White Elephant Gift Exchange

[BMAD]

This question comes from a modified white elephant gift exchange game (and is a tip of the hat to the many swapping puzzles out there lately).

This version of 'White Elephant' is rather simple: There are six gifts and five people. Each gift is a check worth either $10, $100, $1,000, $10,000, and $100,000. No one knows which check is which as they are hidden in sealed envelopes. One person gets to select first. They open the gift and reveal the prize to the group. The person who goes second has a choice, they can take the gift from the first person or select another mystery envelope. If they select a new envelope, then like the first person they reveal the gift to the group and allow the next person an option to take any gift previously opened or open a new envelope. If a person elects to take an opened envelope from a previous person then that item is now permanently theirs and no one can take it from them. The person who lost their envelope now must select a new envelope from the unopened pile. Each person continues until they have an envelope (with one envelope left unopened). The order of selecting the gifts must be finalized before any picking starts.

What would be the best position to go in if you get to decide when you pick? What would be your expected winnings?

[bubbled]

This question comes from a modified white elephant gift exchange game (and is a tip of the hat to the many swapping puzzles out there lately).

This version of 'White Elephant' is rather simple: There are six gifts and five people. Each gift is a check worth either $10, $100, $1,000, $10,000, and $100,000. No one knows which check is which as they are hidden in sealed envelopes. One person gets to select first. They open the gift and reveal the prize to the group. The person who goes second has a choice, they can take the gift from the first person or select another mystery envelope. If they select a new envelope, then like the first person they reveal the gift to the group and allow the next person an option to take any gift previously opened or open a new envelope. If a person elects to take an opened envelope from a previous person then that item is now permanently theirs and no one can take it from them. The person who lost their envelope now must select a new envelope from the unopened pile. Each person continues until they have an envelope (with one envelope left unopened). The order of selecting the gifts must be finalized before any picking starts.

What would be the best position to go in if you get to decide when you pick? What would be your expected winnings?

I only see five gifts. This looks like an interesting puzzle, but I's like to know if there's a $1 gift or a $1M gift.

[BMAD]

This question comes from a modified white elephant gift exchange game (and is a tip of the hat to the many swapping puzzles out there lately).

This version of 'White Elephant' is rather simple: There are six gifts and five people. Each gift is a check worth either $1,$10, $100, $1,000, $10,000, and $100,000. No one knows which check is which as they are hidden in sealed envelopes. One person gets to select first. They open the gift and reveal the prize to the group. The person who goes second has a choice, they can take the gift from the first person or select another mystery envelope. If they select a new envelope, then like the first person they reveal the gift to the group and allow the next person an option to take any gift previously opened or open a new envelope. If a person elects to take an opened envelope from a previous person then that item is now permanently theirs and no one can take it from them. The person who lost their envelope now must select a new envelope from the unopened pile. Each person continues until they have an envelope (with one envelope left unopened). The order of selecting the gifts must be finalized before any picking starts.

What would be the best position to go in if you get to decide when you pick? What would be your expected winnings?

**Thank you for catching that

[bonanova]

I'm going to think it through more, but my first thought is picking last is not a bad option.

Clearly if you are one the first four and you select the big check you will lose it permanently, although possibly to another of the first four. Still, the last to pick has his choice of two unopened envelopes and the best of the unclaimed opened ones, including that of the 4th to pick, who may have opened the big check.

Last to pick therefore has at least a 1/5 chance of getting the big check.

[BMAD]

I'm going to think it through more, but my first thought is picking last is not a bad option.

Clearly if you are one the first four and you select the big check you will lose it permanently, although possibly to another of the first four. Still, the last to pick has his choice of two unopened envelopes and the best of the unclaimed opened ones, including that of the 4th to pick, who may have opened the big check.

Last to pick therefore has at least a 1/5 chance of getting the big check.

Since we can assume that each player will play the best strategy possible, i don't know if you can state that the last person would have a 1/5 chance of picking the 'big money'

[bonanova]

You're right - it's at least a 1/6 chance.

[bonanova]

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.

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

These expectations average out to be

<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

These expectations combine to:

<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.

[BMAD]

In this section...

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 (wouldn't this step and other similar ones need more delineation?)*** see below
• 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

These expectations average out to be
<p1> = 5.5 <p2> = 301.375 <p3> = 526.375

***If p2 gets the 10, then p3 gets 100 or 1000 = 550

If p2 gets 100 then p3 gets 10 or 1000 = 505

if p2 gets 1000 then p3 takes 1000 and p2 gets 10 or 100 = 55

[bonanova]

In this section...

