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T, J and B work for a company but the chairman has decided to fire one person randomly chosen through 3 cards. The chairman decides to fire with unequal probabilities -- T with probability of 15%, B with 5% and J with probability of 80%.

A who is a senior manager knows the chairman's decision. A and T are friends, so T decides to ask A: 'Since I know that either B or J will keep their job, If you tell me the name of one person who will keep their job, I will still be in the dark with regards to my own fate. If both keeps their job, just chose one of their names at random'.

A thinks for a second and says: 'J will keep his job'.

Does the likelihood of t being fired change after her conversation with A?

$P(T) = 0.15$ (Probability of T fired)

$P(B)= 0.05$ (Probability of B fired)

$P(J)= 0.80$ (Probability of J fired)

Let j be the event that Therese is told that J will keep his job.

$P(j|T) = 1/2$ since A could mention either J or B to T.

$P(j|J) = 0$ since if J is fired, A won't tell T that J’s job is safe.

$P(j|B) = 1$ since if B is fired, Arlene has no choice but to tell T that J will keep his job.

Setting Bayes theorem, to calculate T’s probability of being fired given A’s information, we can use Bayes theorem to get posterior probabilities $$P(T|j) = \frac{P(j|T) P(T)}{P(j|T) P(T) + P(j|J) P(J) + P(j|B) P(B)}$$ $$P(T|j) = \frac{1/2 \times 0.15}{1/2 \times 0.15 + 0 \times 0.8 + 1 \times 0.05}$$ $$P(T|j) = 0.6$$

However my intuition says that new information should not change T's probability from 0.15 at all, so am I wrong?

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    $\begingroup$ Since there's nothing wrong with your calculation, it's more about the intuition. Maybe having a look at the Monty Hall problem helps, which has a similar paradox. $\endgroup$ – CIAndrews Mar 24 '19 at 12:37
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First of all, your calculations are correct. New information of course changes the final situation. Jacob has a high chance of being fired, and if Theresa learns that he keeps his job, the remaining chance will be split between Theresa and Boris, with sum being equal to $1$ again. Surely, something must change.

The case is more clear if you consider the probability split, $(0.15,0.85,0)$, instead. When Theresa learns that Jason keeps his job, then her chance of being fired becomes $1$. Using your calculation again, this probability is also found as $1$: $$\frac{1/2\times 0.15}{1/2\times 0.15+0\times 0.85+1\times 0}=1$$

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  • $\begingroup$ So the bit that confuses me about your answer is the probability split (0.15,0,0.85) why does the split happen in a way that Theresa's is (0.15) remains same but Boris's probability split goes up from 0.05 to 0.85? $\endgroup$ – Anya Roy Mar 24 '19 at 11:37
  • $\begingroup$ Oh, I’ll change the ordering, $0.85$ is actually Jason’s probability. But, the concept is making Boris’s probability $0$, and having the information that Jason keeps his job. $\endgroup$ – gunes Mar 24 '19 at 11:43
  • $\begingroup$ Sorry, I don't mean to be annoying but can you run me through the logic of this probability split? Given that Theresa learns that Jacob will be safe, the remaining chance will be split between Theresa and Boris. So why does Jason's probability go up to 0.85 and Boris's go down to 0 ? $\endgroup$ – Anya Roy Mar 24 '19 at 12:31
  • $\begingroup$ Of course.Assume your initial distribution was $P(T)=0.15, P(B)=0, P(J)=0.85$. So, only you and Jason has probability of being fired.This is actually my main idea.And, this split could also be $(0.2, 0.8), (0.3, 0.7)...$ or something else. It doesn't matter if Theresa's probability remains constant (i.e. 0.15) or not. I just tried to make minimal changes to the original problem, while making $P(B)=0$. So, here, only T or J can be fired. And, if T's friend says that J keeps his job, then T is sure that she's fired, i.e. $P(T|j)=1$, which means new info. changed T's probability of being fired. $\endgroup$ – gunes Mar 24 '19 at 12:40
  • $\begingroup$ Why would P(B)= 0 as that's the probability of Boris being fired which is 0.05 $\endgroup$ – Anya Roy Mar 24 '19 at 13:35
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Actually, there's an important detail that's hidden in this statement: "just chose one of their names at random". I assume "at random" means 50-50%.

This is the crux of the problem. If T is fired, under the assumption stated above, T will be told B or J with equal probability. On the other hand, conditional on T not being fired, there is probability $\frac{5}{85}$ (this is the probability B is fired conditional on T not being fired) that he will be told J and $\frac{80}{85}$ probability he will be told B. So if he is not fired, he is much more likely to be told B than J. Thus, being told J should make him think his odds of being fired are worse than if he was told B.

But note that if by "choosing one of their names at random" when T is fired, we allow for unequal probabilities, we could say that we will tell him J will probability $\frac{5}{85}$ and B otherwise. Note that this is still choosing one name at random (but I'm guessing not what the OP meant) and in this case, the odds of being told J is exactly the same whether he will be fired or not. Therefore, being told J will keep their job does not change his probability of being fired!

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  • $\begingroup$ Can I confirm that you mean the following will change Let j be the event that Therese is told that J will keep his job. • P(j|T) = ½ since A could mention either J or B to T. P(j|J) = 0 since if J is fired, A won't tell T that J’s job is safe • P(j|B) = 1 since if B is fired, Arlene has no choice but to tell T that J will keep his job If yes are you saying P(j|T) = 5/85 and P(j|B) = 80/85? $\endgroup$ – Anya Roy Mar 24 '19 at 22:15
  • $\begingroup$ @AnyaRoy: that is correct $\endgroup$ – Cliff AB Mar 24 '19 at 22:22
  • $\begingroup$ So P(T/j) = {5/85 x 0.15}/ {5/85 x 0.15 + 0 x0.8 + 80/85 x0.05} doesnt give me 0.15 $\endgroup$ – Anya Roy Mar 24 '19 at 22:45
  • $\begingroup$ {5/85 x 80/85 x 0.15}/ {5/85x 80/85 x 0.15 + 0 x 0.8 + 1 x0.05} - I think this might be the right way to arrange it as only P(j|T) changes as per your logic. Please could you let me know if I am on the right track? $\endgroup$ – Anya Roy Mar 25 '19 at 2:00
  • $\begingroup$ @AnyaRoy: I think you are talking about my last paragraph? Don't worry too much about it. I'm sure when the original problem says "chooses one of the their names at random", then mean "chooses either name with equal probability". $\endgroup$ – Cliff AB Mar 25 '19 at 3:52
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An interesting question. Right from the start I think you are correct and it is just your intuition that "let's you down".

I guess the only 2 case, I can think of, when you do not get any information is when the initial distribution is uniform or when it is 20%, 40%, 40%. In all the other cases T will get an extra piece of information from the A's answer just because he is not allowed to disclose to T he is actually the one to be fired. There is some sort of inevitable "asymmetry" in the setting.

I encourage you to watch this 3 card problem from a David MacKay's lecture about information theory (the whole series is a gem imho): https://youtu.be/9w4LnXIip5A?list=PLruBu5BI5n4aFpG32iMbdWoRVAA-Vcso6&t=1655. He also rightfully points out why the Monty Halls problem is not very good educational example, since "everyone" has heard the answer but does not know why it actually is what it is :).

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