# Help me figure out what's up with this probability questionApril 9, 2018 9:44 PM   Subscribe

Someone posted a probability question on Mastodon: "A, B and C toot the truth with a probability of 1/3 and lie with a probability of 2/3. A makes a toot and B toots an observation on whether or not A was tooting the truth or not. C toots that B confirmed A was truthful. What is the actual probability that A's toot was truthful?" Either I made it more complicated than it needs to be, or a lot of other people are wrong, and I suspect the former. Help me get to the bottom of this!

A link to the original question (with responses).

My first guess, and the obvious "this is a trick question" answer, was ⅓, on the assumption that the only relevant information is in the first sentence. The original poster says this is the correct answer. But I second guessed myself, because the question seemed naggingly similar to the infamous Monty Hall problem, where the obvious answer is also wrong. And it seems like Bayesian inference means we should be able to get a more accurate probability based on the additional information. But my probability theory is pretty rusty!

I worked out a truth table: of the 8 possible combinations of A, B and C telling the truth or lying, only 4 actually match the scenario.
1. `TTT` - All three are telling the truth. The probability of this is 1/27: (⅓)³.
2. `TFF` - A is telling the truth; B and C are lying. (B lied and said A lied; C lied and said B confirmed A.) Probability 4/27 = (⅓)(⅔)².
3. `FTF` - Only B is telling the truth. (A lied, B claimed correctly that A lied, C lied and said B confirmed A.) Probability 4/27 = (⅓)(⅔)².
4. `FFT` - Only C is telling the truth. (A lied, B lied and said A was telling the truth, C reported accurately that B confirmed A.) Probability 4/27 = (⅓)(⅔)².
Total probability of any of these: 13/27. Probability that A was telling the truth AND one of these occurred: only the first two of those four cases, 5/27. So the conditional probability that A was telling the truth given the scenario described: (5/27) / (13/27) = 5/13.

Am I right or is ⅓ right? If the latter, where did I go wrong: was my general approach correct and I just missed a case or something, or was I completely off base, and why? If the answer is ⅓, is it reasonable to ignore the information about B and C, and why?
posted by valrus to Science & Nature (10 answers total) 1 user marked this as a favorite

A... toot(s) the truth with a probability of 1/3... What is the actual probability that A's toot was truthful?

1/3
posted by OnefortheLast at 10:06 PM on April 9, 2018 [1 favorite]

I would side with you. I think this kind of setup, where all the actors are making independent random choices, but the "after the fact" statements in the problem allow you to eliminate some scenarios, is pretty common in knights-and-knaves puzzles.

Or another way to think about this: the question "what is the actual probability that A's toot was truthful?" is a little ambiguous-sounding, let's try to rephrase it unambiguously.
1. "When A tooted, what was the probability function he used to decide whether to tell the truth?"
2. "If you eliminate all random outcomes that could not have produced the given statements from A, B, and C, and consider all non-eliminated outcomes equally likely, in what fraction of those outcomes was A telling the truth?"

#1 wouldn't be much of a puzzle. On the other hand there certainly exist "trick" probability puzzles with irrelevant information, perhaps that's what the author was going for.

But like I said, having seen similar formulations in other puzzles where something like #2 was meant, I like that better.

Any idea whether the poster is in fact the original author or just grabbed the puzzle from somewhere else?
posted by equalpants at 10:22 PM on April 9, 2018

Or here's another similar way to phrase it.

"A flipped a 2/3 unfair coin to decide whether to make a true or false statement. Then B flipped the same coin to decide whether to make a true or false statement about A's previous statement. Then C flipped the same coin to decide whether to make a true or false statement about A's and B's previous statements. Given that C's statement was XYZ, what can you conclude about how the coin flips must have turned out?"

IMO this rephrasing is clearly asking for your answer. Whether that's what the original author meant, who knows...
posted by equalpants at 10:29 PM on April 9, 2018

Ah, this is a nice puzzle, it's 1/3 for either interpretation (I think). I believe you're wrong about the FFT case. I think C's statement is meant to be "both A and B were truthful" which eliminates FFT as a possibility. That leaves a total of 1/3 for TTT + TFF + FTF. (I'll take it for granted that you're right about the other four cases.)

