Results: Can You Outsmart Everyone Else?

Congrats to Mathh Mathh and Ahaan Rungta for being the closest!
They will be receiving a Brilliant T-shirt for outsmarting everyone else.

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

For the rules + actual game breakdown, see Can You Outsmart Everyone Else?

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

Name	Entry	Absolute Difference from W
Math Man	8.539	8.510
Sharky Kesa	22.722	5.672
Daniel Liu	9.869	7.180
Mietantei Conan	10	7.049
Yannick Yao	31.006	13.956
Anthony Susevski	13.413	3.636
Tan Li Xuan	21.415	4.365
Ahaan Rungta	16	1.049
Aneesh Kundu	19.911	2.861
Victor Song	38.660	21.610
Andy Hayes	14.771	2.278
Mathh Mathh	16.581	0.468
Raj Magesh	14.106	2.943
Victor Martin	29.1	12.050
Enrique Naranjo Bejarano	42	24.950
Samuraiwarm Tsunayoshi	42.377	25.327
Zhijie Goh	28	10.950
Bogdan Simeonov	2.685	14.364
Tan Wee Kean	20.678	3.628
Daniel Ploch	0	17.049
Pranshu Gaba	15.154	1.895
Justin Wong	1	16.049
John Muradeli	0.616	16.433
Chung Kevin	10	7.049
Ajala Singh	45	27.950

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

Here is a follow-up discussion:

1) No one voted above 45. Why?
2) Few people voted above 40. Does this make sense? Why, or why not?
3) Why didn't the rational strategy of "vote 0" win? It turned out to be the 5th worst performer.
4) If this game occurred again, how would your strategy change?
5) Is there an "ideal number" to submit?
6) What would be your best strategy in approaching this game?

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Math

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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`italics` or `_italics_`	italics
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I am happy I wasn't horribly off! Can we do another one of these? It was awesome! :D

Sharky Kesa - 6 years, 9 months ago

My answers:

1) In my opinion, when the first few numbers were put in, most people saw the scale of these numbers and went for smaller numbers.

2) Same as first.

3) I think that everyone wanted their own answer, unique and different. We had plenty of quirky as well as standard numbers. So really, we all overcame conformity.

4) I would have picked a slightly smaller number but who knows how if this game would be played if played again. A strategy everyone could have used, however, is to keep al the numbers roughly equivalent.

5) In my opinion, there really is no 'ideal number'. People picked completely random numbers which all happened to be mainly small.

6) Equivalence means that the average is closer to your number. However, Randomness means that you will get an average closer to you anyway. So really, there is no best strategy. They all work fine if you work in the same scale.

Your opinions? :D

After playing this game, I Googled it and stumbled upon this interesting Wikipedia article on the Keynesian Beauty Contest, in which, similar to this game, A attempts to reason what B would believe C would think, ad infinitum (ouch!). Here's the link.

Why didn't "Vote 0!" win? Possibly because you phrased the question as "Can you outsmart everyone else?", so the contest had two aims: to get the number closest to 0.9 times the average, and to make sure no one else got any closer than you, making 0 an irrational choice. If the aim had been simply to get the number closest to 0.9 times the average, sans any competition, the rational choice would have been zero, in which case everyone would have won and you would have run out of T-shirts!

Also, since we were able to see the other participants' entries, we could guesstimate the value we would need to enter to win. I wonder if the results would have been more different if no one knew anyone else's entry (i.e. assuming everyone else was perfectly rational). Then this becomes analogous to the Keynesian beauty contest.

What if, on the other hand, you gave us, say, 50 random nonzero values and told us that these would be included in calculating the average, in addition to our entries? What would the optimal strategy be then?

After a lot of random Wikipedia surfing, I chanced upon the Monte Carlo method, and began wondering if it would make sense to apply it here. After all, if we repeatedly use large, random sets of data, eventually a small range of numbers would emerge as the clear winners, making it optimal to select these. But then again, if everyone starts doing this, the pattern is upset once more... We can't attempt to hack the system without messing up the system.

Raj Magesh - 6 years, 9 months ago

The "official" name of this game is P-beauty, where "P" stands for the proportion of the average. In this case, we played a 0.9-Beauty game.

Typically, the game involved blind bidding by everyone, so I had to tweak it slightly to suit our current system. I think that the random time cutoff added an interesting element to this game, where withholding your entry gave you more information, but you could potentially lose on the ability to submit an entry.

In almost all simulations run by game theorists, they were unable to find a scenario where "Many people bid 0-1 (close to 0)", and there were few cases where a 0 to 1 bid actually won. There are various explanations for this, and it's worthwhile to find a "rational" explanation why a bid of 10 could make sense.

Possible explanation: You may only assume that you are rational. You do not know to what degree someone else is rational.

Possible explanation: Some people are just out to screw everyone else.

