# Probability for one of two options when one option cannot be examined

Here's a problem:

There's an on/off switch in one specific location. Switch is in off position, waiting to be flipped to on. Only two people can flip the switch, person A and person B. One of them will flip the switch, and once the switch is flipped, it will stay that way. Up to an hour after the switch is flipped we don't know anything about any of the two people. One hour after the switch is flipped we'll get some information about person A, like location, means of transportation, etc. Nothing we'll get to know about person A can confirm that person A flipped the switch, while some things can prove that it was impossible for her to do it, like being too far away. At the same time we'll get no information about person B.

Question is - what is the probability that person B flipped the switch?

To note, we are calculating probability just before we get the information about person A, knowing that we will get to know something about that person within defined limits, but nothing about person B.

My answer is that there is minimum 50% probability that person B flipped the switch.

1) there are only two options

2) without understanding either option, there is 50% chance that either person A or person B flipped the switch

3) since we will get some information about person A, but none that confirms if that person flipped the switch, nothing we'll know can diminish probability that person B flipped the switch, while something we'll know can diminish probability that person A flipped the switch

4) that means that there is minimum 50% chance that person B flipped the switch

Is that correct? Even if this is correct, is this correct way to answer? If this is not correct answer, what's the answer?

• I would argue that as we have no information at all there is no basis for answering the question. "person $X$ flips a coin that might or might not be biased. What is the probability that it comes up $H$?" is pretty much equivalent to "what number am I thinking of?" – lulu Nov 7 '17 at 12:21
• @lulu If I tell you I have a coin that is 70% biased, but I don't tell you which way it is biased and tell you nothing else, then the probability you assign that it comes up heads is 50/50. This is easiest to see from a Bayesian perspective. The information you have about the coin is symmetric with respect to heads or tails outcomes, so you should assign them equal probabilities. Of course, knowing the coin is 70% biased will affect the probability you assign to the second flip given the results of the first, e.g. that it is heads, say. At that point, the information is no longer symmetric. – Derek Elkins Nov 7 '17 at 12:43
• At what time do we calculate the probability? Do we know that the switch have been used? If we do not know anything we could claim to have 33% not used, 33% A, 33% B. If we know it has been used less than an hour ago, then the extra information we "may" get doesn't matter. If I throw a dice, look at it secretly and say that "in an hour I tell you if this is not a 6" this does not change the probability of it being a 6. If it has ghone more than an hour, we can calculate as 100% B or, 50% B and 50% A. – Ove Ahlman Nov 7 '17 at 12:47
• @DerekElkins You are, of course, free to guess that unspecified information is symmetric, but it is just a guess. If that's what is wanted here, then so be it. I agree with you that at least it has the advantage of being an unbiased guess. – lulu Nov 7 '17 at 12:49
• @Ove Ahlman - Switch is in off position, waiting to be flipped to on by either person A or B. We calculate probability just before we get to know information about person A, knowing that we will get to know something about that person, but nothing about person B. – HenryP Nov 7 '17 at 13:03

Probability theory is the process of reasoning from incomplete information. There is always some background information, which I'll denote with $I$, that we use. I'll write $A$ for the proposition "Person A will flip the switch" and similarly for $B$. The background information we have seems to contain the facts: "Only one of Person A or Person B can flip the switch" and "The switch is flipped an hour from now", and everything else we know such as facts about people and switches as well as clouds and rocketships but, by assumption, nothing specific about Person A or Person B.

Since $I$ is unchanged when we consistently swap Person A and Person B everywhere, $P(A|I) = P(B|I)$, that is, the probability that Person A will flip the switch given background information $I$ is equal to the probability that Person B will flip the switch given background information $I$. Since, given the background information $I$, these options are exhaustive, then they must sum to $1$, and thus both will be $1/2$.

You also add some extra details about possibly having some unspecified additional information $I'$ in the future. What $P(A|I'\land I)$ is can't be determined without knowing $I'$. If instead we add to $I$ the fact that we'll learn some unknown extra information in the future is not relevant to our prediction. Call $K$ the fact "In the future, I may learn more about Person A". Then $P(A|K\land I)=P(A|I)$. That is, our prediction given $I$ of whether or not Person A will flip the switch is independent of whether we will learn some unspecified additional information in the future. So even though $K\land I$ is not invariant with respect to swapping "Person A" for "Person B", the extra information $K$ is irrelevant, and thus our prediction remains symmetric.* Your prediction for the outcome of a coin toss doesn't change due to the knowledge that you will find out the outcome in the future. Nor does it change if I tell you that I will only inform you of the outcome if it comes up heads and perhaps not even then.

A key point in the above is symmetry. The $1/2$ result isn't just because there is only two options. If I added the fact that Person B was dead to $I$, the background information would no longer be invariant with respect swapping "Person A" and "Person B" and we would no longer have $P(A|I)=P(B|I)$. Indeed, you'd presumably have $P(B|I)=0$. (This would be based on the additional background information in $I$ that dead people don't flip switches.)

To be a bit more formal about what I mean by symmetry, let $Q[x\leftrightarrow y]$ stand for the proposition $Q$ with all occurrences of $x$ replaced with $y$ and vice versa. We have $$P(Q|I)=P(Q[x\leftrightarrow y]\mid I[x\leftrightarrow y])$$ which just expresses that the probability of some proposition doesn't depend on what we label things. $I$ is symmetric in $x$ and $y$ if $I\equiv I[x\leftrightarrow y]$. In your scenario, the relevant facts are "The switch is flipped on an hour from now" which is clearly symmetric when we swap "Person A" and "Person B", and "Only one of Person A or Person B can flip the switch" which becomes "Only one of Person B or Person A can flip the switch" when we swap "Person A" and "Person B", but this is logically equivalent to the original fact. In symbols, $$P(A|I)=P(A[\text{Person A}\leftrightarrow\text{Person B}]\mid I[\text{Person A}\leftrightarrow\text{Person B}]) = P(B|I)$$ where we can verify that $B\equiv A[\text{Person A}\leftrightarrow\text{Person B}]$.

* You talk the "minimum probability" presumably meaning something like "whatever our probability is knowing just $I$, it will be at least as much knowing also $I'$". There are a lot of issues with this. First, there's no reason to think $I'$ can't make it more likely that Person A will flip the switch. You say that this additional information won't "confirm" that Person A flipped the switch. This seems intended to rule out $P(A|I'\land I)=1$ but it doesn't rule out $P(A|I'\land I)=0.99$. At any rate, the problem statement starts to become incoherent/circular here.