Selfish Trolley Car Debate So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.
His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.
My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.
So, what's the right answer? Is there even a right answer?

Update

People keep downvoting the correct answer, so I'll put it here. The correct answer is neither one of the suggestions.
The assumption of no information about my placement means there is no prior information from which to draw a probability of survival. That means there is only one strategy where this probability is even defined: to pull the lever with probability $1/2$. This trivially maximizes the survival probability ($1/2$), since it is the only probability.
 A: His answer is the right answer. If he acts according to his answer, you survive with probability $\frac56$. If he acts according to your answer, you survive with probability
$$
\frac56\cdot\frac56+\frac16\cdot\frac16=\frac{25+1}{36}=\frac{13}{18}\lt\frac{15}{18}=\frac56\;.
$$
A: This answer is meant to summarise and clarify my viewpoint as it developed in the discussion under the OP’s answer, a discussion I found fruitful; I honestly believe OP is smart, but has gotten on the wrong track with this and sees no way out; I’m trying to switch the lever here.
Firstly, unless we set up a mathematical model and give mathematical definitions to the concepts we are asking for, one cannot make any mathematical statement. OP claimed for a while that "symbols in math are just shortcuts for words", but that’s not exactly true: We are in a situation akin to the Bertrand paradox, where somebody keeps asking "but what is the probability for a randomly chosen chord", and a mathematician (OP among them, I presume) says "well that depends on what you mean by random". In our situation, the words which need a mathematical definition are "survival probability under a given strategy", and, once these are defined, what does it mean for one of them to be "optimal". OP has finally given in to at least proposing a mathematical formulation of the thought experiment, but has not given a mathematical meaning to either of these concepts, which are the ones they ask for.
Both models OP finally proposed (Aug 16) are equivalent to the one I proposed (Aug 3):

Model. Call $p \in [0,1]$ the "prior" probability that you are positioned on the first track; call a "strategy" a $q\in [0,1]$ such that you have a method which lets the trolley go on the second track with probability $q$.

Now everyone seems to have noticed that for most strategies $q$ we cannot compute a survival probability without knowing the "prior" probability of your position, and OP has a right to insist we do not know this prior. 
But what does that mean mathematically? There is a perfectly well-defined function giving out the survival probability, but it’s a function in two variables $p$ and $q$, and which OP also has finally put down in some special cases and different formulations:

$$P_{surv}(p,q) = pq + (1-p)(1-q) = 1-p-q+2pq  \qquad (*)$$

This is the master formula one should look at here, and from which we can derive everything of interest about the model. Indeed, note that joriki’s answer just amounts to computing $P(1/6, 1/6) < P(1/6, 0)$.
Now OP complained that that answer assumed $p=1/6$, and poses the question for an "optimal" $q$ "without assuming anything about $p$". But this question is as meaningful as asking "what $x$ maximises $f(x,y) = (9-x^2)y-(3-x)y^2$"? For a function in two variables, there is no (at least I’m not aware of, I’m happy to learn otherwise) standard meaning for "optimal value of first (or second) variable". 
 NB (OP does not doubt this, but I think it’s worth writing down to show the power of the model and the function $P_{surv}(p,q)$): For fixed $p=p_0$, there is a clear meaning of "optimal" $q$: Find, if it exists, a $q$ such that the function $f(q) := P(p_0,q)$ has a maximum at $q$. Just noting that this function is linear in $q$ with slope $2p_0-1$ shows, in vast generalisation of joriki’s answer:
 If we assume $p_0 < 1/2$, then $q=0$ is the unique optimal $q$ in the sense that $P_{surv}(p_0,0) > P_{surv}(p_0,q)$ for all $q \neq 0$.
If we assume $p_0 > 1/2$, then $q=1$ is the unique optimal $q$, in the sense that $P_{surv}(p_0,1) > P_{surv}(p_0,q)$ for all $q \neq 1$.
If we assume $p_0=1/2$, all $q$ are "optimal" in the sense that $P_{surv}(1/2,q)=1/2$ for all $q\in [0,1]$.
But OP wants an "optimal $q$" without further qualifications, i.e. not for a fixed $p_0$. As said, I am not aware of a standard mathematical meaning these words might have. A function in two variables can have maxima at specific points $(p_0,q_0)$, not "at a line" $q=q_0$. Of course, if need be, I can think of some ways to define such "optimal lines":
O1. First possible meaning of "optimal $q$":

$q_0$ is called "optimal" if $P_{surv}(p,q_0) \ge P_{surv}(p,q)$ for all $(p,q) \in [0,1]^2$.

