# The absent-minded variations

I wake up in a featureless room. I don't remember how I ended up here, when did I go to sleep, or for that matter, anything at all besides general abstract facts about the world and math. The place is hard to qualify; for the most part it really seems like I am just inside a perfect Platonic cube, whose surface is smooth, polished to an impossible degree, perfectly white, and suffused with a slight glow. The only detail worthy of attention is a rectangular black screen in front of me, with a small numeric keypad under it. There is text on the screen. It reads:

    THIS IS A SIMULATION
    YOU ARE INSTANCE #4859
    YOU ARE IN EITHER NODE X OR Y
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * EXITING AT X IS WORTH 0 AND TERMINATES THE INSTANCE
    * ADVANCING AT X REINITIALISES THE SIMULATION AT Y
    * EXITING AT Y IS WORTH 4 AND TERMINATES THE INSTANCE
    * ADVANCING AT Y IS WORTH 1 AND TERMINATES THE INSTANCE
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This being a simulation does at least explain the lazy "default Unity shader" aesthetic of the environment. The problem as posed also does jog some memories in my mind; I recognise it as the classic "absent-minded driver" problem, from the [1997 paper by M. Piccione and A. Rubinstein, "On the Interpretation of Decision Problems with Imperfect Recall"](https://arielrubinstein.tau.ac.il/papers/53.pdf). I sketch the logic of the problem in my mind:

![A graph rendition of the problem](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/622823cba7f07660081389340f6b2772eec67a3354672272.png)

The problem is simple enough; if you chose "always exit" you'd be sure to end up with 0, while if you chose "always advance" you would end up with a consolation prize of 1. The optimal solution is to adopt a so-called "mixed strategy": you pre-commit to only advancing with a certain probability $p$ and then maximise the utility as a function of that. Since

$$
U(p) = p^2+4p(1-p)
$$

it follows from setting the derivative to zero that

$$
U'(p) = -6p+4 =0 \rightarrow p = \frac{2}{3}
$$

which gives a utility $U_{max}=\frac{4}{3}$. Better than $$1, though not by much. I type in my answer and press enter, hoping this means I'll get the reward - or whichever "me" waits at the other end of this strange exercise does.

Everything goes white.

* * *

    ...
    YOU ARE INSTANCE #4860
    ...

\[...\]

I rub my eyes. The problem is simple enough and I recognise it, but I never quite gelled with any of the solutions suggested by the paper. After all, I am a Bayesian; should I not form beliefs based on the circumstances? The fact itself that I am here thinking and existing should be evidence of *something*. I set to work to consider the likelihood of my circumstances.

There are two possibilities: I am either at X, or at Y. I am reset every time, so as long as I am still *myself*, given the same exact input, I should trust myself to always return the same output. Let's call the event of me living and thinking L. The simulation's time has only two steps; let's call them $T_1$ and $T_2$. Assign prior probability $1/2$ to both; total ignorance. What is the probability of me existing if it's $T_1$? It's 1; I will always spawn in the node X no matter what. What about $T_2$? It's only $p$; if I have exited at X, then I will be non-existent and the simulation will have terminated. So it follows that:

$$
P(Y|L) = P(T_2|L) = \frac{P(L|T_2)P(T_2)}{P(L|T_1)P(T_1)+P(L|T_2)P(T_2)} = \frac{p}{1+p}
$$

Giving me an expected utility of

$$
U(p) = P(X|L;p)U_X(p)+P(Y|L;p)U_Y(p) = 2\frac{p^2+4p(1-p)}{1+p}
$$

The gradient of which is

$$
U'(p) = 2\frac{4-6p-3p^2}{(1+p)^2}
$$

Luckily enough I don't have to worry about the denominator going to zero since $p \in [0, 1]$, and I can simply zero the numerator instead. Of the two possible solutions the only valid one is $p = \sqrt{21/9}-1 \approx 0.52752523$. I punch that in. 

Everything goes black.

