Show HN: Learn When to Quit

https://github.com/victorqribeiro/googol

by atum47 • 7 years ago

147 44 7 years ago

github.com

shaftoe• 7 years ago

This is the Secretary Problem.

https://en.wikipedia.org/wiki/Secretary_problem

lqdc13• 7 years ago

Wouldn't this be different since we know the maximum number and the distribution?

The solution wouldn't be to still skip 1/e if you got something like googol - 1 on the first try.

larnmar• 7 years ago

Sometimes when I’m bored and out walking, I play this game with cars. I pretend I’m allowed to keep any one of the next ten cars that go past, but I need to accept or reject it the moment I see it.

If the second car is a Volvo do I pick it, or hold out waiting for a Mercedes? If I pick that Mercedes, am I going to regret it in thirty seconds when a Rolls rolls past? Am I going to get stuck with an old Hyundai due to my inability to settle?

cjauvin• 7 years ago

I have written a small analysis of that same problem using simple Python code, a couple of years ago:

https://cjauvin.blogspot.com/2012/12/find-true-love-on-datin...

leetrout• 7 years ago

Nice work

trisiak• 7 years ago

The distribution of numbers is not uniform across the 0..googol here. The implementation first chooses the length of the number, then fills in the digits. This makes it biased towards smaller numbers.

Does it still have the same optimal strategy as for uniform generation?

atum478• 7 years ago

"The implementation first chooses the length of the number, then fills in the digits. This makes it biased towards smaller numbers."

Why?

bentona3• 7 years ago

Semi-relevant: https://en.wikipedia.org/wiki/Benford%27s_law

fourthark2• 7 years ago

Consider a two digit number. With this algorithm, half of the numbers generated will be less than 10, and half will be 11-99.

Likewise with the full 100 digits this algorithm has a 1% chance of generating a number less than 10, while the correct proportion would be 1e-100.

The algorithm also does not generate numbers with zero in them (presumably to avoid leading zeroes, but this will skew results somewhat).

I think the correct algorithm would be to generate 100 digits and then remove the leading zeroes.

teraflop• 7 years ago

Yep. The optimal strategy depends only on the relative ordering of elements, not on their magnitudes or any other property. So the strategy is the same for any continuous distribution.

(With a discrete distribution, the possibility of ties slightly affects things. But in this particular game, ties seem to be very improbable, so they can be ignored.)

repsilat2• 7 years ago

It isn't optimal for known distributions. If you're sampling from a normal distribution with known parameters, and the first of 100 samples is 5 standard deviations above the mean, you take it.

GistNoesis• 7 years ago

The distribution chosen here is ~ uniform for the number of digits. (https://github.com/victorqribeiro/googol/blob/master/js/main... )

This is not the same hypothesis as the problem stated in the video. The video talks about the "secretary problem". In this problem you know nothing about the distribution. You just know that the number are ordered.

Here because you know the distribution you can do better by using a strategy which choose to stop depending on the value of the number you have so far. For example if at the 15th pick you get a number with 99 digit, you can safely assume that it is better to stop now (every time you pick you have only a ~1% chance of finding a 99-digit number needed to do better so keeping it means you win (99/100)^(50-16) = 70%, whereas the non-exploitable strategy (i.e. optimal strategy for the secretary problem) 50/e=18 will tell you to keep going and you would lose 70% of the times).

For the secretary problem the theoretical way to sample the number is to set it up as a game where the first player plays the game and the second pick the numbers adversely so that the first doesn't win. A good strategy for the second player is to pick a distribution from a big set of varied distributions, and sample the number from this distribution.

curo• 7 years ago

How does this change when you're heavily punished for ending up with a much lower number (e.g., in finding a partner, biz deal, etc)?

Among a set of losing plays, I presume it's 50/50 on whether you turn over every card (and end up with the last one). So a third of the time you end up in the optimal case, another third you end up doing pretty well, and another third you're subject to absolutely random chance. Guess this is an argument for satisficing vs optimizing in most real world applications.

anonymousiam• 7 years ago

The conditions of this "game" are that you must decide then and there which option to choose. In the game, you do not have the luxury of viewing all of the data before you make your choice. Real life can be different (as was mentioned in the video re: Kepler's wife).

repsilat2• 7 years ago

You're right, this algorithm optimizes for "1 point for picking the best option, 0 for anything else", not knowing the distribution.

If you do know the distribution, and have some objective function, and samples are "free", you get a different model that you can solve inductively backwards from the case of "the last sample".