Calvin Lin Staff - 6 years, 9 months ago

That's much more a game of luck than of math! There could be some game changer person who doesn't want a Tshirt (already has one) could post an extreme value! Then that could change the winner from one person to the other! You need luck if you want a Tshirt, but moreover, you need luck to see that something like this has started ! For example me, I saw this discussion of results before I saw that original post i.e., after it has been closed ! $:($

Aditya Raut - 6 years, 9 months ago

Well, you need luck to win the lottery, but if you do win, it's not like you have to divide it evenly between the people who were a number off.

Daniel Liu - 6 years, 9 months ago

$\Huge{\color{#D61F06}{\textbf{LOL}}}$

It is partially luck, but remember, this is a game after all and it's for fun. You haven't lost anything so there's no need to sweat about it. I bet it would be less enjoyable if it was less luck-based. Also, such a game poses some nice questions, which you can't get if you make it extremely simple and 0% luck.

Ahaan Rungta - 6 years, 9 months ago

Truly said, I agree ! You're right !

There were no stakes involved initially, other than "bragging rights" and the fun of participation. See my reply to Justin's comment.

Check back often on Brilliant! You get exciting problems posted by your friends, and who knows what else you may see :)

@mathh mathh , @Ahaan Rungta Can you email me your mailing address + T-shirt sizes? Thanks!

Wow. Just wow.

I'm horrible at math, but this generally is my opinion.

1) And 2) The first few posts were around pi and e, thus nobody wanted to post anything much higher than that (since the aim was to get 0.9x the mean)

3) People are selfish. They don't want everyone to win, they want the whole thing to themselves

4) And 6) Hmm. I don't know, check the trend, if it happens to be around 10, play about 10 too, as not many people would want tout something much higher than the average

5) Hmm. Not really. At least I don't think so.

ZhiJie Goh - 6 years, 9 months ago

Wait what... Oh it autocorrected the headings Should read: First line for 1 and 2 Second line for 3 Third line for 4 and 6 Fourth line for 5

When people type 1., 2., 3., we automatically assume that they want it to be a sequential list and rename it as such. I've edited your response to 1), 2), 3), to fool our system.

ooh darn I thought it was every person posts 0.9 of what you post or something WHOA I would NOT have posted $\frac{\pi^2}{16}$ ! I would've posted 25. Oh well.

John M. - 6 years, 9 months ago

:( Aww @Calvin Lin you overlooked my entry... (I thought it followed the rules?) Anyway it was terribly, terribly off by about $8.574$ , so never mind :D

Yuxuan Seah - 6 years, 9 months ago

Sorry, I decided to ignore your entry as it violated the rule of "you may not edit your entry".

1) 45 was the first submission. In an attempt to be lower than the average at first sight (which started at 45) each consecutive person voted lower than the highest vote, creating a cascade of numbers securing the average around 15-20ish. I guess people weren't brave enough to vote higher because they feared there wouldn't be enough support in the higher tier - maybe that strategy could be easily foiled. Maybe early voters didn't predict future averages and followed the popular strategy of immediate averaging (at least I think that's what happened).

Justin Wong - 6 years, 9 months ago

Hmm... I saw the first few posts and thought people would post things related to pi and e. So I thought the average would be around 16. Then, I remember seeing someone posting a number ~40 and I thought it would increase the average a lot.

I guess the best strategy would be to get as much information as possible? Perhaps if we knew when @Calvin Lin is usually online, we could then guess when the discussion would be locked XD. I went to this website before entering my results. The average then was around ~14.

Tan Wee Kean - 6 years, 9 months ago

Assuming that everyone is rational, it would make sense that they would take the average of everyone else's answers multiplied by 0.9 as their own answer, to have a higher chance of winning. However, when I did this simulation on excel ,(assuming that there were 25 people, and the first answer was 45),I found that the average was 33.85101, and when this was multiplied by 0.9, the result was 30.46591. This is way off from the results of the experiment , which shows that not everyone is rational.

Alita Toh - 6 years, 5 months ago

I didn't participated in the contest but why "For those of you who were waiting to snipe at the last possible moment, tough luck."?

Arulx Z - 5 years, 1 month ago

Are these the only way to get Brilliant t-shirts?

Julian Liu - 6 years, 9 months ago

As for now, winning this competition is the only existing way to win Brilliant t-shirts; and now that it's done, there are no existing ways to win Brilliant t-shirts. I guess you'll have to wait for the next competition.

@mathh mathh @Ahaan Rungta Congrats! But Ahaan, don't you already own a Brilliant.org t-shirt? I would expect you have already.

Oh, I just forgot I had claimed a brilliant t-shirt around 8-9 months ago, but I haven't gotten it yet, ohh well.

Beakal Tiliksew - 6 years, 9 months ago

LOL, thanks. And yes, I owned a Brilliant shirt but I'm one size larger now. I was a baby when I got the previous one. =P

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$