Basically meaning that there is an entire line $q=q_0$ which consists of points which all are maxima in the classical sense.
Exercise: With this definition, no optimal $q_0$ exists.
O2. Second possible meaning of "optimal $q$":
Idea: Somehow project the two-variable function $P_{surv}(p,q)$ to a one-variable function $P_{surv}(q)$, and then use the standard definition for maxima of this. Since we have no information about $p$, we could average over them. Since nothing else is given, the most natural way to do this would be to weigh all $p \in [0,1]$ equally, i.e. set $P_{surv}^{avg}(q) := \int_0^1 P_{surv}(p,q) dp$ and say that

$q_0$ is called "optimal" if  $P^{avg}_{surv}(q_0) \ge P_{surv}^{avg}(q)$ for all $q \in [0,1]$.

Exercise: With this definition, all $q \in [0,1]$ are optimal.
O3. OP’s apparent meaning of "optimal $q$":

$q_0$ is called "optimal" if $P_{surv}(p,q_0)$ is maximal among those $q$ for which $P_{surv}(p,q_0)$ does not depend on $p$; a set which turns out to consist of the singleton $q=1/2$.

No exercise needed: With this definition, $q_0 =1/2$ clearly is the unique optimal strategy. 
The idea seems to be vaguely related to the one behind O2. OP keeps talking about a "survival probability under a given strategy", and a strategy which optimises this survival probability, so they must find a way to give meaning to these words. But these words suggest a function depending only on $q$, thus they also want to extract from the model a function $P’_{surv}(q)$ in one variable $q$, so that there would be a clear meaning of "optimal $q$". OP sees that in a very narrow subset of the model, such a thing could be defined; so they restrict to that very narrow subset. In other words: Instead of projecting the information of the model as in O2, OP wants to slice out the one line $q=1/2$ because restricted to this line, the function $P_{surv}(p,q)$ is independent of $p$ (i.e. constant). After narrowing down like this, there is nothing left to do, OP’s intended answer just pops out.

I claim, meta-mathematically, that O3 is not a reasonable use of the word "optimal". Whether it is, is a philosophical question not to be solved on this site, but let me give my metamathematical arguments:
O1 and O2 allow for some non-trivial math, whereas O3 solves the question by a definition.
Both O1 and O2, via said non-trivial math, come to a conclusion that agrees with a heuristic idea: that without any knowledge about the priors, no quantitative comparison of any strategies is possible (O1), respectively any strategy is as good as any other (O2). Whereas O3, via a very narrow definition, comes to the very unintutive idea that one specific strategy has a better outcome than others.
O1 and O2 make use of information from the entire model. O3 throws out almost the entire model in a first step. In a very vague analogy, O3 to me as an algebraist feels like looking at a subset of representatives when one should be looking at a quotient; modelling parity via $\lbrace 1,2 \rbrace \subset \mathbb Z$ instead of $\mathbb Z/2$. Or like looking at the subset of symmetric tensors instead of passing to the quotient symmetric algebra.
Related to this "slicing out" (O3) instead of "projecting" (O2), or just leaving the model as is and accepting that the question was ill-posed (O1): OP seems to insist for whatever non-mathematical reason that "not assuming a prior" must not be modelled by letting $p$ be a variable, but it must mean that no formula is allowed to contain a $p$, and declares that we are only allowed to look at $P_{surv}(p,1/2) \equiv 1/2$ (albeit they only get this by already using the two-variable function $P_{surv}(p,q)$ evaluated at $q=1/2$), and ignore all $P_{surv}(p,q)$ for $q \neq 1/2$. But what is a good model for something which contains both known and unknown quantities: something where you call the unknown quantity x (or $p$) and write down equations relating the known quantities and x, and then see what you can derive? Or: only allow formulas which do not refer to the unknown quantity at all? I think most mathematicians and scientists would go for the first option.
Added in response to comments: In a comment (Aug 16) to their own answer, OP wrote