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 (wouldn't this step and other similar ones need more delineation?)*** see below

• 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

These expectations average out to be

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

***If p2 gets the 10, then p3 gets 100 or 1000 = 550

If p2 gets 100 then p3 gets 10 or 1000 = 505

if p2 gets 1000 then p3 takes 1000 and p2 gets 10 or 100 = 55

Let p23 fight over 10 100 1000.

• p2 draws 10 <p2> = 10

  p3 draws from 100 1000 <p3> = 550

• p2 draws 100 <p2> = 100

  p3 draws from 10 1000 <p3> = 505 (better than stealing p2's 100)

• p2 draws 1000

  p3 steals 1000 <p3> = 1000

  p2 draws from 10 100 = 55

These expectations average out to be

<p2> = 55 <p3> = 685

[BMAD]

In this section...

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 (wouldn't this step and other similar ones need more delineation?)*** see below

• 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

These expectations average out to be

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

***If p2 gets the 10, then p3 gets 100 or 1000 = 550

If p2 gets 100 then p3 gets 10 or 1000 = 505

if p2 gets 1000 then p3 takes 1000 and p2 gets 10 or 100 = 55

nevermind... lol

[bubbled]

This question comes from a modified white elephant gift exchange game (and is a tip of the hat to the many swapping puzzles out there lately).

This version of 'White Elephant' is rather simple: There are six gifts and five people. Each gift is a check worth either $1,$10, $100, $1,000, $10,000, and $100,000. No one knows which check is which as they are hidden in sealed envelopes. One person gets to select first. They open the gift and reveal the prize to the group. The person who goes second has a choice, they can take the gift from the first person or select another mystery envelope. If they select a new envelope, then like the first person they reveal the gift to the group and allow the next person an option to take any gift previously opened or open a new envelope. If a person elects to take an opened envelope from a previous person then that item is now permanently theirs and no one can take it from them. The person who lost their envelope now must select a new envelope from the unopened pile. Each person continues until they have an envelope (with one envelope left unopened). The order of selecting the gifts must be finalized before any picking starts.

What would be the best position to go in if you get to decide when you pick? What would be your expected winnings?

**Thank you for catching that

I'm going to code this up later to see how these strategies do. But here is how I believe the five players should play:

#1: Just draws

#2: If player 1 drew 100 or higher, steal. Otherwise draw.

#3: If the 1st, 2nd or 3rd highest available values are possible to steal, steal. Otherwise draw.

#4: If the 1st or 2nd highest available values are possible to steal, steal. Otherwise draw.

#5: If the highest available value is possible to steal, steal. Otherwise draw.

By available, I mean it has either been drawn but not stolen yet, or it hasn't been drawn yet.

My intuition tells me that player 5 is best positioned, because he can win if 100,000 is drawn right before him or he draws it. The only way anyone else gets 100k, is if it's drawn right ahead of him. So 5 has two chances to win the big one.

Sadly with these strategies, player 1, never does better than 10.

[bubbled]

This question comes from a modified white elephant gift exchange game (and is a tip of the hat to the many swapping puzzles out there lately).

This version of 'White Elephant' is rather simple: There are six gifts and five people. Each gift is a check worth either $1,$10, $100, $1,000, $10,000, and $100,000. No one knows which check is which as they are hidden in sealed envelopes. One person gets to select first. They open the gift and reveal the prize to the group. The person who goes second has a choice, they can take the gift from the first person or select another mystery envelope. If they select a new envelope, then like the first person they reveal the gift to the group and allow the next person an option to take any gift previously opened or open a new envelope. If a person elects to take an opened envelope from a previous person then that item is now permanently theirs and no one can take it from them. The person who lost their envelope now must select a new envelope from the unopened pile. Each person continues until they have an envelope (with one envelope left unopened). The order of selecting the gifts must be finalized before any picking starts.

What would be the best position to go in if you get to decide when you pick? What would be your expected winnings?

**Thank you for catching that

I'm going to code this up later to see how these strategies do. But here is how I believe the five players should play:

#1: Just draws

#2: If player 1 drew 100 or higher, steal. Otherwise draw.

#3: If the 1st, 2nd or 3rd highest available values are possible to steal, steal. Otherwise draw.

#4: If the 1st or 2nd highest available values are possible to steal, steal. Otherwise draw.

#5: If the highest available value is possible to steal, steal. Otherwise draw.

By available, I mean it has either been drawn but not stolen yet, or it hasn't been drawn yet.

My intuition tells me that player 5 is best positioned, because he can win if 100,000 is drawn right before him or he draws it. The only way anyone else gets 100k, is if it's drawn right ahead of him. So 5 has two chances to win the big one.

Sadly with these strategies, player 1, never does better than 10.

I ran a simulation of 10,000,000 games to check bonanova's work.

5.500594

18535.0704661

18622.6823404

19570.8699793

35840.8914184