You work in an office of 90 employees, all geniuses! One day the boss corrals all ninety of you and says that she has a prize she'd like to give out to one (or more) of you. But to award it, this is what all of you need to do. Each one of you, WITHOUT consultation with anyone else, will write on your business card a number from the set {1,2,...,90} and drop the card into a basket. Afterward the numbers on all the cards will be read and the average of the numbers computed. The prize will go to that person (or those persons) whose number is closest to (4/5) or (5/4) of the average of all the numbers. What number do you write on your business card (if you must have the prize)? (This problem is a variation on a simpler problem that was recounted to my by Paul. I take credit only for the variation part of the problem).

Last edited by quantyst on November 30th, 2008, 11:00 pm, edited 1 time in total.

could you define genius? (and i guess 5/4 is your variation?)

QuoteOriginally posted by: wileyswcould you define genius? (and i guess 5/4 is your variation?)Do I have to?Think of it 'operationally'.

Is it 1? Here's my (probably flawed) reasoning. I wouldn't pick a number higher than 72, because 5/4*72=90, which is the highest possible average. Assuming nobody else picks a number higher than 72, I wouldn't pick a number higher than 57.6, because 5/4*57.6=72. Keep this up, and you get to 1.

Av = Average of cardsx = (4/5) Avy = (5/4) AvQ = Number of cards with x(90-Q) = Number of cards with ySolving Q x + (90 - Q) y = 90 Av we have Q = 50The values for Av for which Av, x, y belong to the set {1,2, ..., 90} are 20, 40 and 60.Scenario 1: They'll converge to 40 (the average of the averages), and randomly choose between 32 (with 4/9 probability) and 50 (with 5/9 probability).Scenario 2: They'll randomly choose between: 16 (with 4/27 probability), 25 (with 5/27 probability), 32 (with 4/27 probability), 50 (with 5/27 probability), 48 (with 4/27 probability) and 75 (with 5/27 probability).Scenario 1 will be chosen, since it would ensure the best chance of being at 4/5 or 5/4 of the average.

rmexico, no it's not flawed. if everyone reasons as what you did, then 1 is the only choice to guarantee you the prize; but if the majority stops after couple of steps, then 1 probably would not get you the prize.i cannot imagine there is a number to guarantee the prize (maybe i read the problem wrong) - i would think it depends on the reasoning behavior of the whole group

Last edited by wileysw on December 1st, 2008, 11:00 pm, edited 1 time in total.

QuoteOriginally posted by: rmexicoIs it 1? Here's my (probably flawed) reasoning. I wouldn't pick a number higher than 72, because 5/4*72=90, which is the highest possible average. Assuming nobody else picks a number higher than 72, I wouldn't pick a number higher than 57.6, because 5/4*57.6=72. Keep this up, and you get to 1.How does your 'reasoning' relate to your decision?Just because what follows your word "because" is true, it does not necessarily follow that your decision is valid. You wrote:"I wouldn't pick a number higher than 72, because 5/4*72=90,..."Yes, it is true that 5/4*72=90. How does that true equality relate to the issue at hand? You need to show connections, close relationships, reasoning, etc. and conclusion. What if I said the Pythagorean Theorem (call this "X") is true because 2+5=7 (call this "Y"). I need to show a connection between X and Y. Without a connection I have proved nothing.What if for some weird reason everyone picked 90? Wouldn't then everyone be the closest to the (5/4) of the average? In which case everyone would win.The challenge here is whether or not there is one number that all will necessarily choose to guarantee a win. Is there? Lacking this, is there a number that all will necessarily choose because that number provides the highest probability of win. Is there?

Sorry. I should've explained better. Maybe I read the problem wrong.I was saying if I choose a number above 72, say 89, then I win if the average is 71.2 or 111.25. The latter isn't possible, so I only win if the average is 71.2. Alternatively, I could choose 71.2*4/5 for my number, in which case I win if the average is 71.2*(4/5)^2 or 71.2, both of which are possible. As a result, I'd say that 71.2*4/5 is always a better choice than 89, because the former gives you more chances to win the game than the latter does.If I thought everyone was going to pick 90, I'd pick 72. Wouldn't I win?I'm not considering how my choice affects the average.

QuoteOriginally posted by: rmexicoSorry. I should've explained better. Maybe I read the problem wrong.I was saying if I choose a number above 72, say 89, then I win if the average is 71.2 or 111.25. The latter isn't possible, so I only win if the average is 71.2. Alternatively, I could choose 71.2*4/5 for my number, in which case I win if the average is 71.2*(4/5)^2 or 71.2, both of which are possible. As a result, I'd say that 71.2*4/5 is always a better choice than 89, because the former gives you more chances to win the game than the latter does.If I thought everyone was going to pick 90, I'd pick 72. Wouldn't I win?I'm not considering how my choice affects the average.The way the question is posed, you'd lose 50% of the time. You could have an equilibrium with a randomized strategy, but it seems there's infinitely many.