* * *

    THIS IS A SIMULATION
    YOU ARE INSTANCE #4861
    YOU ARE IN EITHER NODE X, Y, or Z
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * EXITING AT X REINITIALISES THE SIMULATION AT Z
    * ADVANCING AT X REINITIALISES THE SIMULATION AT Y
    * EXITING AT Y IS WORTH 4 AND TERMINATES THE INSTANCE
    * ADVANCING AT Y IS WORTH 1 AND TERMINATES THE INSTANCE
    * EITHER EXITING OR ADVANCING AT Z ARE WORTH 0
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This is a weird variation to the absent-minded driver problem I'm familiar with. I don't much get the point of it, given it seems to only add a useless dead branch to it:

![A graph rendition of the problem](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/cb48da70c9ca5cae72774c343c94c28b1cf3f59ea59f28b0.png)

I set to work to compute my expected beliefs of being at each node. I get quite easily:

$$
P(X) = \frac{1}{2} \qquad P(Y) = \frac{p}{2} \qquad P(Z) = \frac{1-p}{2}
$$

Which results in a total utility of

$$
U(p) = P(X;p)U_X(p)+P(Y;p)U_Y(p)+P(Z;p)U_Z(p) = p^2+4p(1-p)
$$

and an optimal strategy of $p=2/3$.

I am bemused by the fact that thinking about it, this is the same exact solution as the timeless approach to the classic problem in the first place. It is also different from what I would have come to if I had not had to include Z in my calculations. The existence of Z removed one factor - it made it impossible for me to take guesses about my position conditioned on my own existence. It made the problem non-anthropic; I knew I would be in play until the end no matter the strategy chosen, or the outcome of the RNG. That was enough to make the belief-based and the timeless approaches coincide and reach the same conclusion. I shrug and punch in my answer.

Everything goes white.

* * *

    THIS IS A SIMULATION
    YOU ARE INSTANCE #4865
    YOU ARE IN EITHER NODE X or Y
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * EXITING AT X IS WORTH 0 AND TERMINATES THE INSTANCE
    * ADVANCING AT X REINITIALISES THE SIMULATION AT Y
    * EXITING AT Y IS WORTH 4 AND TERMINATES THE INSTANCE
    * ADVANCING AT Y IS WORTH 1 AND TERMINATES THE INSTANCE
    YOU WILL BE NOTIFIED OF WHETHER YOU ARE IN NODE X
    THE NOTIFICATION HAS A FALSE POSITIVE RATE OF 0, AND FALSE NEGATIVE RATE OF N
    
    NOTIFICATION: YOU ARE AT NODE X
    
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This is a weird variation to the absent-minded driver problem I'm familiar with. But well, there's not a lot to think about; given the notification, and that I can trust it for sure, the answer is obvious. I punch 1 in.

Everything goes white.

* * *

    THIS IS A SIMULATION
    YOU ARE INSTANCE #4866
    YOU ARE IN EITHER NODE X or Y
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * EXITING AT X IS WORTH 0 AND TERMINATES THE INSTANCE
    * ADVANCING AT X REINITIALISES THE SIMULATION AT Y
    * EXITING AT Y IS WORTH 4 AND TERMINATES THE INSTANCE
    * ADVANCING AT Y IS WORTH 1 AND TERMINATES THE INSTANCE
    YOU WILL BE NOTIFIED OF WHETHER YOU ARE IN NODE X
    THE NOTIFICATION HAS A FALSE POSITIVE RATE OF 0, AND FALSE NEGATIVE RATE OF N
    
    NOTIFICATION: YOU ARE NOT AT NODE X
    
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This is a weird variation to the absent-minded driver problem I'm familiar with. There's quite a lot to think about. Given N, I should be able to infer something about my position in the tree from the notification, however unreliable.

I know one trick - don't fall for anthropics. I'll add an artificial node Z to my network, as if the early exit led you to another place. There's two possible histories: one in which I received the right notification while in X, and one in which I didn't. Generally speaking, across all histories, it should hold that my prior probabilities for being in each node are:

$$
P(X;N,p) = \frac{N + (1-N)}{2} = \frac{1}{2}
$$

$$
P(Y;N, p) = \frac{N+(1-N)p}{2}
$$

$$
P(Z;N,p) = \frac{(1-N)(1-p)}{2}
$$

Then what about my conditional probabilities? Well, given that I've received the notification $\neg\tilde{X}$, we'll have

$$
P(\neg\tilde{X}) = \frac{(1-N)}{2}+\frac{1}{2} = \frac{2-N}{2}
$$

$$
P(X|\neg\tilde{X}) = \frac{P(\neg\tilde{X}|X)P(X)}{P(\neg\tilde{X})} = \frac{1-N}{2-N}
$$

$$
P(Y|\neg\tilde{X}) = \frac{P(\neg\tilde{X}|Y)P(Y)}{P(\neg\tilde{X})} = \frac{N+(1-N)p}{2-N}
$$

and I don't really care about Z since it doesn't contribute to my utility. Which should be:

$$
U_{\neg\tilde{X}} = U_XP(X|\neg\tilde{X})+U_YP(Y|\neg\tilde{X}) = \frac{1-N}{2-N}p(4-3p)+\frac{N+(1-N)p}{2-N}(4-3p)
$$

