Hacker News Board Hacker News, browsed like a classic message board
Live data from Hacker News

Choose the smallest number not chosen yet

amolas.dev

Post reply on HN
Print view View original

81–88 of 88 posts

Re: Choose the smallest number not chosen yet

#81

Post by gus_massa »

tromp wrote: post #9
gus_massa wrote: post #4

Earlier quoted context omitted.

There is a nice strategy that is playing always 1, and claiming it. Everyone else should not play 1 because otherwise they would not win. But if no one else plays 1, you win. So they must take turns to lose and make you lose. It's a good strategy to not meet them again ever :)

The other players then always pick 1 and 2, making the latter the winner, and you always the loser. But in this case you lose no matter what strategy you use; you only get to decide which of the other players wins. That's why the article analyses the case where the other players use the same strategy.

It's more fun with more players. They can discuss who is going to lose the round because of choosing 1 to block you. Someone can refuse to collude, so another must take the turn. Someone can decide to become a vigilante and choose always 1 to block you. Humans are very fun!

Re: Choose the smallest number not chosen yet

#83

Post by wikfwikf »

planede wrote: post #50
wikfwikf wrote: post #47

Earlier quoted context omitted.

This is obviously wrong. If you choose one million billion trillion every time you will not win 28% of the time.

You will, your opponents will choose matching numbers 28% of the time if they use the Nash-equilibrium mixed strategy.

You are right.

Choosing a bigger number makes it less likely that you will win by being smaller than the other two, but more likely that you will win by the other two knocking each other out, since there are now more numbers available for them to do so with.

Re: Choose the smallest number not chosen yet

#84

Post by planede »

wikfwikf wrote: post #83
planede wrote: post #50

Earlier quoted context omitted.

You will, your opponents will choose matching numbers 28% of the time if they use the Nash-equilibrium mixed strategy.

You are right. Choosing a bigger number makes it less likely that you will win by being smaller than the other two, but more likely that you will win by the other two knocking each other out, since there are now more numbers available for them to do so with.

I think most importantly your opponents become less likely to knock your number out. The goal is your number to win, not your opponents to lose.

Re: Choose the smallest number not chosen yet

#85

Post by theptip »

HWR_14 wrote: post #77
theptip wrote: post #67

Earlier quoted context omitted.

> This is only true if your goal is to have a score higher than your opponent That is what “dominant” in game theory means, which was the specific claim that’s being discussed.

No you are incorrect in what a "dominant" strategy means. It has nothing to do with dominating your opponent.

A dominant strategy in game theory means it's your best strategy, regardless of what strategy your opponents pick.

> https://en.wikipedia.org/wiki/Strategic_dominance

> In game theory, strategic dominance (commonly called simply dominance) occurs when one strategy is better than another strategy for one player, no matter how that player's opponents may play.

That is what is being described above.

Re: Choose the smallest number not chosen yet

#86

Post by alexmolas »

Sohcahtoa82 wrote: post #64
alexmolas wrote: post #15

Earlier quoted context omitted.

hi! some people told me that they can access the site, but I've never knew why, and I'm not an expert on websites. Thanks for spotting it! How can I solve the problem? What does it mean to put a scheme in a cname?

While we're on the topic of the site... Why do you have links styled to look exactly like the other plain text? At the bottom of the article, it says "This post has been heavily inspired by this question in SO ↩" and I could not for the life of me find the link to said question until I discovered that "question" was a hidden link.

That's weird, because on my computer it looks well. Some people already complained about that, but I don't know why it can happen. Do you have any suggestion?

Re: Choose the smallest number not chosen yet

#87

Post by xnorswap »

rawling wrote: post #75
riversflow wrote: post #71

Earlier quoted context omitted.

> I fail to understand how this can be considered an optimal strategy. Aye, I mean it doesn't even pass the sniff test to me. If all actors are a) informed of the number of participants and b) are trying to win in earnest, I can not understand why a rational actor would ever pick a number larger than the number of participants. It seems an obviously bad strategy that's an artifact of infinite calculus. Really though,…

> I can not understand why a rational actor would ever pick a number larger than the number of participants. If there are two of you and you're both always picking 1 or 2, you're going to both lose half the time when you collide. You'll win 1/4 of the time. If you instead pick from 1-3 you're only going to collide 1/3 of the time, and you'll win 1/3 of the time. 1-4, 1/4 collision, 3/8 win rate. (I think I've got the…

If there's two of you then you are even better off just always picking 1.

You either win, or tie, depending on your opponent's strategy. Your opponent can never win.

That seems extremely optimal.

Re: Choose the smallest number not chosen yet

#88

Post by seanw265 »

credulousperson wrote: post #76
seanw265 wrote: post #66

Earlier quoted context omitted.

Diving into this further, I've found another situation that appears to interfere even more. Our baseline (once again): Agent 1 (nash distribution): .296 Agent 2 (nash distribution): .296 Agent 3 (nash distribution): .296 Another case: Agent 1 (always chooses 1): .489 Agent 2 (even distribution -- equal chance of any number 1-10 being chosen): .411 Agent 3 (nash distribution): .054 In this situation, the nash strategy…

Indeed the Nash equilibrium is not always what we would like to call "optimal", especially in games with more than two players. As you notice, it is possible the Nash equilibrium strategy will be crushed if more than one agent chooses a different strategy (i.e. a situation where players are deviating from a strategy in a non-unilateral fashion). If Agent 1 and Agent 2 work collude beforehand they can completely crush…

This was very helpful. Thank you!
Post reply on HN