A different model is that your objective function is still on order, but it gives some points for "closer to best". Maybe you score minus n for picking the nth best sample. In that case the model need not have a known distribution (It's different if you do or don't.) And the strategy us different: if the 99th sample out of 100 is the second best you've seen so far, you should take it. (Which can't possibly be optimal in the original model.)

iakh• 7 years ago

Consider using a monospace font. It is difficult to tell how large one number is relative to another when each digit has a different width

atum478• 7 years ago

Done! Thanks for the tip, by the way.

cyode• 7 years ago

Having a button to toggle on/off rounded e-notation (e.g. 1.2365479e+45) could make comparison easier as well.

Nice job with this, I learned something new today :)

ronilan• 7 years ago

Madden: "Now with no timeouts I think that the Patriots, with this field position they have to just run the clock out, you have to play for overtime now. I don't think you want to force anything here, you don't want to do anything stupid because you have no timeouts and you are backed up."

Summerall: Brady is in the shotgun and he's gonna throw it ..."

https://www.youtube.com/watch?v=GC4qgrUgF9I&feature=youtu.be...

Always trust the stats ;)

chrisweekly• 7 years ago

^ Favorited

ajushi• 7 years ago

I don't get it sorry. How does this relate with the OP?

smabie• 7 years ago

I don’t understand, could some one explain? What am I trying to do in this game? And how do I win?

atum478• 7 years ago

Did you watch the video? It does a very good job explaining the theory tested on the game.

rkagerer• 7 years ago

Comma's would be friendlier than decimal groupings for North American users; also right-aligning would be nice.

atum478• 7 years ago

right-aligning suggestion taken. =)

OJFord2• 7 years ago

The `.` separation confused me briefly too, being from another country that uses `,`.

But you know what, if I were the author I would have used `,` without bothering to localise to `.` (for some toy demo thing like this anyway) so I'm OK with OP doing the same to me.

flurdy• 7 years ago

For some reason, I thought this was about Vim...

atum478• 7 years ago

:q!

whack• 7 years ago

There is a huge problem with the "optimal stopping strategy" presented in the YouTube video: "sample the first 37% of the options, and then choose the first subsequent option that is better than the best option found in the sample."

By definition, there is a 37% chance of the best option being in your initial sample. You would then futilely browse through all the other options, without ever picking anything.

Ie, if you applied this strategy to "finding the love of your life", as the video suggests, there is a 37% chance of you dying completely alone. That doesn't seem like a very optimal strategy to me.

spencey• 7 years ago

I haven't seen the video but if they are talking about the secretary problem the caveat is that you can't go back to the previous options. You can do that when trying to a find a partner... sometimes.

codetrotter• 7 years ago

Ie, if you applied this strategy to "finding the love of your life", as the video suggests, there is a 37% chance of you dying completely alone. That doesn't seem like a very optimal strategy to me.

I don't think that's correct. You'd set a period of time such as 1 year or 3 years or 5 years or something during which time you will go on n amount of dates total, out of which you reject the first 37%. Then you use that as a basis for the remaining time of that period of time. If you haven't found anyone better by the last date of that time, and you have chemistry with them, you choose that last person to be the love of your life.

And if you don't have chemistry with the last person or you end up breaking up with that person, then you go on dates until you find the next person you have chemistry with and you choose them to be the love of your life.

So it's not like dying completely alone is a probable outcome that would result from the optimal stopping strategy itself.

sneak• 7 years ago

All of these huge numbers are basically un-comparable ergonomically to me. Why not make the distribution 1-10k instead?

hmexx• 7 years ago

Agreed.

Also, there is really no reason to have multiple boxes you have to click. You could just have a single "next" button along with a counter to show how many choices you have left before you run out.

OJFord2• 7 years ago

Yeah that confused me too.

'Next' button along with '# samples left' display would be much clearer.

Also, if the largest drawn so far were highlighted, manual play would be easier.

h1fra• 7 years ago

Very interesting, especially the video :pray:

(actual game does not look at all like the screenshot on mac/chrome)

glaive123• 7 years ago

I just did "Computer" 50 times and tallied the results.

Win = 32% Lose = 68%

But here's the thing, the goal should not be to win, but to get ONE OF the highest numbers with reliability.

Now here's where Optimal Stopping Theory is not optimal.

42% of the time you lose by getting to the end. Meaning you drew all 50 cards and ended with likely a very small card.

The goal of this algorithm should be that you almost never get to drawing the 50th card. You always stop before then and end with a high number.

Therefore, I wouldn't consider this algorithm optimal. Likely you should draw fewer cards in the beginning.

Show HN: Learn When to Quit

Related Stories