What Number Should You Write to Win the 2/3rds Average Math Riddle?

[bjr_jyd15]

On the piece of paper handed out in class please write a number
between 0 and 100. The winner is the one who comes closest to
2/3rds of the average of numbers written by the students of the
class.

Question: What number should you write?

 

[daveb]

The answer depends upon whther or not you get to assume that everyone plays intelligently. If they do, the answer is 0. For example, you assume the average is a number between 0 and 100, say x. Then you choose as a number 0.67x. But since everyone plays intelligently, you have to assume they know this, so they all pick 0.67x which is also the average, and you pick (0.67^2)x. But they also know this...continuing on you get (0.67^n)x as n goes to infinity, so the answer is 0.
If they don't pick intelligently, it depends on the number of people in class.

 

[NateTG]

What happens if people write down uncomputable numbers?

 

[bjr_jyd15]

i really don't know the answer, it was a question from a lecture about numerical methods. but 0 seems logical, as long as everyone knows the rules.

and i don't think people will write down uncomputable numbers...but I'm not quite sure what you mean by that, nateTG.

 

[Mk]

What about the smartasses in class that write pi π, Avagadro's number or $$\sqrt{2}$$

 

[DaveC426913]

What about the smartasses in class that write pi π, Avagadro's number or $$\sqrt{2}$$

Won't affect the results. True, the average will now be non-sero, but the challenge does say "closest to".

 

[NateTG]

Well, it seems like the best answer is going to be something like:

0 if this is the first slip read, 2/3 of the average of the prior otherwise

Regarding uncomputable numbers:
It's concievable to specify a number that is well-defined (i.e.), that is between 0 and 100, but is otherwise incalculable, or alternatively, one that is calculable, but not calculable (or verifiable) in the age of the universe.

I'm not sure that this is example is actually strong, but for example,
consider the last 2 digits of
$$10^{10^{100}}-9$$
LZH compressed
$$10^{100}$$
times.

 

[neutrino]

From when has Avagadro's number been less than 100?

 

[Jimmy Snyder]

consider the last 2 digits of
$$10^{10^{100}}-9$$

You mean 91?

 

[NateTG]

You mean 91?

That's before the LZH algoritm's been applied to it google times.

 

[Jimmy Snyder]

That's before the LZH algoritm's been applied to it google times.

Won't it loop early on?

 

[bjr_jyd15]

this is totally beyond me.

sorry i can't be of help...but I am assuming the answer is 0 then?

 

[Hurkyl]

And once everybody's suckered into picking zero, two people collaborate with one writing 100 and the other writing a sufficiently small positive number!

 

[davee123]

Won't it loop early on?

I think he meant to compress 10^10^100-9 a googol times, THEN take the last two digits. But regardless, I think you're right-- once you compress it ONCE, the compression probably won't get much (if any) different with successive compressions. Unless LZH does something special.

But it's still uncomputable because we don't have computers that can actually represent 10^10^100, because we can't store that many bits, and thusly can't even perform ONE iteration of the algorithm in order to determine the final bytes. You MIGHT be able to compute it by trying successively increasing numbers and looking for a pattern. IE 10^10^1, then 10^10^2, then 10^10^3, etc. Possible that you could logically prove the answer.

As another example of an uncomputable, how about the middle two digits of the googolth prime? Or the googolth prime prime?

DaveE

 

[Jimmy Snyder]

how about the middle two digits of the googolth prime prime?

You mean 42?
Thanks for your insight Dave.

 

[heartless]

I'm not of the great mathematician but I think you should pick 2/3 of 50. THere is high probability that at least two people will choose the same number (birthday paradox) and suppose a class is based upon 30 typical nyc high students. None of them knows about averages, suppose most of them don't think. Let's draw 6 random sets of numbers between 1-100 (that's what they would choose if they don't think, but don't want to disrupt the fun either so all of them are serious about this and they choose whatever comes to mind, there's also high probability that most of them would choose 50 - believe me) all already arranged from smallest to greatest.

