Saw this one on a website, but I won't tell you which one until later (so it'll take a bit longer to find the answer!).
Three people are playing a game with the following rules:
1. Each player picks a natural number (i.e. 1, 2, ...) in secret.
2. After all players choose a number, the numbers are revealed.
3. The player with the lowest unique number wins $3.
4. If all three players pick the same number, each player wins $1.
To clarify the above, if the players pick (2,4,4) or (2,3,5) or (1,1,2) the lowest unique number in each case is 2 and the player who picked it gets $3 (the other two players get $0). If the players pick (1,1,1) then there is no lowest unique number and all three players get $1.
If you are playing this game, what number-picking strategy will give you the highest expected value? You may assume that each of the other players is a "perfect" logician, and will try to maximize their own expected value.
HINT: The optimal strategy must be optimal no matter what strategy the other players use, so all three players will end up using the same one...
I have seen a similar text-messaging (of course, 1 euro per message) action; once two people had text-messaged the same number, that number would be disqualified and the next number would become the lowest. At the end of the week the one with the lowest unique number would win the grand prize, a car or something like that. The best strategy: do not send a text-message.
Originally posted by PBE6First thoughts:
Saw this one on a website, but I won't tell you which one until later (so it'll take a bit longer to find the answer!).
Three people are playing a game with the following rules:
1. Each player picks a natural number (i.e. 1, 2, ...) in secret.
2. After all players choose a number, the numbers are revealed.
3. The player with the lowest unique num ...[text shortened]... what strategy the other players use, so all three players will end up using the same one...
My initial thought was that this probably would requires a mixed strategy. Maybe with players randomizing their choices of 1s and 2s. Intuitively, let's start with a choice of 2 with probability 1/3.
If all do so, then you have a prob. of:
1/3*2/3*2/3 = 4/27 of winning 3 - A (2,1,1)
1/3*1/3*1/3 = 1/27 of winning 1 - B (2,2,2)
2/3*1/3*1/3 = 2/27 of winning 3 - C (1,2,2)
2/3*2/3*2/3 = 8/27 of winning 1 - D (1,1,1)
and 0 for all other combinations.
Expected value of winnings would be 1. Is this an equilibrium? If you choose a number bigger than 2 you would expect to win 3 in case B and still win in case A, so this CANNOT be an equilibrium. Of course if others also increase from 2 to that number then that also cannot be an equilibrium as there would be an incentive for one of the players to reduce back to being the unique 2.
Maybe they need to randomize between three different numbers, but I'm not seeing it yet. I might also need to adjust probabilities, but I foresee a problem with this "high tie" probability. Or maybe I'm just missing an obvious solution here. 😕
I don't see any reason not to pick either "1" or "2", especially with only 3 players.
Why consider choosing anything greater than "2":
If the other players choose "1" and "2", you lose
If both other players choose "1", you should choose "2"
If both other players choose "2", you should choose "1"
If both other players choose something greater than "2", either "1" or "2" would be better.
Originally posted by forkedknightYou don't know ex-ante what they'll choose (draw) if they are using mixed strategies as they also randomize their choices between X options.
I don't see any reason not to pick either "1" or "2", especially with only 3 players.
Why consider choosing anything greater than "2":
If the other players choose "1" and "2", you lose
If both other players choose "1", you should choose "2"
If both other players choose "2", you should choose "1"
If both other players choose something greater than "2", either "1" or "2" would be better.
With 2, you only win over any 1 if the third also has draws 1. But in this case 3, 4, or 50000 would also give you 3 points.
BUT if they both get 2s you win one point if you've chosen 2 (and get it randomly) and 3 points if you've chosen 3 or higher for your high choice.
Originally posted by PalynkaHa! Very nice work. I just did the calculation that you did so I could confirm the answer given on the website, but I got the surprising (to me, at least) result that given a choice between 1 and 2, it doesn't matter what your probability distribution is! This is contrary to the answer on the website, which I now expect is wrong (and I think I know why). However, this doesn't make life easier. It made me want to cry a little, actually. :'(
First thoughts:
My initial thought was that this probably would requires a mixed strategy. Maybe with players randomizing their choices of 1s and 2s. Intuitively, let's start with a choice of 2 with probability 1/3.
If all do so, then you have a prob. of:
1/3*2/3*2/3 = 4/27 of winning 3 - A (2,1,1)
1/3*1/3*1/3 = 1/27 of winning 1 - B (2,2,2)
2/3*1/ em with this "high tie" probability. Or maybe I'm just missing an obvious solution here. 😕
After some more calculations and a heckuva lot more scribbling, I came up with the following hypothesis:
First off, there is only one way to tie this game, and that is to have all three players pick the same number. There are also two ways to win - you can either: (1) pick a relatively high number, but get lucky when the other two players tie with a lower number and eliminate each other (call this a high-win); or (2) pick the lowest number (call this a low-win).
Let's say you're player A and you pick a number "x" with probability Px, and sort out your chances of tying and winning. (The optimal strategy should be adopted by each player so PAx = PBx = PCx = Px).
TIE
There is only one way to tie (all three players pick x), so the probability of a tie is Px^3
HIGH-WIN
There are x-1 numbers lower than x which the other two players can tie on and eliminate themselves with, so the probability of a high-win is Px*(P1^2 + P2^2 + ... P(x-1)^2) = Px*SUM[i|1..(x-1)](Pi^2) using summation notation.
