Guessing game -how low can you go?
A game similar to Guess 2⁄3 of the average,
Choose a number below 1000.
Unique number closest to it wins. (People with same answer are eliminated)
What is your pick and the reason for your choice?
A game similar to Guess 2⁄3 of the average,
Choose a number below 1000.
Unique number closest to it wins. (People with same answer are eliminated)
What is your pick and the reason for your choice?
I think this should be in an open thread.
1000, because nobody here seems to have an interest in actually participating.
1000 - (random 6-digit integer)*(10^-(the XKCD number))
This game really only make sense if a) you specify that only integers are allowable and b) everyone’s choice is revealed at the same time, and winner discovered then.
I’ll go with 999. If I’m the first to say it, then why would anyone else play unless they’re planning collusion with anybody else?
I also pick 999. I sabotage rational agents for the lulz.
My thought process would go like this, 999 is the number that will win this contest. This is common knowledge, others might also be choosing the same and hence likely to be eliminated. The next number I should choose is 998. But this is common knowledge too, others might also be thinking on similar lines. 998 is likely to be eliminated. So, I should go back to 999. But, what if this is common knowledge too? So, to win I need to eliminate both, 999 and 998, and go with the next number 997. But, what if? This can spiral down all the way to 1. By this reasoning, is it better to stick with 999?
About 10% of people will troll with 1000, and other people will have estimates of trollishness between 1% and 50%, centering around 5%.
Also, this should probably go in discussion or in a poll.
As I read it, the problem seems impossible to win? You’re choosing a unique number closest to your chosen to win, which means you can’t win.
I think by “closest to it” he means “closest to 1000″.
A simpler version would be “pick a natural number, the smallest unique natural number wins”. There’s should be a nash equilibrium for each version, but it would be function of the number of players (and of what happens if nobody picked a unique number—play again? stop?).