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Do you switch boxes? (Glitch in the matrix)
Do you switch boxes? (Glitch in the matrix)
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08-15-2021, 01:36 PM
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#1
- numberguy12
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Do you switch boxes? (Glitch in the matrix)
(jk the world isn’t a simulation or matrix and that’s all nonsense, but still if it were one, this puzzling situation might just be the glitch).
There are two closed boxes in front of you, each containing just a stash of money in the form of $ bills. All you are told is one of the boxes has twice the amount of money the other box has. You are to receive the contents of exactly one of the boxes.
So, you pick one box at random, open the box, and it has $40 in it. You are given the option to keep that, or to switch boxes, and walk away with whatever sum is in the other box instead. Would it be in your best interest to switch? (assume you want to maximize the expected value of the amount you are walking away with).
There are two closed boxes in front of you, each containing just a stash of money in the form of $ bills. All you are told is one of the boxes has twice the amount of money the other box has. You are to receive the contents of exactly one of the boxes.
So, you pick one box at random, open the box, and it has $40 in it. You are given the option to keep that, or to switch boxes, and walk away with whatever sum is in the other box instead. Would it be in your best interest to switch? (assume you want to maximize the expected value of the amount you are walking away with).
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08-15-2021, 01:40 PM
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#2
Check the next one.
I’ll be ahead regardless and there is no cost to play? So go home with either +20 or +80 over what I had to start with. 50/50 odds are good in gambling
I’ll be ahead regardless and there is no cost to play? So go home with either +20 or +80 over what I had to start with. 50/50 odds are good in gambling
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08-15-2021, 01:42 PM
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#3
- infinityplus1
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- infinityplus1
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Switvh boxes, before you walked in you had exactly $0 so leaving with either $20, $40 or $80 you are still up
08-15-2021, 01:46 PM
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#4
- Tanerian
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Originally Posted By Anachron⏩
expected value wise, this.If you switch, you either lose $20 from what you already have, or get $40 more. It is in your interest to switch.
The better thought experiment is the 3 box one.
You choose between 3 boxes. 1 box has a million bucks, the other 2 are filled with dirt.
You choose yours at random. Before opening it, one of the other boxes is opened to reveal dirt. You are then given the choice to switch your box with the last remaining box. Do you do it?
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08-15-2021, 01:46 PM
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#5
Originally Posted By Anachron⏩
This. The box I have is worth x and the value of the other box is either .5x or 2x, an average value of 1.25x. You should switch.If you switch, you either lose $20 from what you already have, or get $40 more. It is in your interest to switch.
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08-15-2021, 01:55 PM
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#6
08-15-2021, 01:55 PM
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#7
- numberguy12
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Originally Posted By Tanerian⏩
Ah, that’s the Monty Hall problem, which is readily solved with standard conditional probability considerations (you should switch, if a few assumptions are added to that statement).expected value wise, this.
The better thought experiment is the 3 box one.
You choose between 3 boxes. 1 box has a million bucks, the other 2 are filled with dirt.
You choose yours at random. Before opening it, one of the other boxes is opened to reveal dirt. You are then given the choice to switch your box with the last remaining box. Do you do it?
The better thought experiment is the 3 box one.
You choose between 3 boxes. 1 box has a million bucks, the other 2 are filled with dirt.
You choose yours at random. Before opening it, one of the other boxes is opened to reveal dirt. You are then given the choice to switch your box with the last remaining box. Do you do it?
I’d argue this problem is actually much more interesting.
So to Anachron, who reasonably concludes that you should switch because the options are 1)you gain $40 or 2)lose $20, is the $40 even important here?
Wouldn’t that be the case also if it were, say $100, or $1000, or $A, you should switch? (Assuming you want to maximize expected value).
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08-15-2021, 01:58 PM
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#8
- elterrible987
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- elterrible987
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if you are panicking over whether you get 20, 40 or 80 dollars then you failed in life
08-15-2021, 02:01 PM
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#9
Originally Posted By MiscMathematician⏩
Similar thing goes for gambling. Regardless of how many times you play, your odds of winning are the same. Playing 100 times does not affect your chance to win.This is why many people can play hundreds, if not thousands of times and never win, but someone can come along and place one bet and win big.
One party system; Most Republicans are Democrats, but no Democrats are Republicans.
Hayek and Mises were right; they're all socialists.
"To Call something fair or unfair is a subjective value judgment and not liable to any verification" Ludwig Von Mises
08-15-2021, 02:06 PM
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#10
- elterrible987
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Originally Posted By numberguy12⏩
that was the begining of modern simpdom when ann landers solved the problem with freshmen level math and people claimed she was a super genius for it and men werent giving her credit because muh misogyny. she never invented anything and her claim to fame was running a news paper column, not even being a full time journalist. i guess the standards to be called a male genius are much higher like inventing something or discovering something newAh, that’s the Monty Hall problem, which is readily solved with standard conditional probability considerations (you should switch, if a few assumptions are added to that statement).
I’d argue this problem is actually much more interesting.
So to Anachron, who reasonably concludes that you should switch because the options are 1)you gain $40 or 2)lose $20, is the $40 even important here?
Wouldn’t that be the case also if it were, say $100, or $1000, or $A, you should switch? (Assuming you want to maximize expected value).
I’d argue this problem is actually much more interesting.
So to Anachron, who reasonably concludes that you should switch because the options are 1)you gain $40 or 2)lose $20, is the $40 even important here?