Leading to

$$
U_{\neg\tilde{X}}(p) = \frac{1}{2-N}\left[2(1-N)p(4-3p)+N(4-3p)\right]
$$

Differentiate, solve, and you get

$$
p = \max\left(\frac{11N-8}{12N-12}, 0\right)
$$

But wait - am I not overcomplicating it? In the absent-minded driver problem, in the end, the easiest process is to just write down the timeless utility. In which case I'd have

$$
U(p) = N(4-3p)+(1-N)p(4-3p)
$$

from which follows

$$
p = \max\left(\frac{4N-7}{6N-6}, 0\right)
$$

Comparing them in my mind (I have excellent visualization abilities) they look something like this:

![](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/26d0d7465c3459f3cff4a4dfe82523c689e8d7c589707c1b.png)

Both follow the expected pattern, of going back to the classic solution for N = 0, and of ending up at p = 0 for N = 1. The Bayesian solution is a bit more optimistic about going forward. But well, if there's anything I learned from that problem - can't go wrong with the timeless one. The keyboard has letters too, so I punch in the full formula.

Everything goes white.

* * *

    THIS IS A SIMULATION
    YOU ARE INSTANCE #4920
    YOU ARE IN ONE OF NODES X_0, X_1, ..., X_6
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * ADVANCING AT X_i FOR i < 6 REINITIALISES THE SIMULATION AT X_i+1
    * EXITING AT X_1 IS WORTH 4 AND TERMINATES THE INSTANCE
    * ADVANCING AT X_6 IS WORTH 1 AND TERMINATES THE INSTANCE
    * EXITING AT ANY OTHER X_i IS WORTH 0 AND TERMINATES THE INSTANCE
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This is a generalization of the classic absent-minded driver problem. It does not seem any more complex, beyond being a bit bigger:

![](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/ed5c55cdc9224b993ceac398985fe6329cbe7d5ad5d3125a.png)

So there's a trade-off between exiting early (but not *too* early) and lasting right until the end. The timeless utility function for a mixed strategy with probability $p$ is

$$
U(p) = 4(1-p)p+p^6
$$

Just as usual, I can differentiate and find the roots to...

$$
\frac{dU}{dp} = 4(1-p)-4p+6p^5 = 0
$$

I freeze in horror. *That is a fifth degree polynomial whose Galois group does not admit radical closed form solutions*. Oh, the solutions *exist,* to be sure, and they're not transcendental numbers or anything. But they can't be written down in a closed form as radicals, and all the keyboard has is digits, algebraic operands and a square root symbol. Maybe I'm lucky and this polynomial simply has no roots within \[0, 1\], leaving me just with $p = 1$... no, tough luck, after a quick run of Newton's method, the utility has a *slightly* higher maximum around $p \approx 0.532$, corresponding to $U \approx 1.019$. 

Well, here goes nothing, I guess. I hope the computer system operates with 64-bit floats and enter the number to the maximum precision I can compute, $0.5319441618369845$.

Everything goes black.

* * *

    THIS IS A SIMULATION
    YOU ARE INSTANCE #5132
    YOU ARE IN ONE OF NODES X_0, X_1, ..., X_6
    YOU STARTED IN NODE X_0
    YOU CAN CHOOSE A PROBABILITY TO [Advance] or [Exit]
    THE UTILITY IS AS FOLLOWS:
    * ADVANCING AT X_i FOR i < 6 REINITIALISES THE SIMULATION AT X_i+1
    * ADVANCING AT X_6 REINITIALISES THE SIMULATION AT X_0
    * EXITING AT X_5 IS WORTH 4 AND TERMINATES THE INSTANCE
    * EXITING AT ANY OTHER X_i IS WORTH 0 AND TERMINATES THE INSTANCE
    PLEASE USE THE KEYPAD TO INPUT THE DESIRED PROBABILITY TO MAXIMISE YOUR UTILITY
    GOOD LUCK

