Best Answer bonanova, 06 May 2013 - 02:50 AM

Spoiler for looks like

The expectations for the five players, in order of first choose are:

$5.50 $18534.25 $18624.25 $19591.75 $35836.75

Earlier is worse, later is better; first player will always end up with one or the other of the two lowest checks (equal likelihood); the middle choosers will all have about the same expectation; that expectation is about half of the expectation of the last to choose.

So ... request to choose last.

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The expectations for the five players, in order of first choose are:

$5.50 $18534.25 $18624.25 $19591.75 $35836.75

Earlier is worse, later is better; first player will always end up with one or the other of the two lowest checks (equal likelihood); the middle choosers will all have about the same expectation; that expectation is about half of the expectation of the last to choose.

So ... request to choose last.

Spoiler for analysis

Let two players fight over 1 10 100.

<p1> = 5.5 <p2> = 68.5

Now let three players fight over 1 10 100 1000

<p1> = 5.5 <p2> = 301.375 <p3> = 526.375

Similar analysis for p1234 fighting over 1 10 100 1000 10000 proceeds with p1 keeping 1 or 10 and losing 100 1000 10000 to p2, and the remaining three players carrying out the above scenario:

These expectations combine to:

<p1> = 5.5 <p2> = 2241.1 <p3> = 2376.1 <p4>= 4266.1

Similar analysis for p12345 fighting over 1 10 100 1000 10000 100000 proceeds with p1 keeping 1 or 10 and losing anything else to p2. Then the remaining four carry out the above scenario.

These expectations combine to:

<p1> = 5.5 <p2> = 18534.25 <p3> = 18624.25 <p4> = 19591.75 <p5> = 35836.75

The sum of the expectations are in every case the sum of the checks multiplied by (n-1)/n.

That is, the expectations are approximately:

(sum of checks) x {0, 1/n, 1/n, 1/n, ... 1/n, 2/n} x (n-1)/n. The sequence is strictly increasing.

The recursive analysis works for different selections of the checks because they differ enough (factors of ten) that p2's best strategy is in all cases to take p1's check if and only if it is not one of the two smallest.

Let two players fight over 1 10 100.

- p1 draws 1 <p1> = 1

p2 draws from 10 100 <p2> = 55 - p1 draws 10 <p1> = 10

p2 draws from 1 100 <p2> = 50.5 (better than stealing p1's 10) - p1 draws 100

p2 steals 100 <p2> = 100

p1 draws from 1 10 <p1> = 5.5

<p1> = 5.5 <p2> = 68.5

Now let three players fight over 1 10 100 1000

- p1 draws 1 <p1> = 1

p23 fight over 10 100 1000 <p2> = 55 <p3> = 685 - p1 draws 10 <p1> = 10

p23 fight over 1 100 1000 <p2> = 50.5 <p3> = 683.5 (better than stealing p1's 10) - p1 draws 100

p2 steals 100 <p2> = 100 otherwise <p2> = 5.5

p13 fight over 1 10 1000 <p1> = 5.5 <p3> = 668.5 - p1 draws 1000

p2 steals 1000 <p2> = 1000

p13 fight over 1 10 100 <p1> = 5.5 <p3> = 68.5

<p1> = 5.5 <p2> = 301.375 <p3> = 526.375

Similar analysis for p1234 fighting over 1 10 100 1000 10000 proceeds with p1 keeping 1 or 10 and losing 100 1000 10000 to p2, and the remaining three players carrying out the above scenario:

These expectations combine to:

<p1> = 5.5 <p2> = 2241.1 <p3> = 2376.1 <p4>= 4266.1

Similar analysis for p12345 fighting over 1 10 100 1000 10000 100000 proceeds with p1 keeping 1 or 10 and losing anything else to p2. Then the remaining four carry out the above scenario.

These expectations combine to:

<p1> = 5.5 <p2> = 18534.25 <p3> = 18624.25 <p4> = 19591.75 <p5> = 35836.75

The sum of the expectations are in every case the sum of the checks multiplied by (n-1)/n.

That is, the expectations are approximately:

(sum of checks) x {0, 1/n, 1/n, 1/n, ... 1/n, 2/n} x (n-1)/n. The sequence is strictly increasing.

The recursive analysis works for different selections of the checks because they differ enough (factors of ten) that p2's best strategy is in all cases to take p1's check if and only if it is not one of the two smallest.