QuoteOriginally posted by: rmexicoSorry. I should've explained better. Maybe I read the problem wrong.I was saying if I choose a number above 72, say 89, then I win if the average is 71.2 or 111.25. The latter isn't possible, so I only win if the average is 71.2. Alternatively, I could choose 71.2*4/5 for my number, in which case I win if the average is 71.2*(4/5)^2 or 71.2, both of which are possible. As a result, I'd say that 71.2*4/5 is always a better choice than 89, because the former gives you more chances to win the game than the latter does.If I thought everyone was going to pick 90, I'd pick 72. Wouldn't I win?I'm not considering how my choice affects the average.Let's remember that all of you are geniuses!

How's this?Well, by symmetry my optimal strategy is the same as my neighbor's. because we're all geniuses, and assuming (for now) there is an optimal pure strategy (pick A with probability 1) we will all find it. if A >= 2, the nearest integer to (4/5)A isn't A and the nearest integer to (5/4)A isn't A ((5/4)2 = 2.5 -> 3, if everyone chooses 2 it pays to switch to 3), so choosing A >= 2 isn't optimal as switching will win the prize. so if there's a pure optimal strategy it is to choose 1. if we all choose 1 we all win the prize with probability 1 as there is no better choice than 1 because it is closest to 0.8 and to 1.25.

QuoteOriginally posted by: beero1000How's this?Well, by symmetry my optimal strategy is the same as my neighbor's. because we're all geniuses, and assuming (for now) there is an optimal pure strategy (pick A with probability 1) we will all find it. if A >= 2, the nearest integer to (4/5)A isn't A and the nearest integer to (5/4)A isn't A ((5/4)2 = 2.5 -> 3, if everyone chooses 2 it pays to switch to 3), so choosing A >= 2 isn't optimal as switching will win the prize. so if there's a pure optimal strategy it is to choose 1. if we all choose 1 we all win the prize with probability 1 as there is no better choice than 1 because it is closest to 0.8 and to 1.25.i'm not convinced. if you are sure that everyone will choose the same number, then whichever number does the trick - you don't have to be the only one to win the prize, no? the problem does not explicitly say that you have to beat everyone else. also what if after your reasoning above and write down 1, you discovered that the geniuses for whatever reason somehow believe 10 is a better choice?

Where exactly is the problem? I think you may have misunderstood me.Assume there is a 'best number' A. That is, all the geniuses recognize that picking A will give them the best shot at the prize, and therefore they all pick it. So that's 89 who choose A. If the closest integers to (4/5)A or (5/4)A are pickable and isn't A, then I can choose one and win the prize, while everyone else loses. That means A can't possibly be the best choice. The only possible best number is one where the closest choosable integer to (4/5)A or (5/4)A is A itself. For example, if for some reason all the geniuses think 10 is the best bet, then I can choose 8 and expect to win (I'm a genius too!). The geniuses should know this and therefore won't choose 10. The same reasoning invalidates all of the numbers but 1. No one person can steal the prize if everyone else chooses 1, so assuming that none of the 90 communicate a plan, its not worth it for any person to play anything but 1. It's the prisoner's dilemma. also assuming we only want the title of 'winner' (it doesn't matter how many people win, as long as I do), then everyone should conclude that everyone will conclude that 1 is the best bet, and when we all play 1, we all win.

i understand your reasoning. and i'm wondering why you believe there is a best number.just use the 10 example, why do you prefer number 8 (or 12, 13) to number 10? don't you expect to win by 10 as well? note you believe everyone else chooses same number. isn't 10 also an equilibrium?

QuoteOriginally posted by: wileyswi understand your reasoning. and i'm wondering why you believe there is a best number.just use the 10 example, why do you prefer number 8 (or 12, 13) to number 10? don't you expect to win by 10 as well? note you believe everyone else chooses same number. isn't 10 also an equilibrium?It's been a long time, but thanks to Beero, I remember some game theory now.10 isn't an equilibrium. If I knew everyone else would choose 10, I could choose 8 (or 12.5) and win. Alternatively, I could be a nice guy and choose 10, so that everyone wins, but there's no guarantee that I would. Everyone else would see this coming, so nobody would choose 10.1 is an equilibrium. If everyone else chooses 1, and I choose something different from 1, I don't win. None of the other numbers in the interval [1, 90] has that property.