What is the number you should pick for you to have the highest

>>8662393
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

>>8662393
if they're all logicians, 0
otherwise 40

>>8662421
>>8662455
These. The Nash equilibrium is everyone picking zero.

>>8662421
then if i'd go against 4 intelligent ppl and pick 1, i'd win.
you stupid.

>>8662393
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

>>8662457
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?

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>>8662421
Alright, you're playing against these opponents. You can only infer their intelligence from their looks and occupation. What do you pick?

>>8662393
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.

>>8662482
mb im not enuff intelligent sry senpai

>>8662525
4 brainlets and 1 mathematician, I'd go with 10.

>>8662393
It can only be calculated if you assume that the other players are picking at random.

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>>8662557
There is such a thing as being too logical. Looks like you've lost the first round :^)

>>8662624
>the handsome looking character wins

wow what a surprise

>>8662624
After this round, the players will realize they need to go lower to win. After a few rounds they will all choose 0.

>>8662653
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?

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>>8662693
Forgot pic

>>8662696
Obviously if there was collusion this changes the entire problem.

>>8662624
Bullshit. The writer thinks this game is cleverer than it is.

Post all five rounds pls

>>8662624
>nobody was playing to win

NORMIES GET OUT
REEEEEEEEEEEE

>>8662696
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

>>8662696
29

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>>8662726
>>8662710
Whoops, forgot pic again.

>>8662726
Nobody would pick the highest number unless everyone agreed to. The choice of 100 is unexplainable.

>>8662727
But what is the question?

>>8662737
I think the two bitches were working together. Take this shit back to /a/ OP

>>8662737
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?

>>8662733
22

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>>8662721
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.

>>8662455
>>8662456

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

>>8662777
Obviously 40.

>>8662777
>If everyone else picks a random number then which number should you pick to have the highest chances of winning
800/21

>>8662784
You have to take into account that your own number will be part of the average.

>>8662777
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

>>8662793
And now calculate the actual chance of winning that you have with that pick.

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

>this triggers the mathematician

>>8662784
Obviously you're a dumbass

>>8662801
Too hard I give up

>>8662393
Areally you going to tell us the source or not?

>>8662810
Alright senpai, source is Alice in Borderland.

>>8662393

32 will maximize your winnings.

>>8662801
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|<|w-4(38+w+x+y+z)/25|
or:
|38-4(38+w+x+y+z)/25|<|x-4(38+w+x+y+z)/25|
or:
|38-4(38+w+x+y+z)/25|<|y-4(38+w+x+y+z)/25|
or:
|38-4(38+w+x+y+z)/25|<|z-4(38+w+x+y+z)/25|
Then I have won.
Assuming the average is below 38, we can get rid of the absolute value signs on the left side. If the average is above 38 then we'd need to multiply the entire expression by -1. But let's just assume the average will be below 38 at first. Also, let's assume the average is below w
950-152-4w-4x-4y-4z<25w-152-4w-4x-4y-4z
950<25w
w>38 So whenever w is greater than 38, (if the average is less than 38) I will win. This occurs 61 times (if the average is less than 38). Now we will assume the average is above w:
950-152-4w-4x-4y-4z<-25w+152+4w+4x+4y+4z
646<-17w+8x+8y+8z
646-8(x+y+z)<-17w
w<38-.47(x+y+z)
The same argument can be said for any variable, so I will win (61)^4 different times (if the average is smaller than 38). But what is the probability of the average being smaller than 38? That's a job for a computer not me.

>>8662792
I did. The number will be 40 for almost any amount of players.

>>8662803
I guessed 22 so I win.
>>8662753

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>>8662881
10/10 Pretty good

>>8662878
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

>>8662905
Ah, I forgot the 0.8 factor in the win criterion. Rerunning gives a win percentage of 51.123%.

>>8662879
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.

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>>8662879
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

>>8663054
>The number will be 40 for almost any amount of players.
>If the other four players
...

>>8663095
Jesus fucking Christ, can't any of you read? >>8662879

>>8662393
What would happen if everyone picks the same number? Does everyone lose, or win?

>>8663513
yes

>>8663503
But OP was only asking about the case where there are five players.

>>8663540
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 >>8662525 and you.