A new puzzle

[QuikSand]

Got this idea... I'm hopeful that this fairly simple puzzle is provocative enough to be worth solving, and there are any number varieties that seem like they could pop up from the general construct.

Here goes:

You are faced with three opaque boxes. One contains all white chips, worth $1. One contains all red chips, worth $5. The other is split 50/50 with the two colors, and has a 50/50 chance of releasing a white or red chip (as if the chip were replaced after being taken out).

The game is to draw whatever number of chips you want from whatever combination of boxes, and correctly identify the contents of each box, with your one guess. A wrong answer earns you zero and ends the game.

Your prize for a correct answer is $100 minus the dollar value of all the chips you drew from the boxes. So, if you draw three whites and three reds, they add up to $18, and you would win $82.

What is your strategy in this game?

[Warhammer]

Not sure I follow, do I get draws before my guess? Do I guess the individual box, and if correct get to keep guessing what the next item is going to be?

[JPhillips]

Obviously, my "strategy" is to make two guesses, get one red and one white, then make one guess, then know all three boxes.

But, I guess this comes down to what's your comfort level for "knowing" which single color you have? I'd say if I get three consecutive in one color I'd take those odds and guess that color. That should give me a maximum loss of $20, three red and a red on box two which would mean a mixed guess after one draw.

[tyketime]

The first thing I do is take one chip from each of the three boxes, and the result will be one of the following two outcomes (for simplicity, I refer to the boxes as #1/2/3 since I don't think the order would matter):

1. Box 1 = white, Box 2 = red, Box 3 = white

2. Box 1 = white, Box 2 = red, Box 3 = red


Whichever result gives me two chips of the same color tells me that the 3rd box with the opposite color is correct. And I would now focus on the two remaining boxes. I keep alternating opening those two boxes until a chip of the other color appears. I can then confirm that is the 50/50 two-colored box.


Here is an example to make it easier to illustrate:

I open up all three boxes and get the following result:
Box 1 = white
Box 2 = red
Box 3 = white

I can now automatically state that Box 2 will always be red.

I then alternate opening Box 1 & Box 3 until a red chip appears. The box with the red chip must be the two-colored 50/50 box, and the remaining box is always white.

[AnalBumCover]

If after the first two pulls (1 from each of the first 2 boxes), I get 2 whites, I do not pull the third box. That will be the red box (and I just saved $5 from not having to pull from that box).

[tyketime]

Originally Posted by AnalBumCover

If after the first two pulls (1 from each of the first 2 boxes), I get 2 whites, I do not pull the third box. That will be the red box (and I just saved $5 from not having to pull from that box).

Good point!

[AnalBumCover]

So how to minimize money lost? We need think of the strategy after what the first 2 pulls give us. There are three results.

Red/White (or White/Red)
White/White
Red/Red

[QuikSand]

In theory, we should be willing to draw more $1 chips in the white/white outcome than we would be willing to in the red/red outcome, right? We can only be certain once we get a mix of chips from one box, but it seems it would be worth it to keep drawing whites longer than reds...without yet doing the math here.

[AnalBumCover]

Originally Posted by QuikSand

In theory, we should be willing to draw more $1 chips in the white/white outcome than we would be willing to in the red/red outcome, right? We can only be certain once we get a mix of chips from one box, but it seems it would be worth it to keep drawing whites longer than reds...without yet doing the math here.

Yeah, basically it's a question of how many pulls off the white box before we decide to just throw out a guess and risk whatever balance we have left?

[JPhillips]

Originally Posted by AnalBumCover

So how to minimize money lost? We need think of the strategy after what the first 2 pulls give us. There are three results.

Red/White (or White/Red)
White/White
Red/Red

You could know with only a 7$ loss, white, white, red. But you could also be 20$ down, red, red, red, red and still be 50/50 on a guess.

I think this eventually comes down to what level of risk is acceptable to you. How many same color in row do you need to feel confident of a guess if you still hadn't solved the problem?

[QuikSand]

Originally Posted by JPhillips

I think this eventually comes down to what level of risk is acceptable to you.

No, I think this is solvable. If you assume constant value of dollars and no bias for/against taking a risk, I think you can maximize EV in this game perfectly. It's not a judgment call.

At some point down each decision tree, the cost of getting one more chip and losing either $1 or $5 in winnings is more than the added certainty you'd get by drawing again. It's just a bit of brute force to get there.

[JPhillips]

Originally Posted by QuikSand

No, I think this is solvable. If you assume constant value of dollars and no bias for/against taking a risk, I think you can maximize EV in this game perfectly. It's not a judgment call.