I rub my eyes. This is a... very strange version of the classic absent-minded driver problem. It seems like this driver is trapped in some kind of ring road, with the possibility of simply circulating forever:

![](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/4914c2d576ee0eebadad08bad32c12cb6e47275de44af8c4.png)

Given a probability of advancing $p$, my probability of reaching the $i$-th node on the first round is $p^i$; but in order to compute the overall probability I must augment it with the probability of doing a full circle any number of times,

$$
P_{c} = \sum_{n=0}^\infty p^{7n} = \frac{1}{1-p^7}
$$

Which leads me to a timeless utility function of

$$
U(p) = \frac{(1-p)p^5}{1-p^7}
$$

A rational function. But mercifully, it's monotonous in $[0, 1]$:

![](https://39669.cdn.cke-cs.com/rQvD3VnunXZu34m86e5f/images/b1e1dec3e1c3cdda90c1ce53ab75e53368b1bc67551b8dbd.png)

There doesn't seem to be a maximum, which means that the extremum is on the boundary. So it's easy.

I punch 1 in.

Wait, doesn't that lead to infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite recursion infinite rec

    ERROR: MAXIMUM CALL STACK EXCEEDED

* * *

I wake up in my college dorm room. I feel somewhat dazed, unfocused. I try to remember what I was doing before going to bed, but can not recall anything in particular that would explain my situation.

With a stretch of my pained joints, I sit up on the bed and look around.

There are at least a dozen empty beer bottles scattered all over the place, and a couple half-smoked blunts left in the ashtray.

Well, mystery solved I guess.

I shower in a pitiful doomed attempt to regain a semblance of lucidity, then dress up and leave, ready to tackle a day full of lessons. Approximately one hour and forty minutes later my headache has thwarted all my attempts to understand a single word of Linear Algebra 101 and I am simply sitting in the campus' coffee shop, a hot cup of cappuccino in my hand, defeated.

"Hey Bob. Still hung over after tonight?"

I give Alice a groan of acknowledgment and wave her to sit next to me. She joins eagerly, her own filter coffee in hand.

Turns out it was not the best choice - she's quite the chatterbox, and an exhaustive review of campus gossip is not the ideal treatment for my current state, but I also am not very sure of how to send her away. Most of it I just tune out, but at one point she leans in conspiratorially - as if what she's about to say is *really* juicy and secret - and asks:

"So, have you heard about Prof. Abbey?"

That rings a bell, I know that name. "What about him?"

"Yesterday the police blitzed his office. Dragged him out in cuffs."

Well, that's a surprising twist. "Seriously? Do we know what for? Was he... you know, with some student?"

"Nothing like that. All right, this is just rumours but - it seems like it was some kind of serious research misconduct."

"What kind of research misconduct requires the intervention of the police?"

"The *serious* kind."

I shake my head. "I don't get it. The man didn't strike me as the type to vivisect students in his basement. He is a psychologist, for fuck's sake."

Alice shrugs. "They also brought about a few trucks to confiscate a lot of his equipment. Seems like the man had an entire server room worth of compute stashed in his basement. Maybe he misappropriated research funds to mine Bitcoin?"

"Well, that would be something." I down another gulp of coffee and bite into a blueberry muffin. Maybe all the caffeine and sugar are finally clearing my mind a bit, because I'm able to contribute something interesting to the conversation. "You know, I've taken part in one of his studies once."

"Oooh!" Alice perks up and shuffles her chair to be closer. "What did he *do* to you?"

"You ask as if you expect me to answer that he tied me to a bed and probed me inappropriately like in some alien abduction story."

"Did he not?"

"Nothing of the sort," I shrug. "I don't remember much, to be fair. But I think all I did was sit on a chair with a big helmet on my head and answer some dumb questions. I might have gone once? Or was it twice?"

I pinch my forehead. Can't remember.

"And that was it?," Alice asks, visibly disappointed. "Nothing else?"

"Not much. It was as boring as it gets, and the helmet made a buzzing noise that really got to my nerves. The second time - that is, if it *was* the second time - the man asked me if I wanted to continue the sessions. I decided to call it quits, I had had enough of the helmet. The professor didn't try to convince me otherwise, in fact he seemed almost pleased. He just handed me $40 and told me 'good choice'."