Best Strategy for Dragon's Dilemma (April Fools Event)

[wobster109]

Hi everyone, I've been thinking about what's the "best" way to do the April Fools event that gives upside-down eggs. I propose that everyone flip a coin, and if it's heads then blame, and if it's tails then say nothing. Bear with me, this is a math post. But it's truly fascinating! Here we go!

 

First, we have to be specific about what we mean by "best". I think "best" means that the maximum possible upside-down eggs are handed out by the site, and everyone who plays has an equal chance of getting one. Getting eaten is a bad outcome, and I'll count getting a normal egg as a bad outcome too. What would be really perfect is, if every pairing resulted in an upside-down egg given out. Then 1/2 of players would receive one. However that's impossible, since we don't know what the other player will choose. We can see it's BAD if everyone blames, since then you everyone gets eaten. It's also bad if everyone says nothing, since everyone gets a normal egg.

 

So the "best" outcome is somewhere in the middle, where some people blame and some people say nothing. If we're lucky, then a blamer and a say nothing player get paired together, and someone gets an upside-down egg. And there will still be some pairs where two blamers or two say nothing players get paired, but that's unavoidable.

 

The question is, what's the best ratio? In other words, how many people should blame, and how many should say nothing, so that we maximize the chance of them getting paired together? Let's say that the fraction of people who blame is p. Then everyone else says nothing, meaning the fraction of people who say nothing is 1-p. For example, if there are 100 players and 25 of them blame, then p=1/4 and 1-p=3/4.

 

So we have player A and player B. There are four possibilities* here:

1. Both players blame. Probability of this happening is p*p.
2. Player A blames, player B says nothing. Probability of this happening is p*(1-p).
3. Player A says nothing, player B blames. Probability of this happening is (1-p)*p.
4. Both players say nothing. Probability of this happening is (1-p)*(1-p).

 

There are two good outcomes here (2 and 3). When we add together their probabilities, we get p*(1-p) + (1-p)*p. We want this value to be as big as possible. Simplifying gives:

p*(1-p) + (1-p)*p = (p-p^2) + (p-p^2) = 2p - 2p^2.

 

There are a couple ways to find the maximum. The algebraic way is to say that this is a quadratic equation, and the squared term is subtracted, so it opens down. That means the highest point is the vertex. The calculus way is to take the derivative and set it equal to 0. If you don't know either of those ways, you can go to wolframalpha.com and type in "maximum of 2p-2p^2", and this comes up: https://www.wolframalpha.com/input/?i=maximum+of+2p-2p^2.

 

Any way we solve it, the maximum happens at p=1/2, meaning we have the best odds when 1/2 the players blame and 1/2 the players say nothing. In this case, 1/4 of pairings will have the bad outcome of both players blaming, another 1/4 of pairings will have the bad outcome of both players saying nothing, and the remaining 1/2 of the pairings will have the good outcome where the site gives out an upside-down egg.

 

However, in good pairings, only one player gets to keep the special egg. This raises the question of how we decide who gets to keep it. And we each have to decide without knowing anything about the other person, such as what choice they picked or whether they already have an upside-down mint, etc. So, if we decide with a coin flip, then everyone has the same chance. (This strategy only works if everyone listens to their coin flip. For example, maybe your coin flip says to say nothing, and you decide to blame anyway. But if everyone did that, then everyone would blame, and no special eggs would get handed out.)

 

Summary: Flip a coin to decide whether to blame or say nothing. Then 1/2 of pairings will result in the site giving out an upside-down egg, meaning each player has a 1/4 chance to receive an upside-down egg. This is the best we can do, and there is no way to make your odds of getting an upside-down egg better than 1/4.

 

This strategy was inspired by articles about the Prisoner's Dilemma in the book Metamagical Themas by Douglas Hofstadter.