At some point down each decision tree, the cost of getting one more chip and losing either $1 or $5 in winnings is more than the added certainty you'd get by drawing again. It's just a bit of brute force to get there.

Yeah. I'm not enough of a gambler or mathematician to do that. I'd just have to go on comfort level or read what you guys eventually figure out.

[tarcone]

I draw 4 from each box.
If I get all whites Im pretty certain its the white box. Same for the red. And most likely, I will get a combo of red and white from the 3rd.

I spend $40 at most to find out if Im correct.

[Radii]

I'm pretty sure that someone is going to tell me my reasoning/math is horribly flawed, but I'm going to go ahead and hop in anyway.

Lets assume a worst case scenario. We pull one from each box, we get a white and two reds. We have $89 left. We're now going to alternate drawing between the boxes that contained the red chips until we see a white chip and can solve the puzzle, or until its not worthwhile to do so anymore.


Is it correct to say that for every draw from now on there is a 25% chance of drawing a white chip and solving the puzzle with certainty, and a 75% chance that we draw another red and lose $5 and we still have a 50/50 guess? Is that thinking flawed? This is what everything is based on.

So when we draw with $89 left we have a 25% chance of drawing a white chip and winning $88. The value of this draw is $22 (88/4), but we're only spending $5 on it. Its worth the draw. We can continue this exercise with every red draw for quite some time until we get to the following:

With $24 left from continually drawing $5, one more draw:

25% chance of winning 23. The value of this draw is $5.75. If we draw red, we're down to $19, the value of our next draw would be $4.50 (a 25% chance to win $18). We're better off guessing once we hit $19.


It feels like there's got to be a better approach? But there's my brute force attempt to examine the value of each individual draw in a worst case scenario. I look forward to seeing what the better approach is that leaves you with a much higher floor.


Also, the opportunity for this is just too great:


You better sticking to something else. if you draw red chip this many times in a row. Opaque box game might not be your game.

[Radii]

dola, for the purposes of this puzzle I'm assuming an infinitely large box so we are not able to change our odds based on our results so far, or at least a sufficiently large enough box that the odds don't change in the number of draws required to reach our limit where it becomes better to guess than to waste another draw.

[Carman Bulldog]

So I think there are three different strategies to employ and this is dependent on what is drawn with the first two chips. For every strategy, I think the first move is to pull a chip from Box 1 and Box 2, leaving you with an outcome of either two reds, two whites, or one of each.

Strategy A - Red Chips in Box 1 and 2

If left with two reds, this obviously means the third box is white. In this instance, I think the best move is to alternate pulling chips from each red box (Box 1 and Box 2). For example, let's say you pull a second red chip from Box 1. Instead of potentially pulling a third red chip from Box 1 in an attempt to confirm that it's the red box, I think you are better off pulling a second chip from Box 2 in the hopes that it is a white.

The reason being is that at that point the pull from Box 1 has a greater likelihood of costing you $5 (as this may be the "red" box) while the pull from box 2 should generally give you the same information. Keep alternating boxes until you get a white. This is generally the worst case scenario but I think I like the alternate boxes strategy better here as you are generally just throwing your money away pulling repeated attempts from a box that keeps giving you reds.

Strategy B - Both White Chips in Box 1 and 2

Similar to Strategy A, but instead of alternating pulls from Box 1 and 2, keep pulling from the same box repeatedly. You want to keep pulling whites, as this only costs you $1. Pull from one box until you are satisfied. This is the best case scenario.

Strategy C - One White Chip and one Red Chip in Box 1 and 2

The common consensus seems to be pull one from Box 3. I'm not sure that I agree with this. In this example, let's say that Box 1 was White and Box 2 was Red. I think I would target the "white" box 1 with a few more pulls. Sure, if we pull from box 3 then we have one box "confirmed" but if we instead pull a red chip from the "white" box, we have now solved the puzzle. Box 1 being Red/White, Box 2 being Red and Box 3 obviously being white. Essentially, you have the potential to solve the puzzle with your third pull, whereas by drawing from Box 3, you are forced to draw at least once more.

I'm open to suggestions on this however.

[cuervo72]

I think that's a good setup but I'm not sure alternating boxes in strategy A is the right way to go - I think you should keep pulling from the same box. If you pull red/red, you've pulled three reds from that box - 12.5% chance of that happening if that's the mixed box. So you're at a 12.5% chance of getting $0 and 87.5% of $79 at that point? So on average you make $69.13.

If you pick another from box 1 and it is red again, there's a 6.25% chance of that happening...so 6.25% @ 0, 93.75% of $74 = $69.375 EV? That's probably your stopping point as you could only make a max of $69 if you pull another red.