Edit: believe I'm wrong, my interpretation of C's statement means FFF is a legit possibility instead. Sorry about that. Well, off to bed, hopefully this will be settled by tomorrow :).
posted by equalpants at 10:40 PM on April 9, 2018

Best answer: I was able to confirm your answer. Let X be the event that A is truthful, and Y be the event that C says that B says that A is truthful. We are looking for the probability of X given Y, or P(X|Y). By Bayes' Theorem, this is equal to P(Y|X)P(X) / [P(Y|X)P(X) + P(Y|~X)P(~X)], where ~X denotes "not X".

By assumption, P(X) = 1/3, and P(~X) = 2/3. Given X (that is, A is truthful), there are two ways that C can say that B says that A is truthful: either C is telling the truth, so B is also telling the truth, or C is lying, so B is also lying. Therefore, P(Y|X) = (1/3)*(1/3) + (2/3)*(2/3) = 5/9. Given ~X (that is, A is lying), either C is telling the truth, so B is lying, or C is lying, so B is telling the truth. Therefore, P(Y|~X) = (1/3)*(2/3) + (2/3)*(1/3) = 4/9.

Putting it all together, we have P(X|Y) = (5/9)*(1/3) / [(5/9)*(1/3) + (4/9)*(2/3)] = 5/13.

I think the OP of the question may have intended "actual probability" to mean the probability that A is truthful regardless of any other information, but in my opinion this is a misunderstanding of probability. In fact, the wording "What is the actual probability that A's toot was truthful?" specifically refers to the toot we have information about. In any reasonable interpretation of probability, having additional information changes how we calculate the probability of an event. It doesn't make sense to disregard the condition that C says that B says that A is truthful.
posted by J.K. Seazer at 12:07 AM on April 10, 2018 [3 favorites]

By the way, if we follow equalpants's interpretation that Y is the event that C says that both A and B are truthful, then the only thing that changes in the analysis above is that P(Y|~X) = 2/3, so P(X|Y) = 5/17.
posted by J.K. Seazer at 12:17 AM on April 10, 2018

Best answer: For those who think that the only relevant part of the problem setup is
"A... tells the truth with a probability of 1/3... What is the actual probability that A's statement was truthful?"
I invite you to consider the similarly structured but much less challenging problem:
A tells the truth with probability 1/3 and lies with probability 2/3. B and C tell the truth with a probability of 100% and never lie. A makes a statement and B makes an observation on whether or not A was telling the truth or not. C says that B confirmed A was truthful. What is the actual probability that A's toot was truthful?
Only in the case where A's statement is the only information you are given is it reasonable to assign probability 1/3. But in the case where you are given extra information beyond A's claim (in the case of B and C's comments) it is necessary to adjust your estimate of the probability to include that additional information (as long as we can assume from the structure of the puzzle that B and C actually know the truth or falsity of A's statement -- I am presuming we can take that as a given.)
posted by Nerd of the North at 12:19 AM on April 10, 2018 [11 favorites]

The probability of the answer being anything but 1/3 is also 1/3.
There are 3 ways to interpret the question as the way it is presented is not made clear:
1. A, B and C toot the truth collectively...(1/3)
2. A, B and C toot the truth independently...(1/3)
3. A, B and C toot the truth in sequence (as described above by another poster.)

I'd say the actual problem is a language usage one with the writer, who in all probabilty is unqualified to compose such questions.
posted by OnefortheLast at 1:32 AM on April 10, 2018 [1 favorite]

The asker also fails to make clear whether B and C always respond in the supplied way (it appears so the way it is worded), and so probabilty is assigned to A only, or whether their responses also carry probabilty.
This is written... quite stupidly. No wonder the confusion.
posted by OnefortheLast at 1:43 AM on April 10, 2018

I took a shot at the problem using only the above-the-fold part of your question, before looking at the [more inside], or the link, or any of the comments. I came up with 5/13.
posted by DevilsAdvocate at 8:01 AM on April 10, 2018 [1 favorite]