Can you outsmart everyone else?

This discussion is now locked. No further entries will be considered. \color{#D61F06} { \text{This discussion is now locked. No further entries will be considered.}}

Thanks everyone who participated. For those of you who were waiting to snipe at the last possible moment, tough luck.

You can discuss strategies and results in this discussion.


The goal is to post a (real) number from 0 to 100 (inclusive) that is closest to 0.9 times the average of every number posted.

Clearly, the later you post the more information you will get. However, the catch is that this discussion will be closed at some time within the next 24-48 hours, and no further entries will be accepted.

Participation rules:
1) Your entry should be a reply to my comment "Post your entries here".
2) Only 1 entry per person.
3) You can post whenever this note is live. It will be locked within the next 24-48 hours.
4) You may not edit your entry.
5) Have fun!

Feel free to make comments discussing this topic, or advocating people to post a certain number.


Results

Congrats to Math Math and Ahaan Rungta for being the closest!

Number of participants: 25
Average: 18.944
0.9 * ave : 17.0499

Raw results: Values have been rounded down to 3 decimal places

NameEntryAbsolute Difference
Math Man8.5398.510
Sharky Kesa22.7225.672
Daniel Liu9.8697.180
Mietantei Conan107.049
Yannick Yao31.00613.956
Anthony Susevski13.4133.636
Tan Li Xuan21.4154.365
Ahaan Rungta161.049
Aneesh Kundu19.9112.861
Victor Song38.66021.610
Andy Hayes14.7712.278
Math Math16.5810.468
Raj Magesh14.1062.943
Victor Martin29.112.050
Enrique Naranjo Bejarano4224.950
Samuraiwarm Tsunayoshi42.37725.327
Zhijie Goh2810.950
Bogdan Simeonov2.68514.364
Tan Wee Kean20.6783.628
Daniel Ploch017.049
Pranshu Gaba15.1541.895
Justin Wong116.049
John Muradeli0.61616.433
Chung Kevin107.049
Ajala Singh4527.950

Entries which did not follow the participation rules have been ignored.

#ComputerScience #GameTheory

Note by Calvin Lin
6 years, 9 months ago

1 vote

  Easy Math Editor

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Comments

Post your entries here

Calvin Lin Staff - 6 years, 9 months ago

piepi*e

math man - 6 years, 9 months ago

pi is phi

math man - 6 years, 9 months ago

You mean ϕ×e \phi \times e ?

Hmm... I work that out to be about 4.3982723895 4.3982723895

Yuxuan Seah - 6 years, 9 months ago

@Yuxuan Seah pi = 3.141.......

math man - 6 years, 9 months ago

@Math Man so it is 8.53...

math man - 6 years, 9 months ago

π2\pi^2

Daniel Liu - 6 years, 9 months ago

25eϕ\dfrac {25}{e - \phi} just because I can.

Sharky Kesa - 6 years, 9 months ago

45

Ajala Singh - 6 years, 9 months ago

10

Chung Kevin - 6 years, 9 months ago

@John Muradeli said π216 \frac{ \pi^2}{ 16} .

Calvin Lin Staff - 6 years, 9 months ago

1

Justin Wong - 6 years, 9 months ago

eee^e

Pranshu Gaba - 6 years, 9 months ago

0

Daniel Ploch - 6 years, 9 months ago

My number is rr such that rr is equal to the maximum value of xxxx..x^{x^{x^{x^{..}}}}

Rindell Mabunga - 6 years, 9 months ago

20.67824563

Tan Wee Kean - 6 years, 9 months ago

K=2.685...

Bogdan Simeonov - 6 years, 9 months ago

Whoops sorry Calvin 28.

ZhiJie Goh - 6 years, 9 months ago

24+ππe 24 + \pi^{\pi - e}

Yuxuan Seah - 6 years, 9 months ago

42.377942.3779

Samuraiwarm Tsunayoshi - 6 years, 9 months ago

The answer to life the universe and everything, and also this game, is 4242

A Former Brilliant Member - 6 years, 9 months ago

29.1

Víctor Martín - 6 years, 9 months ago

π2+ϕeϕ10=14.106...\pi^2+\phi \cdot e-\dfrac{\phi}{10} = 14.106...

Raj Magesh - 6 years, 9 months ago

1.2i(πeϕi)-1.2i(\pi\cdot e\cdot\phi \cdot i)

mathh mathh - 6 years, 9 months ago

14.771883793201

Andy Hayes - 6 years, 9 months ago

38.6603773585

Victor Song - 6 years, 9 months ago

19.91161761

Aneesh Kundu - 6 years, 9 months ago

16 \mathbf{16}

Ahaan Rungta - 6 years, 9 months ago

Cos(i)×iiCos(i) × i^{i}

Aman Sharma - 6 years, 9 months ago

21.415926535...........

