Bertrand's Box Paradox

[magicalhats15]

I think you're missing an important point here.

Say I put 50G medallions in the first drawer, 1G and 1S in the middle, and 1 S in the right one. Now, when I reach in to grab a medallion, and I feel 50 of them, and pull out one, then I already know that the drawer I pulled it out of is the left one. Similarly, if I feel only one medallion, I know it was the right one. Now, there's the obvious objection that I may not have felt around before grabbing, but with sufficient care, it's foolproof.

[marlboroman1985]

The "paradox" is the answer changes how you phrase the set-up. I assume you're right here, but I thought Betrand's Box Paradox actually asks "what the probability is of selecting another gold coin from the same box you chose the first gold coin" (not the probability of a box containing gold etc).

[xyaqua]

A paradox can never physically exist. Heck, even Leonard Nimoy (as Spock) said it:

"Nothing unreal exists"

It's so much fun though, how great the used of language is in telling an untruth...

I'd put more faith in the kid who questions 2+2=4 than the kid who blindly accepts the square root of -1 is i (or j). Intelligence required intllectual aggressiveness. Mind you- not beligerance, but a demanding of proof until the open mind of a skeptical child can see the reason of the noumena.

[Zaltor2]

If it can be explained so easily then why are we still calling it a paradox?

[xyaqua]

@Zaltor2 --If it can be explained so easily then why are we still calling it a paradox?

EXACTLY!

I have real trouble understanding Bertrand Russel's philosophy, but his logical computation is very impressive, when i can understand it.

[TheAlebandro]

stick or switch



[MariluLane]

also, also cousinomacal, i have a question for you on the monty hall problem.

what if, two people played the same monty hall game, at the same time. Three doors: A, B, C. Person X always chose door A then swapped. Person Y always chose door C then swapped. Only ONE person could see the 66% success rate. the other would HAVE to see the 33% sucess rate, even though they both swapped, right???!!!! ;)

how can we then say the 66% rate applies to EVERY game????

interested to hear your thoughts

[MariluLane]

even if they didn't always choose the same door (A for Z and B for Y), the point would be that they both chose a door different than the other, and then swapped over

[CousinoMacul]

There are two ways that this simultaneous game could be played. One is that neither player X nor Y sees which empty door is shown to the other. In that case, it is identical to two independent games. ⅓ of the time the prize will be behind door A so X will lose when he switches and Y will win, ⅓ it will be behind door C and Y will lose and X will win, and ⅓ of the time it will be behind door B and both X and Y will win when they switch. They will each win ⅔ of the time.

[CousinoMacul]

The other way is that each sees which empty door the other is shown. ⅔ of the time, the prize will be behind either door A or C and they will both (presumably) be shown door B to be empty. Each has a chance of ½ of winning by switching. ⅓ of the time the prize will be behing door B and each will be shown doors A and C to be empty and they'll be 100% sure to win by switching. Total probability = ⅓*1 + ⅔*½ = ⅔

[MariluLane]

Im not a mathmo so equation is difficult! Lol but you said, in the second way the game could be played, Each has a chance of ½ of winning by switching. Doesnt that mean the percentage changes from 66% to 50% if two people play at the same time?? This I guess is only relevant if X and Y both leave an empty door after theyve chosen (so that could be affecting the statistics) but take a single game, where they just make one choice...

[MariluLane]

... An empty door is left after X and Y chose, the correct door must be A or C. If they both switch, you say the 66% chance for each player falls to 50%, although if they made exactly the same choices on their own it would remain as 66%?

:)

[CousinoMacul]

It's not the same. In the second scenario there are two possibilities: Monty opens one door to show that it's empty (door B, this happens ⅓ of the time), and Monty opens two doors to show that they're empty (doors A & C, this happens ⅔ of the time). In both cases, the odds change. They change to ½ in the former, and to 1 (100%) in the latter.

The first scenario I gave is identical to each player on his own.