$$\indent\(\pmb{A=\{2,8,12,12,16,16,18,19,24,29,30,43,54,}\\ \pmb{55,59,59,60,60,62,69,75,80,82,82,82,86,}\\ \pmb{87,89,94,99\}}\)$$

$$\indent\(\pmb{B=\{1,4,8,19,20,23,25,25,27,38,38,39,44,}\\ \pmb{46,55,57,60,60,61,64,65,67,69,74,84,88,}\\ \pmb{88,94,94,98\}}\)$$

$$\indent\(\pmb{C=\{2,9,9,11,17,17,21,21,23,23,26,30,37,}\\ \pmb{37,39,41,41,49,52,54,54,56,58,60,65,68,}\\ \pmb{69,80,81,99\}}\)$$

$$\indent\(\pmb{D= \{2,2,5,10,11,26,32,43,43,44,44,47,47,}\\ \pmb{49,55,55,61,62,64,64,70,75,79,80,81,84,}\\ \pmb{86,92,97,100\}}\)$$

$$\indent\(\pmb{E=\{11,12,14,15,19,19,26,30,32,36,38,46,}\\ \pmb{47,48,53,55,59,61,62,69,72,74,75,75,75,}\\ \pmb{76,78,81,85,97\}}\)$$

$$\indent\(\pmb{F=\{2,3,3,14,14,20,28,40,40,45,54,54,62,}\\ \pmb{63,67,68,71,73,73,75,76,80,80,81,83,86,}\\ \pmb{88,90,97,98\}}\)$$now let's find avarages of all of them

Avg. Set A = 521/10 ~ 52
Avg. Set B = 307/6 ~ 50
Avg. Set C = 1249/30 ~ 42
Avg. Set D = 1610/30 ~ 53
Avg. Set E = 1540/30 ~ 51
Avg. Set F = 1728/30 ~ 57

Avg. of Sets ~50

As I said It's all about ordinary high school students who tend to choose numbers at random. You can also study the theory of human brain and behavior. If for example the prize would be a computer, most of them would choose 50, doesn't need an explanation I think why. However if you are to ask thinking nyc high school students like from Bronx high school of science or something, you would have to choose about 2/3 of 2/3 of 50 and again if you would be to ask harvard college students and give them some time you would have to choose 2/3 of 2/3 of 2/3 of 50 probably since most of them would come to the conclusion of 2/3 of 50.

Thanks, and I'm looking for improovement on this nice problem.

 

[moose]

On the piece of paper handed out in class please write a number
between 0 and 100. The winner is the one who comes closest to
2/3rds of the average of numbers written by the students of the
class.

Question: What number should you write?

it all depends on how many people there are in the class. The closer the class is to infinity, the closer you should pick to 34. This is assuming that everyone picks an absolutely random number. Wait, is that 0 - 100 inclusive or exclusive?

 

[bjr_jyd15]

Wait, is that 0 - 100 inclusive or exclusive?

Good question, I don't know the answer, but for our purposes let's assume either.

Either way, assuming everyone know's how to play the game, the answer should be 0, correct?

 

[LarrrSDonald]

In principle, zero is probably the correct guess if everyone plays perfectly - this would also yield a tie between everyone because everyone will be correct.

In reallity though, I'd probably venture to guess slightly more then zero, perhaps even one. As eluded to, all it takes is one person going with 37 and boom, you're up to a little over one in a class of thirty and someone going a little over zero would be the sole winner. In a sense, this tendency might then drive the number up further, should everyone apply it. It's not something that's easy to quantify and come to a solid conclusion about. Surely unless one is in a class of complete morons no one would write anything over 66 barring collusion or some form of simple mischief (screw this, I'm just going to write 100-2pi just to be odd) since it would make it impossible to be right and in theory I can see where repeated application of this would drive the number quite far down (since no one will write that, surely it will be this, but if everyone does that, then it'd be less, etc, until everyone is writing zero) but I would think it wise to compensate for those who, in one way or another, decided to not write zero as well as the others who guessed some wouldn't and went slightly above zero to compensate.

Had I been asked, I would in a class of 30 probably go with four people are gonig to average about 20 amongst them and the rest will average around 1 and go with 3.53333... or perhaps just 3.54 (in reality I'd prolly clock in at pi+(4/10) - when someone asks for a number in a range I feel it my duty to have my guess be irrational). Then, I would suffer the ridicule of those with most in-depth theories in stride claiming to take a more human nature approach since there is no solid answer.