Game I Made Up

[jimdmcevoy]

I was bored, so I made up this game to play with my friend the other day:

we each pick a whole number and write it down

then we compare numbers, call the larger of the two numbers L, and the smaller of the two S

the player who picked the smaller number pays the other player 100-L+S cents (negative meaning he recieves money instead of paying)

if we both pick the same number it's a tie of course

it's a pretty fun game, can anyone devise an unbeatable strategy? (as in guarantee that your EV is not negative)

[Paul2432]

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it's a pretty fun game, can anyone devise an unbeatable strategy? (as in guarantee that your EV is not negative)

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Randomly select 0-200.

Paul

[jimdmcevoy]

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it's a pretty fun game, can anyone devise an unbeatable strategy? (as in guarantee that your EV is not negative)

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Randomly select 0-200.

Paul

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hmmm...

according to my calculations if you do that and i always pick 100 you'll have -EV

[Paul2432]

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it's a pretty fun game, can anyone devise an unbeatable strategy? (as in guarantee that your EV is not negative)

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Randomly select 0-200.

Paul

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hmmm...

according to my calculations if you do that and i always pick 100 you'll have -EV

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Pick 0-199 randomly.

Pretty sure this is right.

Paul

[irchans]

If player one randomly chooses between 0 and 200, and player 2 chooses 100, then

50% player one has the higher number and pays 50 (average).
50% player two has the higher number and pays 50 (average).

zero expecctation.

[LouisMatthews]

just say 200 every time

[kyro]

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just say 200 every time

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better hope you're not playing with somoene remotely intelligent and decides to start picking 1 each time...

[Paul2432]

0-200 is not defeated by picking 100 every time, but it is a -EV strategy. If you your opponent picks 0-99 everytime you will lose, with the worst result coming from picking 0 every time.

Paul

[irchans]

What happens when player 1 chooses 1 and player 2 chooses 200?

According to the rules as stated,

player 1 pays player 2 100-200+1 = -99.

Does that mean player 2 pays player 1 99?

[jimdmcevoy]

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What happens when player 1 chooses 1 and player 2 chooses 200?

According to the rules as stated,

player 1 pays player 2 100-200+1 = -99.

Does that mean player 2 pays player 1 99?

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yep

[irchans]

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What happens when player 1 chooses 1 and player 2 chooses 200?

According to the rules as stated,

player 1 pays player 2 100-200+1 = -99.

Does that mean player 2 pays player 1 99?

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yep

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Then I could not afford to play the game. Suppose I bid 121 and my opponent bids 1,000,000. Then

He pays me 100-(1000000)+121 = -999779.

Thus I pay him 999779.

Basically, if one player bids x then the other player can win 1000 or more by bidding x+1100 or more.

It seems that if you were forced to play the game, then you should bid the largest number you can imagine.

Am I missing something here?

[jimdmcevoy]

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He pays me 100-(1000000)+121 = -999779.

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nah small pays big

[jedi]

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Then I could not afford to play the game. Suppose I bid 121 and my opponent bids 1,000,000. Then

He pays me 100-(1000000)+121 = -999779.

Thus I pay him 999779.


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the player who picked the smaller number pays the other player 100-L+S cents (negative meaning he recieves money instead of paying)


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In your example, the smaller player (121) would pay the bigger player (1,000,000) 100-1,000,000+121 = -999,779 Negative number means the smaller player would receive $999,779. Right?

[Paul2432]

Right. If you pick is less than 200 you can never lose more than 99 cents.

Paul