# Pick a number, any number

PBE6
Posers and Puzzles 09 Sep '08 17:46
1. uzless
The So Fist
12 Sep '08 02:10
Originally posted by Palynka
FAIL
explain.

If everyone has optimal strategy and realizes 1 is the best number, then what, they are supposed to come up with a different number? Well, if they are going to come up with a different number, then you should just pick 1 because it will be the lowest!

Hell, earlier you proposed the players should pick random numbers.

pfft, random number picking...Thanks for that brilliant strategy palynka!
2. 12 Sep '08 03:02
I still like my way best.
3. 12 Sep '08 08:542 edits
Originally posted by uzless
explain.

If everyone has optimal strategy and realizes 1 is the best number, then what, they are supposed to come up with a different number? Well, if they are going to come up with a different number, then you should just pick 1 because it will be the lowest!

Hell, earlier you proposed the players should pick random numbers.

pfft, random number picking...Thanks for that brilliant strategy palynka!
you have to vary if you want to win the most points. Basically, you have to find the strategy that will have you picking 1 when the two others pick 2 and 2 when they pick 1
4. wolfgang59
invigorated
12 Sep '08 10:22
Originally posted by uzless
explain.

If everyone has optimal strategy and realizes 1 is the best number, then what, they are supposed to come up with a different number? Well, if they are going to come up with a different number, then you should just pick 1 because it will be the lowest!

Hell, earlier you proposed the players should pick random numbers.

pfft, random number picking...Thanks for that brilliant strategy palynka!
OBVIOUSLY one must pick numbers randomly else your logical opponents will know your number!

The question is what frequency do you pick the numbers.

1 will be your most oft used but you cannot overuse it.
and so on

How often to choose 1? 90%? 50%? 33%?
5. Palynka
Upward Spiral
12 Sep '08 10:55
Originally posted by PBE6
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.
You also win if they tie with numbers higher than x. So the probability of that type of win must be Px*SUM[i|1..(x-1)](Pi^2) + Px*SUM[i|(x+1)...Inf](Pi^2).

If they have the same draw, you always win as long as your number is different than theirs.
6. PBE6
Bananarama
12 Sep '08 15:321 edit
Originally posted by Palynka
You also win if they tie with numbers higher than x. So the probability of that type of win must be Px*SUM[i|1..(x-1)](Pi^2) + Px*SUM[i|(x+1)...Inf](Pi^2).

If they have the same draw, you always win as long as your number is different than theirs.
That calculation is included in the "low-win" part, as if they tie with a higher number your number isn't "high" anymore.

Did anyone read the analysis on the website? With the given parameters, the answer is that any number x should be picked with frequency (1/2)^x. Any thoughts?
7. 12 Sep '08 17:41
I think the solution is as has been stated already. If your opponents use the same strategy you will win equal number of times. Also the fact that it's a kind of "zero sum game" makes it so no one wins in the end.

Basically, play with morons and win :p
8. irontigran
Rob Scheider is..
13 Sep '08 01:49
8?
9. 13 Sep '08 15:291 edit
Just a note here, it is possible to have a situation in this game impossible to win.

If you are the third person, and the other two split thier picks between 1 and 2, no matter what you pick you're not going to win. The most you can do is to decide who does.

I don't think there is ever an occasion to pick 3 or more with only 2 opponents, because either you were going to lose anyways, or else you have a lower pick which will work just as well.

Will have to take a look at this one, but I do think a random mixture between 1 and 2 will end up optimal, but there is a chance that opponent strategy could change the optimal strategy.

Imagine if you played the same game, but the losing players each paid the winning player \$1.

In this case, a win would be worth \$2, a 3-way tie worth nothing, and a loss costs you \$1. The game could be repeated indefinitely, without having any outside source of funds, and patterns would thus become much more visible..
10. wolfgang59
invigorated
17 Sep '08 09:44
Originally posted by geepamoogle
Just a note here, it is possible to have a situation in this game impossible to win.

If you are the third person, and the other two split thier picks between 1 and 2, no matter what you pick you're not going to win. The most you can do is to decide who does.

I don't think there is ever an occasion to pick 3 or more with only 2 opponents, because e ...[text shortened]... without having any outside source of funds, and patterns would thus become much more visible..
If your two opponents both randomly select from 1 and 2 then they DRAW 50%.

Which means you win 50% by simply choosing 3 or 1,000,000 or anything else.

So alternating between 1 and 2 is obviously NOT the optimal strategy.
11. 17 Sep '08 18:10
The goal of game theory in this case is as follows: to find a mixed strategy to follow for the game which optimizes overall wins.

An optimal strategy in a reciprocal game such as this is one which wins more often against any deviation from itself.

It is important to note that for mixed strategies, you have to choose as randomly as possible with the frequency of each pick being in line with the ratio found as being optimal. Predictablility is your enemy, as if an opponent knows what you are going to do, he has a heads up on countering it effectively.

Randomness insures that even if he knows your strategy, he will have a harder time countering a specific instance, because he can't accurately predict your exact course of action.

I can't recall exactly how one goes about determining the ratio for the mix of pure strategies ("Pick 1", "Pick 2", etc), but I do note that 1 has better overall odds then 2 or 3 in the scenario that the enemy is equally likely to pick 1-3 randomly.

If both limit themselves to 1 or 2, then 3 becomes a winning pick, oddly enough, while 1 or 2 tend to break even. (I guess 3 does have a use in a 3-person game..)

Both pick 1-3 randomly
1 gains 4/9 on average.
2&3 lose 2/9 on average.
4+ tends to break even..

Both pick 1 or 2 randomly
1&2 tends to break even.
3+ gains 1/2 on average.

Of course, your opponents might also behave based on this information, and could change their strategy AND the resulting effectiveness of the pure options. The whole thing loops back on itself, which is why mixed strategies with randomness come into being..