$p(S)=p(L)p(D)+p(R)p(P)$ is the value we seek.

This is the same as my formula $(*)$ with $p :=p(L)$ and $q:=p(D)$. (OP then notes that $p(R) = 1-p(L)$, and I assume they can also notice $p(D) =1-p(P)$, hence be able to write $pq+(1-p)(1-q)$ too.) The plain math is now that this expression is e.g.
$=p$ for $q=1$, or
$=1-p$ for $q=0$, or
$=3/4-p/2$ for $q=1/4$, and no-one doubts it is
$=1/2$ for $q=1/2$.
So e.g. my answer to "what's the survival probability if you pull the lever with certainty ($q=0$)" is "Well, if we agree on that model, it is $1-p$, and since I don't know what $p$ is, I don't know what that is." However, OP seems to know some ontological property of $p$ called "existence". This property is related to other big words, and somehow has the crucial effect that, for the case $q=1/2$, OP's derivation of their survival probability $1/2$ as $=\frac12 p + \frac12 (1-p)$ (comment to their answer, Aug 3) is allowed, but my derivation, e.g. for the case $q=0$, of $1-p$ as $=0p+1(1-p)$ makes no sense. Where naive me is mathematically tempted to say that the formula $(*)$ contains all information about the model, and simplifies to something independent of $p$ for $q=1/2$, but depends on $p$ for other values of $q$, OP writes a lot about "destruction of information" which under "maximal entropic strategies" can "force [a probability] to exist", as "the information" gets "replaced" and "resolves [the formula] into a real number" so that their

$1/2=1/2p + 1/2(1-p) \qquad (**)$

is a profound insight resp. something totally obvious which I supposedly don't get, but my (e.g.)

$1-p =0p+1(1-p)$

amounts to "calculating a probability with information you do not have" and gives too much credit to some "auxiliary function" which "exists" only "if $p$ exists", which it does not. (But apparently, OP's equation $(**)$ "exists" even though $p$ "does not exist".) Since I do not know enough about that ontology, have never learned to tell existing from non-existing variables, have no clue which algebraic methods are and are not allowed in the case of non-existing variables, and my metaphysics are quite rusty in general these days, I cannot argue with that any further.
Dear user144527: It’s as if you ask "what $x$ maximises $f(x,y) = (9-x^2)y-(3-x)y^2$", and while I insist that question is ill-posed, but try to come up with interpretations of it which at least involve information about the entire graph of $f$, you insist that the true answer is $x=3$, "because that’s the only $x$ for which the function does not depend on $y$, hence the only one for which my question makes sense, so that’s the solution".
Final conciliatory remark: We’re not that far apart actually. We both agree that with the strategy $q=1/2$, you survive with a probability of 50%. For any other strategy $q$, I say you survive with a probability of $pq +(1-p)(1-q)$, which is unknown because $p$ is unknown; and I cannot say whether that’s better or worse. You say, for those strategies no probability “exists”; and say that dying with a probability of 50% is better than dying with a non-existing probability. If that makes you happy, fine.

Update: In further discussion, faced with my challenge "wherever you get your $1/2$ from, that’s where I get my $pq+(1-p)(1-q)$ from", OP has found a stunning way out. Maybe granting that optimality definition O3 is ridiculously narrow, as it throws out most of the information of the model, OP has decided to -- scrap the model and instead propose the

New Model. A "survival probability depending on a strategy" is defined as follows: It is $1/2$ for the strategy "flipping a coin", and undefined for every other strategy.