Tan Li Xuan - 6 years, 9 months ago

pi^pi / e

Anthony Susevski - 6 years, 9 months ago

I would try pi^3, slightly larger than 31

Yannick Yao - 6 years, 9 months ago

28.16

Adrian Stefan - 6 years, 9 months ago

-6!

swapnil rajawat - 6 years, 9 months ago

10

mietantei conan - 6 years, 9 months ago

Interesting idea. I think the answer depends on whether the value of victory is a constant-value 'prize', or a divided-value 'lottery' (In the lottery, the more winners there are, the less money/utility each winner gets). In the constant-value 'prize' model, there exists a Nash-equilibrium at 00 for 22 or more players, so the only rational outcome is for everyone to post 00.

However, I'm not sure this applies to the lottery case. I think it does, but it's a fair bit harder!

Daniel Ploch - 6 years, 9 months ago

But doesn't that assume that everyone else is equally as "rational" as you? As seen by the other 2 entries, neither of them said 0.

Even if you said 0 right now, you would not win.

Chung Kevin - 6 years, 9 months ago

Yes, Game Theory and Nash-equilibriums rely on the assumption that all players behave rationally. If we don't have any way to reason about our "opponents'" behavior, we have nothing, and so the best we can do is make a wild guess.

Daniel Ploch - 6 years, 9 months ago

@Daniel Ploch You can do some algebra whatever you want, maybe the mean, median, or stuffs. And then you can do some more estimations for the future.

Samuraiwarm Tsunayoshi - 6 years, 9 months ago

@Samuraiwarm Tsunayoshi You only get one guess, and anyone who posts after you can always make a guess that's at least as close to the 910\frac{9}{10} average as yours is, if not closer, so I don't see how 'estimating the future' can help. As soon as you post on a non-zero board, you forfeit your ability to win to the next poster.

However, were one to look at the game thus far, without seeing the rules, I would be forced to conclude that the objective is "Post the most arbitrary and clever combination of constants you can think of", and certainly not "Guess as close as possible to 910\frac{9}{10} the average of all guesses", and it turns out the former is nigh impossible to reason about with the second objective in mind. Social structures have a knack for forging their own goals, regardless of what the coordinators have dictated :).

Daniel Ploch - 6 years, 9 months ago

I am somewhat highly amused that you are the only person who voted 0. What do you think is your probability of winning with this strategy?

I think that there is a reasonable strategy to help guide your thinking. You may not guarantee that you will win, but you could increase the chances of winning. For example, anyone who voted > 60 is unlikely to win.

Calvin Lin Staff - 6 years, 9 months ago

I feel like this is tough... and does the closest person get a prize?

Also, is there some way to see the numbers listed chronologically (in order of submission) after the discussion is closed? That would be interesting to look at how, if at all, the guesses changed over time.

Justin Wong - 6 years, 9 months ago

The closest person gets bragging rights. I'd see what else I can do.

Hm, I agree that having a record of entries is good. I might edit that in.

Calvin Lin Staff - 6 years, 9 months ago

Wait so do we start posting already? If so, I'm going for π216\frac{\pi^2}{16}.

John M. - 6 years, 9 months ago

Sorry, I had technical difficulties and was unable to make the comment. I'd record your entry down though.

Calvin Lin Staff - 6 years, 9 months ago

Some many pis and so many Es. I'm going for 28. I don't know why.

ZhiJie Goh - 6 years, 9 months ago

For your entry to be accepted, please post it as a reply to my comment. This allows for easy bookkeeping and consolidation of the entries. Thanks.

Calvin Lin Staff - 6 years, 9 months ago

FunFact: As far as I know, in the whole of the English Language, the only words that have 33 consecutive repeated letters are "bOOKKEEp" and its other forms. (e.g. "bookkeeping" and "bookkeeper".)

Yuxuan Seah - 6 years, 9 months ago

@Yuxuan Seah What about subbookkeeper?

Sharky Kesa - 6 years, 9 months ago

45

Muh. Amin Widyatama - 6 years, 9 months ago

Please read the rules. For your entry to be accepted, please post it as a reply to my comment.
This allows for easy bookkeeping and consolidation of the entries. Thanks.

Calvin Lin Staff - 6 years, 9 months ago

1919

milind prabhu - 6 years, 9 months ago

Please read the rules. For your entry to be accepted, please post it as a reply to my comment.
This allows for easy bookkeeping and consolidation of the entries. Thanks.

Calvin Lin Staff - 6 years, 9 months ago

00

Yuxuan Seah - 6 years, 9 months ago

Please read the rules. For your entry to be accepted, please post it as a reply to my comment.
This allows for easy bookkeeping and consolidation of the entries. Thanks.

Calvin Lin Staff - 6 years, 9 months ago
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