Wouldn’t that be the case also if it were, say $100, or $1000, or $A, you should switch? (Assuming you want to maximize expected value).
08-15-2021, 02:07 PM
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#11
- numberguy12
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Just so it is clear, this is not dependent on what your own gambling preferences are (you feelin lucky?, etc).
It is asking whether the strategy of switching results in a greater expected winning than just keeping the contents of the box.
To MrBourbon. Couldn’t you apply that reasoningbeforeyou even open the initial box, and conclude that the second box has on average 1.25x the amount, so it is the better box to go with. Anyone see how this is absurd?
It is asking whether the strategy of switching results in a greater expected winning than just keeping the contents of the box.
To MrBourbon. Couldn’t you apply that reasoningbeforeyou even open the initial box, and conclude that the second box has on average 1.25x the amount, so it is the better box to go with. Anyone see how this is absurd?
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08-15-2021, 02:09 PM
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#12
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Originally Posted By elterrible987⏩
I believe you are referring to Marilyn vos Savantthat was the begining of modern simpdom when ann landers solved the problem with freshmen level math and people claimed she was a super genius for it and men werent giving her credit because muh misogyny. she never invented anything and her claim to fame was running a news paper column, not even being a full time journalist. i guess the standards to be called a male genius are much higher like inventing something or discovering something new
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08-15-2021, 02:16 PM
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#13
Originally Posted By Kraken⏩
Yep and you can make the "correct" play oddswise and still lose. The idea is if you could play the same hand 1000 times you'd come out ahead.Similar thing goes for gambling. Regardless of how many times you play, your odds of winning are the same. Playing 100 times does not affect your chance to win.
This is why many people can play hundreds, if not thousands of times and never win, but someone can come along and place one bet and win big.
This is why many people can play hundreds, if not thousands of times and never win, but someone can come along and place one bet and win big.
Smooth Seas don't make Strong Sailors. Keep your head up.
MrWhiskey24 for jolly cooperation (PS)
08-15-2021, 02:18 PM
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#14
- numberguy12
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So to be clear, you have a totally symmetric situation in front of you: just two closed boxes. And you came to the conclusion that one of the boxes has a higher expected value.
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08-15-2021, 02:57 PM
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#15
- numberguy12
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Originally Posted By Anachron⏩
Name the closed boxes at the beginning Box 1 and Box 2.$40 is inconsequential.
It's a 50/50 split regardless, since there are only two boxes.
A better question would be, if the other box is of lower value, you get nothing. How would you choose then?
It's a 50/50 split regardless, since there are only two boxes.
A better question would be, if the other box is of lower value, you get nothing. How would you choose then?
So you’d agree with MrBourbon that without anything being opened yet, you could apply this reasoning to Box 1 (and conclude that whatever you open in Box 1, Box 2 has on average 1.25x Box 1, and you should therefore go with Box 2). All this before anything is opened.
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08-15-2021, 03:07 PM
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#16
- numberguy12
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Originally Posted By Anachron⏩
Back to your first post. You noted that if you switch, you gain $40 (with 50% chance), or lose $20 (with 50% chance).Nope.
With only two boxes, it's a pure 50/50, regardless whether or not you open the first box.
I think it becomes more psychological after opening the first box - i.e. is it a life changing amount to begin with.
With only two boxes, it's a pure 50/50, regardless whether or not you open the first box.
I think it becomes more psychological after opening the first box - i.e. is it a life changing amount to begin with.
Whatever is in Box 1, say it is $A: Surely according to this logic you would either gain $A (larger) or lose $A/2 (smaller) by switching, thus making an argument for Box 2.....before even opening a single box.
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08-15-2021, 03:28 PM
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#17
- numberguy12
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Originally Posted By Anachron⏩
It doesn’t seem psychology is at play here when we are merely talking about expected value and maximizing this.There is no Box 1 and Box 2.
The only thing opening the first box does, it gives you an indication of the range of the game. This is not a logical query, it is a psychological one.
The only thing opening the first box does, it gives you an indication of the range of the game. This is not a logical query, it is a psychological one.
MrBourbon (and essentially you in the first post) for example come to the conclusion that Box 2 has an expected value of 1.25A, where A is the amount in Box 1. It doesn’t seem like it matters whether Box 1 is opened or not: you could make that exact argument about expected value of Box 2 regardless of what the A turns out to be....so arguing for Box 2 in a symmetric situation when neither box is opened yet.
This of course is an absurd situation. You could apply the very same reasoning to Box 2, and conclude Box 1 is the better box to go with.
Here is a devils advocate argument.
Let the contents of the boxes in front of you be x and 2x. These are fixed.
If you picked the 2x, switching brings you back down to x (so losing x). Remembering x is the smallest of the two boxes, a fixed amount.
If you picked the x, switching brings you up to 2x (gaining x).
So the amount you could gain by switching would be the same as the amount could lose (contradicting some posts above).
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08-15-2021, 03:36 PM
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#18
- numberguy12
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Originally Posted By Anachron⏩
There isn’t really a gotcha. This is a famous problem, and it surprisingly isn’t fully answered, although a ton has been written on it. Just curious what people think about it.I am disappointed in this entire thread, especially considering you are the OP - if it is a gotcha, get around to it.
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08-15-2021, 03:47 PM
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#19
- numberguy12
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What if the amount in the box you opened was $1125, and not $40 like in the OP? Would you keep?
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