(ok, there's maybe a little more math there because there's still the chance you pull a white and are guaranteed $73 but I'm not dealing with any more math, was up too early)

[AnalBumCover]

Since we're given the option to make a blind guess, maybe we should compare the EV of the guess vs the EV of pulling the next chip.

If the EV(guess) exceeds EV(pull), then at that point we should stop.

[AnalBumCover]

Originally Posted by AnalBumCover

Since we're given the option to make a blind guess, maybe we should compare the EV of the guess vs the EV of pulling the next chip.

If the EV(guess) exceeds EV(pull), then at that point we should stop.

So thinking about this further, after 2 (or 3) pulls, you'll know the color of exactly one of the boxes. You'll need to pull from the remaining two boxes to verify their colors. By pulling from the remaining two, you have 25% chance of pulling the correct color chip (0% from one box, 50% from the other).

The EV of guessing is 50% of the current amount of the prize.
The EV of pulling the correct chip is 25% of the current prize amount minus the cost of that chip.

So, it makes sense to simply guess immediately after determining the color of one box.

[MIJB#19]

There are six possible distributions of the chips:
box A = mix, box B = red, box C = white
box A = mix, box B = white, box C = red
box A = red, box B = mix, box C = white
box A = red, box B = white, box C = mix
box A = white, box B = mix, box C = red
box A = white, box B = red, box C = mix


With no information, any of the six distributions is possible, so guessing correctly is a 1 in 6 chance to be correct. EV: $100/6=$16.666


With information from box A containing red chip, four distributions are possible, so guessing correctly is a 1 in 4 chance to win $95. EV: ($100-$5)/4 = $23.75
With information from box A containing white chip, four distributions are possible, so guessing correctly is a 1 in 4 chance to win $99. EV: ($100-$1)/4 = $24.75
Drawing a single chip has +EV: ($23.75+$24.75)/2-$16.666= +$7.783


Is a second drawn chip profitable?

If we draw from the same box:
If box A contained red and we pull red from box A, there are still four distributions remaining (1 in 4 to get $90). EV: ($100-$5-$5)/4=$22.5
If box A contained red and we pull white from box A, there are only two distributions remaining (1 in 2 to get $94). EV: ($100-$5-$1)/2=$47
This move is still better than stopping after 1 chip drawn. EV ($22.5+$47)/2=$34.75

If box A contained white and we pull red from box A, there are two distributions remaining. EV: ($100-$1-$5)/2=$47
If box A contained white and we pull white from box A, there are four distributions remaining. EV: ($100-$1-$1)/4=$24.5
This move is still better than stopping after 1 chip drawn. EV ($47+$24.5)/2=$35.75

Drawing the second chip from the same box after red has +EV $34.75-$23.75 = +$11
Drawing the second chip from the same box after white has +EV $35.75-$24.75 = +$11

if we draw from box B:
If box A contained red and we pull red from box B, there are two distributions remaining (1 in 2 to get $90). EV: ($100-$5-$5)/2=$45
If box A contained red and we pull white from box B, there are three distributions remaining (1 in 3 to get $94). EV: ($100-$5-$1)/3=$31.333
($45+$31.333)/2=$38.166
Taking the chip is $38.166-$23.75 = +$14.416 EV

If box A contained white and we pull red from box B, there are three distributions remaining. EV: ($100-$1-$5)/3=$31.333
If box A contained white and we pull white from box B, there are two distributions remaining. EV: ($100-$1-$1)/2=$49
($31.333+$49)/2=$40.166
Taking the chip is $40.166-$24.75 = +$15.416 EV

All in all, drawing a second chip, regardless of the box is +EV, noting it's better to draw from box B.



Drawing a third chip?
We'll continue with the 4 scenarios with 1 chip from box A and 1 chip from box B.

In case of two reds, the only way to get better information is by pulling from either of the red boxes. Will it improve EV?
draw white from box A (25% chance), one distribution left. Winnings: ($100-$5-$5-$1)=$89
draw second red from box A (75% chance), two distributions, 1 in 2 chance to guess correctly. EV ($100-$5-$5-$5)/2=$42.5
EV = ($89*1+$42.5*3)/4=$54.125
Still +$9.125 EV compared to the $45 we had.