*The probabilities are not exactly accurate. If there are 100 players, and 1/4 of them blame and 3/4 say nothing, and you are a blamer, then your chance of meeting a say nothing player is 75/99, not 3/4. This is because you can't be matched with yourself. However, when there are many many players, my calculated probabilities are very very close to right, just like 75/99 is very very close to 3/4.

[olympe]

While that may be the best thing for the site, it certainly isn't in any player's best interest to say nothing because it means that *they* won't get a special egg. Prisoner's dilemma.

[dustpuppy]

Do we actually get matched with another real player? I thought I read somewhere that the DC event is basically just a simulation giving you a random outcome (based on what outcomes are possible for your decision), but I can't find the post. If we really do get matched with another player, I'll give the coin flip a go next time (or just say nothing since I already have an upside-down Mint).

[Sazandora]

I don't think the game really matches you up with a random player. But in any case, I already have two upside-down Mints so it doesn't matter whether I "win" or "lose" any more Dragon's Dilemmas. After all, these ARE just regular ol' Mints as the upside-down sprite isn't visible in the lineage view. I don't think they're any more "special" than non-upside down Mints.

[Aqub]

Wow, I didn't realize we weren't with a real person... I was going to stay quiet to try to let someone else get an upside down mint next time, but I guess I shouldn't if we're just paired with fake people?  

[HeatherMarie]

I'm pretty sure it'd be impossible for this specific event to pair users with actual other users, because as soon as you do the game and pick what you want to choose, you get the result. There is no waiting around to find out what another person has chosen, you get the result (and possibly get the egg) right away. Despite DC being relatively popular (? I don't know what counts as popular nowadays!), I very much doubt it's possible for enough users to be on to be paired together at any moment of the day. 

[wobster109]

@olympe If it were another real player, and everyone blamed according to Prisoner’s Dilemma, then everyone’s chance would be 0%. This is worse than 25%. It’s true that if you flip a coin and it comes up tails, your chance becomes 0. Best would be if everyone agrees to push a button that flips the coin and locks in the answer for you, so when you push the button your chance is 25%, and no one can “back out” after the coin flip.

 

@dustpuppy @Sazandora @Aqub @HeatherMarie

Oh wow, I had no idea it wasn’t a real player! That changes everything! In that case the “best” strategy is definitely blame the other player.

 

 

[olympe]

We don't know for sure we aren't matched with a real player. Not at all. Because it could work like this:

Step 1: DC generates 1 random answer.

Step 2: Player 1 logs in and gets to play. Their answer gets matched with the random answer. The game calculates the result instantaneously, as both "player answers" (actual player answer plus the random thing) are already available.

Step 3: Replace the random answer with the answer of Player 1.

Step 4: Player 2 logs in and gets to play. Their answer gets matched with the answer Player 1 gave. The game calculates the result instantaneously, as both player answers are already available.

Step 5: Replace the answer of Player 1 with the answer of Player 2... 

 

I think you get the idea.  

[Shadowdrake]

It seems very likely that it is randomly generated though: if it really were based on what people picked, a lot more people should have picked "lie" and thus there should have been a lot of "both lie" responses after the first year. (Also there was that matter of a person multiscrolling with about 200 scrolls to get 100 mints, which rather suggests 50-50 odds.)

[wobster109]

19 hours ago, Shadowdrake said:

(Also there was that matter of a person multiscrolling with about 200 scrolls to get 100 mints, which rather suggests 50-50 odds.)

 

What! Did this actually happen? Even though multi-scrolling is against the rules, I'm kind of thrilled that there's actual data on this.  

[Shadowdrake]

44 minutes ago, wobster109 said:

 

What! Did this actually happen? Even though multi-scrolling is against the rules, I'm kind of thrilled that there's actual data on this.  

Yep. It was pretty obvious since they let their mints tick down until they were conspicuously low in time, then put them all in valley Sherwood at once. I think someone counted or estimated them and coming up with about 100. Of course the multi-er got mad once they were caught and posted on the news forums, claiming they had over 200 scrolls. Granted, this is somewhat anecdotal so I think gathering data from multiple players will be more accurate.