of suspects).

54. 54 54 Eliezer

I just got the allusion, or is it illusion? Can’t wait till tomorrows post.

55. 55 55 Ben

Another clarification: I interpret the odds statement as meaning that smokers are more likely to be the culprit. It appears that many commenters understand it to mean that smokers are less likely to be the culprit. This is the difference between Alice being the culprit with 2/3 probability versus 1/3 probability.

@Dave and Ben,

I really don’t see how you can disconnect the 2:1 statement from the composition of the suspect pool.
What you are doing when you’re “disconnecting” the 2:1 statement from the existing suspect pool is simply that you assume the ratio of smokers to non-smokers in the general population is 1:1.

The statement that “there is a 2:1 likelihood the killer is a smoker” means absolutely nothing unless you relate it to a population. Imagine the ratio of smokers to nonsmokers in the general population and the suspect pool is 2:1.
What information would the statement convey? None whatsoever.
You can add an unlimited number of totally meaningless statements reflecting the percentage of the population/suspect pool with blue eyes or sexual dysfunction.

57. 57 57 Martin-2

Finally remembered to take into account the initial distribution or something and wrote out the theorem instead of using my heuristics. Anyone with /7 in their comment is correct. Carol with 2/3 probability is the answer.

58. 58 58 Eliezer

correction to Post 52.

Last line should read: 2 to 1 that the killer is a smoker, therefore P(Carol)= 2/3.

My apologies to Alice.

59. 59 59 Ben

@Advo, if the meaning of the 2:1 statement were as you describe in #2 (or as @Martin-2 describes in #33), wouldn’t that align with my reasoning for Alice @ 1/3 (or 2/3)?

I was wrong in #2 :)

61. 61 61 Geoffrey

My understanding.

It is definitely Carol or Alice. You can ignore Bob and Ted and me for that matter. We are not part of the equation.

So Carol is the most likely suspect – at 66.5% probability.

Geoffrey

62. 62 62 Geoffrey

That’s 66.6% (typing error)

63. 63 63 roystgnr

“…my own response is based on the order of independent information, and I don’t see why the order should be relevant.”

You aren’t being given independent information. You’re being given information about two *posterior* probability estimates (with different variables marginalized in each case), and since each posterior depends on all preceding evidence the order of those matters. If you backed out the evidence required to get from your prior to each posterior and then reversed the order of the evidence then you should still get to the same final result.

@ roystgnr
Exactly. That’s what I would have said if I knew math.

65. 65 65 Dave

“I really don’t see how you can disconnect the 2:1 statement from the composition of the suspect pool.
What you are doing when you’re “disconnecting” the 2:1 statement from the existing suspect pool is simply that you assume the ratio of smokers to non-smokers in the general population is 1:1.”

Not true. Here’s a hypothetical scenario to demonstrate:

Let’s say there is a test that can be performed that reveals whether a smoker has been at a location sometime in the previous day by detecting exhaled smoke (even smoke that had been originally inhaled any time in the last 48 hours). The test can narrow down a time window for when the smoker was present. The forensics team finds evidence of this residual exhaled smoke, but can’t say with certainty that it was exhaled at the time of the murder. Their best estimate is that there is a 2/3 chance it was exhaled at the time of the crime. Evidence like this would lead to a statement like, “you’ve concluded that the odds are 2-to-1 that the culprit is a smoker” without any assumptions regarding the population of suspects.

That’s the kind of idea I had in mind when I said it matters whether the evidence pertain to smoking itself, or whether smokers vs non-smokers were just an arbitrary grouping of the suspects.

66. 66 66 Dave

… in case my example was too complicated with hypothetical details, you can think of it as just being a forensic test for whether a smoker was present or not during the time of the crime and giving its output with a probability. Just trying to avoid a debate about a hypothetical forensic test…

@Dave,
I don’t think it would.

Imagine that everyone was a smoker, the entire population (aside from the victim).
Obviously, the chance of the killer being a smoker would then be 100%, would it not?
Similarly, if out of a population of 1 million there is only one non-smoker, and that’s a 102-year-old quadriplegic, the fact that the CSI has imprecise measuring equipment doesn’t suddenly make him a suspect.