In case of two whites, the only way to get better information is by pulling from either of the white boxes.
draw red from box A (25% chance), one distribution left. Winnings: ($100-$1-$1-$5)=$93
draw second white from box A (75% chance), two distributions, 1 in 2 chance to guess correctly. EV ($100-$1-$1-$1)/2=$43.5
EV = ($93*1+$43.5*3)/4=$55.375
Still +6.375 from the $49 we had

In case of the mixed bag (one red, one white), let's consider the options. We could either draw from the unknown box to improve to a 50/50 situation, or we can redraw from the white box with a 25% to pin it down to a single distribution.

draw from box C:
draw red from box C (50% chance), two distributions, 1 in 2 chance to guess correctly. EV ($100-$5-$1-$5)/2=$44.5
draw white from box C (50% chance), two distributions, 1 in 2 chance to guess correctly. EV ($100-$5-$1-$1)/2=$46.5
Average: ($44.5+$47)/2=$45.5
EV $45.5-$31.333 = +$14.166

draw from box white:
draw red from box white (25% chance), one distribution left. Winnings: ($100-$5-$1-$5)=$89
draw second white from box white (75% chance), still three distributions, 1 in 3 chance to guess correctly. EV ($100-$5-$1-$1)/3=$31
Average: ($89*1+$31*3)/4=$45.5
EV $45.5-$31.333 = +$14.166
Short term, this is actually just as good as drawing from box C.

Fourth and further draws will mostly increase EV, but it depends on how many red chips were drawn earlier. More red will more quickly force you to make a 50/50 guess between the all red box and the mixed box. More white will buy you time to have to wait for the guess between all white box and the mixed box.

Once we've drawn similar colors from two different boxes and no other colors from those boxes yet, I see no reason to alternate between boxes or keep drawing from the same box. The odds of drawing the other color will always be 1 in 4.

[QuikSand]

Originally Posted by AnalBumCover

Since we're given the option to make a blind guess, maybe we should compare the EV of the guess vs the EV of pulling the next chip.

If the EV(guess) exceeds EV(pull), then at that point we should stop.

yes

[QuikSand]

Originally Posted by MIJB#19

With information from box A containing red chip, four distributions are possible, so guessing correctly is a 1 in 4 chance to win $95. EV: ($100-$5)/4 = $23.75
With information from box A containing white chip, four distributions are possible, so guessing correctly is a 1 in 4 chance to win $99. EV: ($100-$1)/4 = $24.75
Drawing a single chip has +EV: ($23.75+$24.75)/2-$16.666= +$7.783

I disagree with your math here, but not with your overall conclusion.

The four possible distributions left after pulling one chip are not equally likely, as you assume here. This is, essentially, the Monty Hall problem redux. It's 2/3 likely that the box you just drew the chop from is ALL that color, and only 1/3 likely that it's HALF that color.

[MIJB#19]

Originally Posted by QuikSand

I disagree with your math here, but not with your overall conclusion.

The four possible distributions left after pulling one chip are not equally likely, as you assume here. This is, essentially, the Monty Hall problem redux. It's 2/3 likely that the box you just drew the chop from is ALL that color, and only 1/3 likely that it's HALF that color.

I see your point.

box Red: red, red
box Mix: red, white
box White: white, white

The first red chip is 1/3rd of the time from the mixed box and 2/3rd of the time from the all red box. (Ditto for white).

If box A contained red, there are 4 possibilities:
box A = red, box B = mix, box C = white
box A = red, box B = white, box C = mix
box A = mix, box B = red, box C = white
box A = mix, box B = white, box C = red

My mistake here was to claim all four options are equally likely, which isn't true, because the last two options are half as likely. We should basically approach that as 6 options:
box A = red, box B = mix, box C = white
box A = red, box B = mix, box C = white
box A = red, box B = white, box C = mix
box A = red, box B = white, box C = mix
box A = mix, box B = red, box C = white
box A = mix, box B = white, box C = red

Which means the first two options are 'better' guesses than the last two.

[britrock88]

My one critique of the reasoning I've seen is that it makes less sense to alternate drawing from two boxes that give the same initial chip than to draw repeatedly from one of those boxes. Doing that works toward a solution in the vein of binomial probability--once you've drawn 5 red chips from the same box, there's only a 1/32 (3.125%) chance it's the mixed-chip box.

[QuikSand]

Originally Posted by britrock88

My one critique of the reasoning I've seen is that it makes less sense to alternate drawing from two boxes that give the same initial chip than to draw repeatedly from one of those boxes. Doing that works toward a solution in the vein of binomial probability--once you've drawn 5 red chips from the same box, there's only a 1/32 (3.125%) chance it's the mixed-chip box.

Right, agreed. As we continue to draw white chips, the likelihood that we have the all-white box is greater. At some point, it's no longer worth it to draw another chip, because of the certain loss of $1 or $5 from the eventual prize.

I think the point is that when drawing from a box that we know can generate a red chip, the greater likelihood of a $5 loss in the prize makes that decision point arrive more rapidly.

Indeed, it might not make sense to ever draw a second chip from a box that generated a red chip, unless it's literally the only option remaining (like a 1->R, 2->R start).