[MariluLane]

hey cousino!!! greetings to you!! have you found a good video that explains the mathematical machinations behind the monty hall problem? i watched the one from nate the mathematics guy but he simply iterates the conundrum.... any pointers? :)

[MariluLane]

by the way i must say this is a brilliantly clear explanation, perhaps no other video is needed ;)

[spankbutt123]

Hmm, where's the paradox? Its pretty obvious isnt it?

[CousinoMacul]

I get that question a lot on this video, and I'm not sure why. The problem is called "Bertrand's Box Paradox". I didn't invent the name.

And what is a paradox other than an unexpected result that's difficult to reconcile? You could define it as something that's "seemingly impossible yet true". But isn't that really just a rephrasing of my definition?

Perhaps I didn't explain BBP in paradoxical enough terms. ;-)

[Tolstoievsky]

@CousinoMacul that is deeply not a paradox. i believe the correct term would be "serendipity"

[zz0mfgz]

@CousinoMacul When first watching this I figured it out quite effortlessly and even paused and pointed out to myself that the only way to change the odds would be to change the number of silver or gold medals in the B box, only to later hear you say that. I say this only to point out that I understand the way you and others who've solved this are seeing this, but when you changed it to 2 boxes, i paused and thought about it, and it changed my entire prospective on the entire problem (continued.)

[zz0mfgz]

@zz0mfgz (continued.)Now I'm sure I'm just missing something here so I'm just wondering if you can point out the mistake in my logic. I see the answer to the problem as wrong now, the only way it would have a ⅓ chance is if you were to do something like dump all the medals on to a ground and pick a gold one, then say something along the lines of "There's a 33% chance that this medal came from box B." (assuming you have 2 in box A and not 100.) The thing is if you pulled a gold one out(continued)

[zz0mfgz]

@zz0mfgz (continued) then just in the same way that box C becomes irrelevant, the amount of silver medals in box B would become irrelevant, because the knowledge of it being gold changes the problem into it being there are 2 boxes that have gold in them, regardless of the amount of other medals in the drawers. This changes the answer to a 50% chance of it coming from box B.

[zz0mfgz]

@zz0mfgz Please, CousinoMacul (or anyone really), explain to me the fallacy in my reasoning.

[trndphilliephan]

PhD's and other people much smarter than myself call this a paradox so, i'm going to go with them on it

[alacowboy0]

I watched the whole vidoe to see how it is solved, and it just ended without a solution

[JoshD667]

This video was just an exucse for you to show off those medals wasn't it! Stuff the paradox, your medals were the real purpose of this video!

Great video thanks!

[CousinoMacul]

I've been discovered! Oh noes!!!!

[darkheritic]

very interesting paradox. almost like the monty hall one, but not quite.

[HaleyMary]

Good vid. Math was never my best subject in school, but I understand the odds part of it. I don't understand how this was a paradox, though.

[strouze]

i dont think this is a paradox - its just an interesting problem

[RaistlinJ]

enjoyed this vid, cheers

[bezvezeceda]

Why is it a paradox??? Where is the paradox??? I don't get it.

[MasterGhostKnight]

This realy ain't a paradox.

It in you first experiment the mistake lies on assuming that probability comes from counting the gold medals, it hapens to give the same value, but it is not how it should be calculated.

The problem lies in the selection method. If you are going to pick it out from a drawer, you must first select a drawer and then a medal. It would be a difrent story if I have mixed up all the medals in a pot and tryed to find the probabilit yfrom witch drawer it was originaly from.

[JFZcafe]

Great videos! I used to be pretty sharp on this stuff. Thanks for the refresher. And just knowing some basics about statistics can help people make informed decisions about a lot of practical things. I read a great little book about this years ago that became somewhat popular, it is called Innumeracy, by John Allen Paulos. It's easy to read and I strongly recommend it.

[CousinoMacul]

Thank you JFZ, I'll have to check that book out. In the meantime, if you enjoy this sort of thing, you should watch my three videos on statistical fallacies. I have three more SF videos planned so far (gambler's fallacy, sampling bias, and correlation vs. causation) and maybe more. :-)

[velocityeleven]

wierd, as you were explaining it, i thought it was obvious

[CousinoMacul]

Thank you, Taylor.