68. 68 68 Ken

Letting C be the crime, S be smokers, W be woman, and M be man, we can compute the following (X’ means the complement of X):

P(S|C) = 2/3
P(W|C) = 1

There are some who want to further say the following:

P(SW|C) = 2/9
P(MW|C) = 4/9

which leads to the prior distribution:

P(SW|C) = 2/9
P(MW|C) = 4/9
P(S’W|C) = 1/3
P(S’M|C) = 0

which further leads to the posterior distribution of:

P(SW|C) = 2/9/(2/9+1/3) = 2/5
P(MW|C) = 0
P(S’W|C) = 1/3(2/9+1/3) = 3/5
P(S’M|C) = 0

However, there isn’t any good reason to conclude P(SW|C) = 2/9 and P(SM|C) = 4/9. This assumes a uniform distribution for the three smokers. Since the probability of a smoker vs non-smoker isn’t uniform, why assume that the distribution of the individual smokers is uniform?

If you consider the two by two joint probability matrix:

P(SM|C) P(S’M|C)
P(SW|C) P(S’W|C)

We know only two entries, P(S’W|C) = 1/3 and P(S’M|C) = 0, for the prior distribution. The other two are indeterminant. The only requirement for P(SM|C) and P(SW|C) is that they and non-negative and sum to 2/3, but there are an uncountable combination of numbers that satisfy those requirements.

Then from the posterior we see that the second row marginalizes to 1, but again, we have no idea what the entries of that row is, because we had no idea what P(SW|C) was from the prior (the first row entries are obviously 0).

All that can really be said is that he probability that Alice committed the crime is at least 33%, but there is very little you can say about the probability that Carol committed the crime.

A better example for my point (I think).
Imagine that at one point, fairies existed, sprinkling pixie dust everywhere they went, which would break down into characteristic chemical compounds.
In addition to fairies, a race of nasty sprites existed, which also sprinkled their magic dust everywhere, which would break down into similar (though not quite the same compounds).
Both fairies and sprites exist in equal numbers.
Scientists developed a test which would analyze the magic droppings of the fey creatures and would be able to determine with 80% accuracy whether it’s fairy or sprite dust.
Meaning that if you analyze 100% fairy dust, you will end up with results of 80% fairy dust, 20% sprite dust.

I believe if a scientist then said “I think this is fairy dust”, there is an 80% chance that it indeed is fairy (and not sprite) dust, right?

Now one day, something horrible happened. The sprites decided they had enough of tinkerbell, went to war and exterminated all the fairies.

After the great fairy armageddon, if a scientist examines a sample and comes up with the result “fairy dust”, what are the chances that it indeed IS fairy dust (given that there are no fairies left)? What are the chances if there are just a few fairies left?

70. 70 70 Yancey Ward

Steve has fooled all of you. The murderer is Sharon.

71. 71 71 Ben

Hmm, now I’ve confused myself. Consider this proof:

Let m = number of smokers in suspect pool
Let n = number of suspects in suspect pool
P(culprit|smoker) = P(smoker|culprit)*P(culprit)/P(smoker) [Bayes' theorem]
P(smoker|culprit) = 2/3 [by the 2:1 odds statement]
P(culprit) = 1/n [by assuming each suspect is equally likely to be the culprit in the absence of other information]
P(smoker) = m/n [self-evident]
Therefore, P(culprit|smoker) = 2/(3*m)

So…that means I was wrong about P=2/7 for Carol before the gender information, yet I was right about P=2/3 for Carol (and therefore P=1/3 for Alice) after the gender information.

I suspect @Advo will say I’ve implicitly used the gender evidence before the smoker evidence and I’ll get a different answer if I use the smoker evidence first. I find it hard to believe that order of evidence matters in Bayesian analysis, but I’d be interested to see a clear case in which it does (perhaps this is one of those cases, but I’m not yet convinced)

72. 72 72 Ken B

@71
Nor should you be, until they repeal the commutative law for multiplication.

73. 73 73 JohnC

“Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.”

And there lies the rub: How formally you interpret that statement determines whether you think Alice (40%) or Carroll (30%) is ultimately more likely to be the perp.

(Or, just screw it: arrest them both; sever the two; simultaneous, separate trials.)

74. 74 74 loveactuary

I second the motion for repeal @KenB.

But only with the understanding that my 7th grade Algebra test is re-graded appropriately.

(… as if none of my other classes would change at all …)

75. 75 75 Martin-2

@Ken (68) – “All that can really be said is that he probability that Alice committed the crime is at least 33%, but there is very little you can say about the probability that Carol committed the crime.”

