What is the number you should pick for you to have the highest probability of winning this game?

What is the number you should pick for you to have the highest probability of winning this game?

depend on the intelligence of the opponents, if they are animals in human form 40, if they dumb 30, if they intelligent 0.

Source?

51

if they're all logicians, 0
otherwise 40

These. The Nash equilibrium is everyone picking zero.

then if i'd go against 4 intelligent ppl and pick 1, i'd win.
you stupid.

since 4/5s of 100*5 = 80. Nobody will pick a number greater than 80.

If everyone picks 80 then the new average is 4/5s of 80 = 64. So no one will pick a number greater than 64 etc....

therefore, everyone picks 1 or 0.

if everyone picks 0, then it's a tie
if everyone picks 1 then it's a tie

if 4 people pick 1 and one person picks 0, then it's a 4 way tie.

if 3 people pick 1 and 2 people pick 0, it's a 2 way tie.

so your best bet is to pick 0

If everyone else picked zero and you picked one, 4/5 of the average would be 0.16. Is 0.16 closer to 0 or 1?

Alright, you're playing against these opponents. You can only infer their intelligence from their looks and occupation. What do you pick?

if everyone is just playing randomly, the average value will be around 50 and the answer you want to be close to is 40. So everyone is going to play such that the average is 40, thus the new average becomes 32. After each successive round, the average will be ~50(.8)^n with n being the number of rounds. This average value approaches 0, So every player will play 0 as their number. Even if one person deviates from this number, the average will still be closer to 0. In fact, 4 people or more would have to deviate from 0 in order for 0 not to win.

mb im not enuff intelligent sry senpai

4 brainlets and 1 mathematician, I'd go with 10.

It can only be calculated if you assume that the other players are picking at random.

There is such a thing as being too logical. Looks like you've lost the first round :^)

>the handsome looking character wins

wow what a surprise

After this round, the players will realize they need to go lower to win. After a few rounds they will all choose 0.

You're right. But before that happens, the 2nd person decides to pick 100, sacrificing himself in an attempt to break the Nash equilibrium.
So the next question is: What's a good number to pick next?

Forgot pic

Obviously if there was collusion this changes the entire problem.

Bullshit. The writer thinks this game is cleverer than it is.

Post all five rounds pls

>nobody was playing to win

NORMIES GET OUT
REEEEEEEEEEEE

There was no collusion before this round though, the girl was just a dummy. The collusion only happens after this round precisely because of this round.

The answer is 42

29

Whoops, forgot pic again.

Nobody would pick the highest number unless everyone agreed to. The choice of 100 is unexplainable.

But what is the question?

I think the two bitches were working together. Take this shit back to /a/ OP

You're only right if they want to win just this round, but if the player's strategy is to draw people away from all choosing 0, then round 5's choice makes a little bit of sense, and its purpose was effective, was it not?

22

I'll just post the results. Yep, Nash equilibrium definitely happened.

I'll stop now, since the next few rounds are probably too hard to predict without context.

Party poopers.

The real question is:
If everyone else picks a random number then which number should you pick to have the highest chances of winning

Obviously 40.

>If everyone else picks a random number then which number should you pick to have the highest chances of winning
800/21

You have to take into account that your own number will be part of the average.

Well with everyone else picking randomly, they would average a pick of 50. So if your pick is X, then (X+4*50)/5 is the average. You want your number X to be equal to .8*average. So we have the equation
X=.8*(X+4*50)/5
X=(4X+16*50)/25
25X=4X+16*50
21X=16*50
X=38

And now calculate the actual chance of winning that you have with that pick.

Posting the next 4 rounds for the heck of it. Nash equilibrium happens again.

>this triggers the mathematician

Obviously you're a dumbass

Too hard I give up

Areally you going to tell us the source or not?

Alright senpai, source is Alice in Borderland.

32 will maximize your winnings.

There are a total of 101^4 different games possible where I pick 38.
The average will be (38+w+x+y+z)/5
If:
|38-4(38+w+x+y+z)/25|

I did. The number will be 40 for almost any amount of players.

I guessed 22 so I win.

10/10 Pretty good

After brute forcing 100.000 trials, I get a win percentage of 34.482% for always choosing 38 vs 4 random values. Breddy gud.

50

Ah, I forgot the 0.8 factor in the win criterion. Rerunning gives a win percentage of 51.123%.

If the other four players chose 50 on average and you chose 40, the target value is 38.4. If you chose 200/81, then the target value is exactly 200/81.

Here's some experimental evidence to convince you. 200k trials basis.

Most people are going to pick 40 so the best is to pick 32 or 27.

>not picking 0

>The number will be 40 for almost any amount of players.
>If the other four players
...

Jesus fucking Christ, can't any of you read?

What would happen if everyone picks the same number? Does everyone lose, or win?

yes

But OP was only asking about the case where there are five players.

No he wasn't. He didn't say anything about how many players there were. And if you assume it's the same situation as the thread then there are six players, these five and you.