[condaryt]

Two differents probabilities for two differents experiments. There is no paradox just Conditional Probability

[CousinoMacul]

Most "paradoxes" turn out to not be paradoxes once you understand your mistaken reasoning. ;-P

By the way, were you affected by the earthquake?

[condaryt]

well, I live in Antofagasta City and the epicenter was 150 Km to the northeast. Fortunately nothing happened here except the big big big big movement (7.7 Richter). My family is OK and no one is hurt... and the house is still stand up. :-)

[Vinifera7]

Neat paradox!

[CousinoMacul]

Indeed!

[Mjhavok]

Do a video on the travelling salesman problem. Hasn't been solved yet but its still interesting.

[CousinoMacul]

That's an idea. I was also thinking about doing some game theory videos (prisoner's dilemma, traveler's dilemma, etc.)

[Mjhavok]

That would be cool. Have you read the book "Prisoner's dilemma"?

[AtheistAaron]

Dawkins had a great section on it in "Nice Guys Finish First".

[CousinoMacul]

Dawkins has written extensively about game theory and the prisoner's dilemma in several of his books. However, I think that Mj was referring to a biography of John Von Neumann.

[Mjhavok]

Yes.

[CousinoMacul]

If you're talking about the biography of John Von Neumann, then yes I read many, many years ago. He's always been one of my favorite characters. In fact, my very first post on my blog (thesciencepundit DOT blogspot DOT com) was about him. You can find it at

/2006/07/john-von-neumann-and-­mathematicians DOT html

[oogashakaooga2]

Very good. I didn't know this one.

It's interesting how the order of the procedure affects how we nest the probabilities -- and thus their actual value.

I think that's what throws people off in these paradoxes, we tend to jump to the conclusion without checking every step of the reasoning.

[CousinoMacul]

Indeed! Most of these paradoxes disappear if you just plot everything out. Another example would be the paradox of the three diners and the missing $2. It only seems like a paradox because it is presented in a misleading way.

[ogbash]

I see a mathematical mind :), generalization rules.

Recently, I heard of the next question.

"Monk starts at 8am and climbs mountain at constant speed. I takes him 10 hours to reach the temple on the top. Next morning he leaves at 8am and goes down at constant speed. It takes him 12 hours to go back. Is there a position on the road that he reaches at the same time."

[CousinoMacul]

Let the x-axis be the time in hours and the y-axis be the distance traveled with 100 (for 100%) being the total traveled each day. The line describing the 1st day's travels goes through (8,0) and (18,100) giving y=10x-80. The 2nd day's line goes through (8,100) and (20,0) giving 3y=-25x+500. Those lines intersect at (148/11,600/11) which means at 1:27 each day the monk will pass the spot which was 54.5% of the way the 1st day.

[ogbash]

yeah, the simple solution is to take 2 monks ;)

[CousinoMacul]

Since the original question was just a yes/no question, the two monk method clearly shows that they must pass each other at some point. However that seemed so ridiculously obvious to me that I decided to have some fun and calculate where and when that happens. ;-)

[Muaguana]

I love this stuff - great video, Javier.

[CousinoMacul]

Thank you.

[ryanisfootdrums]

Hey I have been enjoying your math videos :)

[CousinoMacul]

Thanks! I'll keep making them.

[AtheistAaron]

Excellent. I knew of the Monty Hall paradox, but not of Bertrand's Box. Thanks for the great video, Javier.

[CousinoMacul]

Your welcome, Aaron.

[dudev]

Great stuff, Javier. (And everyone should check out Jason Rosenhouse's evolution blog. He's the man. It's great to see evolution from a mathematician's perspective.)

[CousinoMacul]

Jason Rosenhouse is indeed the man!

[rozeboosje]

Applause!!!

[CousinoMacul]

[takes a bow]

[worldlystone]

Great video!

[CousinoMacul]

Thanks!

[TrueNorth15]

I was lost until you showed the math, once again your vids are really cool.

[CousinoMacul]

Thanks, Nathan.

[thequestionmarkofGod]

mmmm I get this one, with not a single stupid arguement. :D

[CousinoMacul]

:-D