The language in most of your post was above my level so I didn’t understand it. But since Alice and Carol are the only two remaining suspects surely P(Carol did it) = 1 – P(Alice did it). So I don’t see how there can be more to say about one probability than the other.

76. 76 76 Martin-2

Just the barebones math. The abbreviations should be clear.

Given the list of suspects:

P(A) = P(C) = 1/4

Now we can update with smoking first:

P(s) = P(A)*P(s|A) + P(~A)*P(s|~A)
= (1/4)*(1/3) + (3/4)*(2/3)
= 7/12

P(A|s) = P(A) * P(s|A) / P(s)
= (1/4) * (1/3) / (7/12)
= 1/7 => updated P(A)

P(C|s) = P(C) * P(s|C) / P(s)
= (1/4) * (2/3) / (7/12)
= 2/7 => updated P(C)

P(f) = P(A) + P(C)
= 3/7

P(A|f) = P(A) * P(f|A) / P(f)
= (1/7) * 1 / (3/7)
= 1/3

P(C|f) = P(C) * P(f|C) / P(f)
= (2/7) * 1 / (3/7)
= 2/3

Or with gender first:

P(A) = P(C) = 1/4

P(f) = P(A) + P(C)
= 1/2

P(A|f) = P(A) * P(f|A) / P(f)
= (1/4) * 1 / (1/2)
= 1/2 => updated P(A)

P(C|f) = same numbers
= 1/2 => updated P(C)

P(s) = P(A)*P(s|A) + P(C)*P(s|C)
= (1/2)*(1/3) + (1/2)*(2/3)
= 1/2

P(A|s) = P(A) * P(s|A) / P(s)
= (1/2) * (1/3) / (1/2)
= 1/3

P(C|f) = P(C) * P(f|C) / P(f)
= (1/2) * (2/3) / (1/2)
= 2/3

Same results! Thank Neptune.

77. 77 77 Mike H

It’s not a very well-posed probability question. It works perfectly as a puzzle on this blog, designed to provoke discussion. It would be lousy as an exam question.

Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker

Now, how did I come to this conclusion? Perhaps it was based on phone records showing that on the night of the murder, Bob was in Paris. Or that Carol was. In either case, I then give 2-in-1 odds that the culprit was a smoker, but the posterior probabilities are quite different.

If Bob has an alibi, my culprit is one of the two smokers Carol or Ted, or the non-smoker Alice.

If, on the other hand, Carol has an alibi, my culprit is one of the two smokers Bob or Ted, or the non-smoker Alice.

We have to make some independence assumptions to make any progress on this question. But which??

More formally, before the crack investigative team arrives, we have

S = “culprit is a smoker”
E = “the evidence about smoking occurs”
A = “Alice is the culprit”
B = “Bob is the culprit”
C = “Carol is the culprit”
T = “Ted is the culprit”

P(S|E) = 2/3
= P(S and E) / P(E)
= P(C and E) / P(E) + P(T and E) / P(E) + P(B and E) / P(E)
= P(C|E) + P(T|E) + P(B|E).

We can’t link P(C|E) or P(B|E) or P(T|E) to P(C) or P(B) or P(T), unless we *assume* that each of T, B and C is independent of E. Besides being a very odd assumption, it contradicts the assumption that the suspects were (initially) equally suspicious:

Assuming independence,

P(S|E) = P(C|E) + P(T|E) + P(B|E) = P(C) + P(T) + P(B) = 3/4.

Hence 2/3 = 3/4

So, we must either:

* abandon the assumption that initially, P(B)=P(C)=P(T)=P(A), OR
* abandon the possibility of using P(S|E)=2/3 in a meaningful way.

Abandoning either assumption makes the puzzle unsolvable.

78. 78 78 Floccina

Before looking at everyone else answer I will guess 60% chance it is Alice and 40% chance it is Carol.

79. 79 79 Alan Watson

I think that this is a trick question which most of the answers have fallen for. Bayes Theorem deals with conditional probabilities, but neither of the two probabilities in this situation, P(smoker)=2/3 and P(female)=1, is conditional. So Bayes Theorem does not apply. There are only two female suspects, Carol who is a smoker and Alice who is not a smoker, so Carol is the most likely suspect at a probability of 2/3, with a 1/3 probability for Alice.

80. 80 80 Ken

Martin,

So I don’t see how there can be more to say about one probability than the other.

Of course, this is right.