LOW-WIN
For the time being, let's constrain our analysis to numbers between 1 and n. There are n-x numbers bigger than x which each of the other players can choose from, so the probability of a low-win is Px*(P(x+1)*P(x+1) + P(x+1)*P(x+2) + ... + P(x+2)*P(x+1) + P(x+2)^2 + P(x+2)*P(x+3) + ... P(n)^2) = Px*SUM[j|(x+1)..n]SUM[k|(x+1)..n]Pj*Pk. Sorry if that looks a little ugly, it's supposed to be a double summation from (x+1) to n. Of course, the real probability over the natural numbers is the limit as n approaches infinity, but I'll include that at the end.
So to recap, we have:
P(tie,x) = Px^3
P(high-win,x) = Px*SUM[i|1..(x-1)](Pi^2)
P(low-win,x) = Px*SUM[j|(x+1)..n]SUM[k|(x+1)..n]Pj*Pk
Adding in the tie and win payoff values, V(tie) and V(win), we can calculate the expected value E(x) when choosing x between 1 and n:
E(x) = V(tie) * Px^3 + V(win) * (Px*SUM[i|1..(x-1)](Pi^2) + Px*SUM[j|(x+1)..n]SUM[k|(x+1)..n]Pj*Pk)
The total expected value E(total) is the sum over all x from 1 to n, being sure to take the limit now as n approaches infinity:
E(total) = lim(n-->inf) SUM[x|1..n]E(x)
I've kept the final expression in short form, but even that looks ugly as hell. And that's not even the end!! Once we calculate this expression, we have to take the derivative with respect to the probability distribution to find all the stationary distributions, which I don't even know how to do, if it's even possible (does it have anything to do with calculus of variations?!?). The whole thing makes me want to fart.
I'm hoping that some clever individual can find some sort of shortcut. I'll post the original website link to see if that helps. If anyone needs me, I'll be recovering from a coma. 😕
Question 198 here: http://mathproblems.info/group10.html
Aha! I think I have something much simpler.
I just checked the expected value for n=3, and came up with a very surprising result - the expected value is always 1!! No matter what distribution you choose, the answer is always the same. Which got me to thinking...
My original assumption that all players will employ the identical optimal strategy makes the problem symmetrical. In addition, the entirety of the prize money is always awarded, either all to one player or a little to each. This means that since it doesn't matter which player you choose (as they all employ identical strategies), on average each player will receive an equal share of the money, or $3/3 = $1.
Another way to arrive at this conclusion is to say that there are four possibilities for payouts (A wins, B wins, C wins, 3-way tie), with probabilities P(tie) and P(A win) = P(B win) = P(C win) = (1-P(tie))/3. The expected value for A is then $1*P(tie) + $3*(1-P(tie)/3) = $P(tie) + $(1-P(tie)) = $1.
I'd get a secret partner. We would make sure that one of us would always pick 1 while the other always picks 2. Perhaps we could each alternate between the two to make it less obvious. Since one of us would always win regardless of what the third person picks, we would definitely come out with more than we put in together. Then we could just split the winnings.
Originally posted by PBE6Interesting problem.
Saw this one on a website, but I won't tell you which one until later (so it'll take a bit longer to find the answer!).
Three people are playing a game with the following rules:
1. Each player picks a natural number (i.e. 1, 2, ...) in secret.
2. After all players choose a number, the numbers are revealed.
3. The player with the lowest unique num ...[text shortened]... what strategy the other players use, so all three players will end up using the same one...
How much will change if (instead of a natural number) we can chose a positive real number?
Originally posted by PBE6Edit - Scratch that, I understand what your strategy is now. I think. :
Ha! Very nice work. I just did the calculation that you did so I could confirm the answer given on the website, but I got the surprising (to me, at least) result that given a choice between 1 and 2, it doesn't matter what your probability distribution is! This is contrary to the answer on the website, which I now expect is wrong (and I think I know why). Ho ing from a coma. 😕
Question 198 here: http://mathproblems.info/group10.html
Edit - I still think it is impossible that it doesn't matter what the probability distribution is. As I've shown above, with P2 = 1/3, then all have E(value) =1, but can increase that with the correct deviation.
Originally posted by PBE6Neat! The thing is that, as I've shown above, that isn't a sufficient condition to ensure that they are using an optimal strategy.
In addition, the entirety of the prize money is always awarded, either all to one player or a little to each. This means that since it doesn't matter which player you choose (as they all employ identical strategies), on average each player will receive an equal share of the money, or $3/3 = $1.
Originally posted by PBE6top of my head i'd say 1 just because of this:
Saw this one on a website, but I won't tell you which one until later (so it'll take a bit longer to find the answer!).
Three people are playing a game with the following rules:
1. Each player picks a natural number (i.e. 1, 2, ...) in secret.
2. After all players choose a number, the numbers are revealed.
3. The player with the lowest unique num what strategy the other players use, so all three players will end up using the same one...
If you pick 2, the only way you win is if they both pick 1 or both pick 3
If you pick 3, the only way you win is if they both pick 1 or both pick 2.
If you pick 1, you win if they both pick 2, they both pick 3, or they pick 2 and 3.
Picking 1 gives you 3 chances to win, picking 2 gives you 2 chances, and picking 3 gives you 2 chances.
Originally posted by uzlessFAIL
top of my head i'd say 1 just because of this:
If you pick 2, the only way you win is if they both pick 1 or both pick 3
If you pick 3, the only way you win is if they both pick 1 or both pick 2.
If you pick 1, you win if they both pick 2, they both pick 3, or they pick 2 and 3.
Picking 1 gives you 3 chances to win, picking 2 gives you 2 chances, and picking 3 gives you 2 chances.