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  #1  
Old 07-03-2005, 04:40 PM
AlphaMeridian AlphaMeridian is offline
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Default Optimum Strategy in this \"game\" (not poker related)

This isn't poker related, but I figured that I might give the forum a shot, it's a lot better at game theory than I am. In a week, there will be a "draft" of game players for a small intrasite league. The order of the draft is determined by who is closest to the number that the administrator of the league picks (between 0 and 100). The numbers do not "wrap" (that is, if i pick 99, and the number is 2, I am not 3 away, I'm 96 away)

Is there a optimum number in this circumstance? If there is, is it 50 defacto because it nearly assures you 2nd place assuming 2 other rational opponents? If i pick 50, and they both go low, and the number is high, I get first. If i pick 50, and one of my opponents picks low, and one picks high, I get 2nd place (the numbers don't wrap). The only time I "lose/3rd pick" is when I pick 50, the number is either high or low, and BOTH of my opponents pick in that direction.

Are there any better numbers?

-Alpha
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  #2  
Old 07-03-2005, 07:57 PM
SheetWise SheetWise is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

How does the administrator "pick" the number?
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  #3  
Old 07-03-2005, 09:54 PM
AlphaMeridian AlphaMeridian is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

Out of his ass, most likely. He might use a RNG, I'm not sure.

-Alpha
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  #4  
Old 07-03-2005, 10:36 PM
SumZero SumZero is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

If all three players are super-logical rational actors who are aware that all players are super-logical rational acotrs (and are aware of the awarness recursively forever) and the number is randomly and picked from the space [0,100] with uniform distribution then you'd expect the strategy each actor to take would be to pick from a probability distribution function that is symetric and at its biggest at 50. Think sort of like the normal distribution (but snipped at 0 and 100).

Figuring out the standard distribution is harder. My guess is the standard deviation should be somewhere around 17, but I don't have a good way of showing that or proving that.

In reality I would think that none of the actors are super rational and that the number picker is probably trying to pick a "random" number which really means he's much more likely to pick numbers like 37 and 73 than he is to pick 0, 100 or 50.

Do you have a read on the other players? Are they likely to naively pick 50 or 51/49? If so I'd probably try picking 44 or 56. If they are likely to think about the meta game I'd likely pick 37.
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  #5  
Old 07-04-2005, 03:42 PM
AlphaMeridian AlphaMeridian is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

My disadvantage here is that the other 2 players have done this before. I'm not sure they're taking it this far, but they might have picked up knowledge of a more correct strategy just by intuition and experience. I'd agree the number picker is not going to pick 0, 50, or 100, but something inbetween. Though, since I don't know which number he's likely to pick or even the method he's going to use to pick it (no "read" on him).

The only other possible useful tidbit is that numbers that we pick will be known to the other players as we pick them - IOW, there will be someone who picks first, and the other two players will know his number. I'm not sure if there's an advantage to picking last in this situation.

http://www.arsclan.net/forum/showthread.php?t=7311

-Alpha
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  #6  
Old 07-04-2005, 08:11 PM
SumZero SumZero is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

[ QUOTE ]
The only other possible useful tidbit is that numbers that we pick will be known to the other players as we pick them - IOW, there will be someone who picks first, and the other two players will know his number. I'm not sure if there's an advantage to picking last in this situation.

[/ QUOTE ]


I'm assuming the random number has been picked already or will be computer picked before anyone picks?

Oh. This is a totally different game then. I think you want to pick second as it is a huge advantage. Assume the number is truly random ([0,100] uniform probability).

Player 1 chooses a number N. This bisects the range into [0,N-1] and [N+1,100] ranges of unpicked numbers. WOLOG say N >= 50 (if it isn't rename so 0 maps to 100, 1 to 99, etc. and then N maps to 100 - N).

Player 2 chooses a number M which depends very much on N.

There are two mistakes that player 1 can make: too close to 50 and too far from 50. If it is too close to 50 then player two chooses a number 1 further away from 50 in the other direction (I.e., 1 chooses 50, 2 chooses 49 or 1 chooses 55 2 chooses 44. This works as long as there until there are twice as many numbers between the numbers 1 and 2 chooses as there are above 1 because now 3 will always choose 1 greater than 1 choose and 2 gets the numbers [0,49].). If 1 chooses too far from 50 then 2 chooses a number about 1/3 of 1. And then 3's choice doesn't really matter as long as it is less than 1 (and 3 can be a king maker between 1 and 2).
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  #7  
Old 07-05-2005, 12:58 AM
AaronBrown AaronBrown is offline
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Default Re: Optimum Strategy in this \"game\" (not poker related)

If we assume the Administrator uses a random number generator that weights all the numbers equally, then the first player should pick 24 or 76, the second player should pick 75 if the first player picks 24 and 25 otherwise. The third player should pick some number in between.

Let's say the first two picks are 24 and 75. The third player could pick 76 and win on 25 numbers (76 to 100), he could pick 23 and win on 24 numbers (0 to 23) or, say, 50 and win on 25.5 numbers (38 to 62, with a tie at 37). So the third player cannot do better than to pick in between, although it doesn't matter what number he picks in that range.

The second player cannot do better than to pick 75. If he picks 74, the third player will pick 75 to get 26 winning numbers (75 to 100). The second player is left with 25.5 winning numbers (50 to 74, with a tie at 49). If he picks 75, he is guaranteed 26 winning numbers if the third player plays correctly, and he could get up to 50.5. On average he'll get 38.25.

The first player also cannot do better. If he picks 25, the second player can pick 74, and the third player will pick 24. By picking 24 he gets at least 25 winning numbers, he could get up to 49.5. His average is 37.25.

Adding the expectations gives 37.25 (first player) + 38.25 (second player) + 25.5 (third player) = 101, the number of numbers picked.

However, this assumes all players play rationally. The great advantage of going last, is you do better if any player makes a mistake. If you go first or second, you can get screwed either by a mistake from a later player or just an unfriendly play that makes no difference to him.
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