Others could argue that this amounts to optimality definition O3 with an extra grain of chutzpah, and that all my meta-mathematical arguments against O3, and more, hold against this entire model, but OP stresses they are happy I finally got their point, and indeed I cannot argue mathematically against the fact that within this model, their answer has a good chance ($\ge 50\%$ I bet) to be correct.
A: Your argument about "no prior information" would be applicable in the following situation:

A trolley car is heading down a track. Your boyfriend has been told by a reliable person that there are one or more people on that track who will be killed by the trolley unless he throws a lever that sends the trolley down a second track. But the same reliable source told your boyfriend that there are one or more people on the second track who would then be killed, moreover that you are one of the people who are standing on one or the other track.

In that case, due to the complete symmetry in the information, there really is no reason to say that there is anything but a $\frac12$ chance that you are on the first track.
But suppose your boyfriend then notices that he can see both groups of people by looking out a window. He looks and sees there are $5$ people on the first track and $1$ on the second track.
That is new information your boyfriend now has that he did not have before.
Can he make use of it?
A frequentist could reason in one of two ways. One line of thought is that since the only further information about the people (other than the number standing on each track) is that you are one of them, you are equally likely to be any one of the six people he sees. Therefore there is a $\frac56$ chance you are in the group of $5$ people on the first track.
The other line of thought for a frequentist is that there is some conditional probability $p$ that you would be standing on the first track rather than the second, given that you are standing on a track. A reasonable assumption is that any other person has the same conditional probability $p$ to be on the first track. Given that in a sample of $6$ individuals we observe $5$ on the first track, the maximum likelihood estimator of $p$ is $\hat p = \frac56.$ Therefore your boyfriend should act as if $p = \frac56.$
A Bayesian might have decided before looking out the window that $p = \frac12.$
Now that he has looked out the window, however, that value is the prior probability $p$ which now should be changed to a posterior probability $p'$ in light of the new evidence. Under any reasonable assumptions, the posterior probability will be greater than $\frac12.$
The point is that while it's fine to say a priori (in the absence of evidence to the contrary) that all distributions are equally likely (assuming you could somehow describe the set of "all distributions" in a such a way that there was an obviously uniform distribution over them), it is not fine to say all distributions are equally likely in the face of evidence that some distributions are more plausible than others.  Seeing $5$ people on the first track and $1$ on the second is evidence that argues in favor of distributions in which you are more likely to be standing on the first track and against distributions in which you are more likely to be standing on the second.

There is further the question whether your the best way for your boyfriend to look after your safety is to deterministically throw the lever or to follow a randomized strategy
(that is, throw the lever with some probability $q$) given that he estimates a probability $p$ that you are standing on the first track and that $p > \frac12.$
Let's consider the case where $p = \frac56.$
Strategy 1 is to throw the lever. You survive with probability
$$ P(\text{survive} \mid \text{Strategy 1}) = p = \frac56. $$
Strategy 2 is to throw the lever with probability $q.$ In the event that your boyfriend throws the lever you survive with probability $\frac56,$ the same as Strategy 1. In the event that he does not throw the lever, you survive with probability $\frac16.$ Since he throws the lever with probability $q$ and refrains from throwing it with probability $1 - q,$ your chance of survival is
$$ P(\text{survive} \mid \text{Strategy 2})
 = q\times \frac56 + (1 - q)\times\frac16 = \frac16 + \frac23 q. $$
Note that since $q$ is a probability, it must satisfy $0 \leq q \leq 1,$ and therefore $\frac16 + \frac23 q \leq \frac56$ with equality only if $q = 1.$
So Strategy 2 is gives you a lower chance of survival than Strategy 1
unless your boyfriend decides to make $q = 1,$ that is, unless he follows Strategy 1 in effect.
In summary, in this particular case your boyfriend's first thought is to act in your best interest, and it is better for you if he would follow that thought.
Of course that does not contradict in any way the idea that he should take your advice in other situations.
