Guess 2/3 of the Average - Is 0 More "Powerful" than 100? I was reading about this game called "Guess 2/3 of the Average" (https://en.wikipedia.org/wiki/Guess_2/3_of_the_average). Players guess an integer between 0 and 100 - the winner is the player whose guess was closest to 2/3 of the average number guessed by all players.
I read that if everyone knows how this game works, the best guess to make is 0. This is because guessing the number 0 will "shrink" the average of all guesses (and 2/3 of the average) closer to 0, thus making your guess closest to the average of 2/3rd's of all guesses and making you the winner. (However, this strategy is only valid if everyone knows how the game works - if players guess random numbers, then 0 is no longer the best guess)
I tried to explain this game to my friend and explain why guessing the number 0 is the best guess. My friend argued that if guessing the number 0 has the ability to "shrink" 2/3rd's of the average guesses towards 0 - wouldn't guessing the number 100 have the equal ability to "expand" 2/3rd's of the average guess towards 100?
I used this analogy: Suppose there are 3 exams - if you score 100/100 marks on two of the exams and score 0/100 on the third exam, your average grade is 66%. But if you score 0/100 on two exams and score 100/100 on the last exam, your average grade is only 33%. My informal reasoning was that the number 0 is much more powerful in pulling an average towards itself  compared to the number 100. But now, I beginning to doubt my own logic and reasoning.

*

*Can someone please explain the logic behind why guessing the number 0 is the best guess for this game (assuming that everyone knows how this game works)?

Thank you!
 A: This game was actually used by the french review "Jeux et stratégies", specialized in math and logic puzzles, and games such as chess or go.
Once a year, there was a big contest with a lot of difficult puzzles to solve.
Despite the difficulty, there were always ties at the end, because some game clubs used to put many people in order to solve everything. And the ultimate tie breaker was... this 2/3 game.
This was in the 80s so I don't remember what were the winning figures. I remenber they were low, but not 0, actually something somewhere around 5. (In this variant you could emit any decimal number between 0 and 100, it was not restricted to integers.)
Why wasn't the winning figure 0? Well, first some people would not make a "correct" reasoning. Then some people would include in their reasoning that some people would not make the "correct" reasoning. Etc. And also some people could cheat, sending answers (by different people in their group) with different figures, including 100.
A nice property of this tie-breaker, is that the magazine could reuse it year after year. The winning "2/3 of the average" decreased year after year, but nevertheless did not reach 0.
Another nice property is that cheating by sending large numbers was not really cheating: you just needed to take that into account in your reasoning.
EDIT: actually the Wikipedia page about this game shows the distribution of the 1983 edition of the "Jeux et stratégies" contest, which seems to be the inventor of the game. And my memory was not very good: integers were asked for, on a range from 0 to 1 000 000 000. And 5 (if on a scale from 0 to 100) was not close to 2/3 of the average, but close to the average itself. I was 15 and I can't remember what figure I sent. :-)
https://en.m.wikipedia.org/wiki/Guess_2/3_of_the_average
A: No matter how many players and no matter what their guesses, the average of their guesses can be no more than $100$ and $2/3$ of that average is always less than or equal to $(2/3)(100)< 67$.  So guessing $67$ will always put you closer to the true average in comparison to guessing any number higher than $67$ (including 100).
Now one way to continue is to argue recursively: If we assume everyone has figured out that their guesses should be no more than $67$ then we can safely assume that the average of all guesses will be no more than $67$ and $2/3$ of that average will be no more than $(2/3)(67)< 45$.  Now if everyone figures this out then no bets will be more than 45 and $2/3$ of that will be $(2/3)(45)< 30$, and so on and so on.
After iterating this reasoning we see our bet converges down (close to) 0. Of course, only the first stage is solid (bet no more than 67) while the other stages assume that all players have figured out the first stage.

Strictly speaking, this recursive reasoning together with the integer constraints concludes only that we should bet no more than 1 (so, either 0 or 1):
\begin{align}
&(2/3)100 = 66.667 < 67 \\
&(2/3)67 = 44.667 < 45 \\
&(2/3)45 = 30 \\
&(2/3)30 = 20 \\
&(2/3)20 = 13.333 < 14 \\
&(2/3)14 = 9.333 < 10 \\
&(2/3)10 = 6.666 < 7 \\
&(2/3)7 = 4.666 < 5 \\
&(2/3)5 = 3.333 < 4 \\
&(2/3)4 = 2.667 < 3 \\
&(2/3)3 = 2  \\
&(2/3)2 = 1.333 < 2 \\  
&(2/3)2 = 1.333 < 2 \\
&(2/3)2 = 1.333 < 2 \\
&...
\end{align}
So we are stuck at 2.  But if we notice that 1.333 is closer to 1 than it is to 2, we can argue that we should either bet 1 or 0.
