[superprismatic]

Consider the following game:

There are 10 opaque boxes with labels

A1, A2, A3, A4, A5, B1, B2, B3, B4, and B5.

One of the boxes is chosen at random and

$14 is placed inside of it. The other boxes

contain nothing. The boxes are then stacked

in 2 stacks -- boxes A1, A2, A3, A4, and A5

are arranged in order into stack A with A1 on

top; similarly, boxes B1, B2, B3, B4, and B5

are arranged in order into stack B with B1 on

top. To play this game, you must pay to

inspect boxes (explained in the next paragraph)

until you find, and get to keep, the $14.

You may only take boxes from the top

of a stack and pay, in dollars, whatever

numeral is on its label. You may then either

take (and pay appropriately for) the next box

in that stack or switch to the other stack.

BUT, if you switch to the other stack, you must

restore the stack you have been using to its

original order before doing so. You must pay

for every box which you take off any stack

even if you had previously paid for it. The

game ends when you find (and get to keep) the

$14.

What is the best strategy for playing

this game? If you follow the best strategy

with every play, how much, on average, would

you expect to win on each play?

[Guest]

Ah, this is quite easy. The top two boxes average the best profit, which are the only two profitable ones. Meaning A1 and B1.

Every box has 10% to contain the riches, meaning on average every box is worth 1.4 dollars. The only boxes worth the risk on prolonged tries are the top ones. Top boxes average a .4 dollar profit each try, so it means, that the top two boxes profit .8 dollars per game on average. That's the best you'll get

[Glycereine]

Ah, this is quite easy. The top two boxes average the best profit, which are the only two profitable ones. Meaning A1 and B1.

Every box has 10% to contain the riches, meaning on average every box is worth 1.4 dollars. The only boxes worth the risk on prolonged tries are the top ones. Top boxes average a .4 dollar profit each try, so it means, that the top two boxes profit .8 dollars per game on average. That's the best you'll get

Agree with everything but went about it a different way:

open box 1 of each stack each try means 20% chance of winning. In theory 1 in every 5 games should be a winner. 10 dollars spent on 5 games and 14 dollars won means a 4 dollar profit.

4/5 = 80 cents per game profit.

[superprismatic]

Ah, this is quite easy. The top two boxes average the best profit, which are the only two profitable ones. Meaning A1 and B1.

Every box has 10% to contain the riches, meaning on average every box is worth 1.4 dollars. The only boxes worth the risk on prolonged tries are the top ones. Top boxes average a .4 dollar profit each try, so it means, that the top two boxes profit .8 dollars per game on average. That's the best you'll get

You can't stop playing the game whenever you want. The statement "you must pay to inspect boxes (explained in the next paragraph) until you find, and get to keep, the $14" was meant to mean exactly that. So, you have to keep playing (and paying) until you find the $14. Sorry.

[k-man]

I like this puzzle! Even though it's not very difficult it has an answer that is not intuitive.

At first I thought that the best approach would be going through one stack and if the money is not found go through another stack. With this strategy you will spend $14.50 on average if you play long enough, so it's a losing strategy. Since the money is randomly placed in one of the boxes the probability of any box containing the money is equal. Here are the possible outcomes with this strategy:

A1: $1

A2: $3

A3: $6

A4: $10

A5: $15

B1: $16

B2: $18

B3: $21

B4: $25

B5: $30

The average of these outcomes is $14.50.

I found 2 other strategies that are equally good and will win $.10 per play on average. In other words on average they cost $13.90 to play. Basically, the idea is that you take to top box (or top 2 boxes) from one stack and if the the money isn't there you go through another stack before going back thorough the first. Here are the outcomes and the sequences:

Try A1, then go through B stack and finally back to A:

A1: $1

B1: $2

B2: $4

B3: $7

B4: $11

B5: $16

A2: $19

A3: $22

A4: $26

A5: $31

Try A1 and A2, then go through B stack and finally back to A:

A1: $1

A2: $3

B1: $4

B2: $6

B3: $9

B4: $13

B5: $18

A3: $24

A4: $28

A5: $33

As I already stated both these strategies average $13.90.

Attempting to continue this pattern and try 3 boxes from A before going to B will cost you $14.20 per play.

[Guest]

Agree with what K-man wrote (not completely though).

I worked slightly differently and found that although both strategies give the same expected benefit of 10 cents, the sequence of looking into A1, A2 then switching to B1 to B5 and back to A1 to A5 gives a positive expected outcome for first 6 boxes checked while the strategy of looking into A1 then switching to B1 to 5 and back to A1 to 5 has only first 5 boxes that give positive expected result. See attached figs.

[superprismatic]

I like this puzzle! Even though it's not very difficult it has an answer that is not intuitive.

At first I thought that the best approach would be going through one stack and if the money is not found go through another stack. With this strategy you will spend $14.50 on average if you play long enough, so it's a losing strategy. Since the money is randomly placed in one of the boxes the probability of any box containing the money is equal. Here are the possible outcomes with this strategy:

A1: $1

A2: $3

A3: $6

A4: $10

A5: $15

B1: $16

B2: $18

B3: $21

B4: $25

B5: $30

The average of these outcomes is $14.50.

I found 2 other strategies that are equally good and will win $.10 per play on average. In other words on average they cost $13.90 to play. Basically, the idea is that you take to top box (or top 2 boxes) from one stack and if the the money isn't there you go through another stack before going back thorough the first. Here are the outcomes and the sequences:

Try A1, then go through B stack and finally back to A:

A1: $1

B1: $2

B2: $4

B3: $7

B4: $11

B5: $16

A2: $19

A3: $22

A4: $26

A5: $31

Try A1 and A2, then go through B stack and finally back to A:

A1: $1

A2: $3

B1: $4

B2: $6

B3: $9

B4: $13

B5: $18

A3: $24

A4: $28

A5: $33

As I already stated both these strategies average $13.90.

Attempting to continue this pattern and try 3 boxes from A before going to B will cost you $14.20 per play.

Nicely done, k-man! It's interesting to note that there are also 2 best solutions when there are 5 stacks of sizes 1,2,3,4, and 5. I wonder if there are always 2 for N stacks of sizes 1,2,3,...,N. Hmmmm.............

[superprismatic]

Agree with what K-man wrote (not completely though).

I worked slightly differently and found that although both strategies give the same expected benefit of 10 cents, the sequence of looking into A1, A2 then switching to B1 to B5 and back to A1 to A5 gives a positive expected outcome for first 6 boxes checked while the strategy of looking into A1 then switching to B1 to 5 and back to A1 to 5 has only first 5 boxes that give positive expected result. See attached figs.

Nice way to think of it, especially if there were some slight bias in which box is chosen for the $14. If there were such a bias, the two strategies would almost surely lead to different expected payoffs!