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spaceotter @spaceotter

Bertrand's Box Paradox: There are 3 boxes. One box has 2 gold coins. One has 1 gold coin & 1 silver coin. One has 2 silver coins. Find a box that has at least one gold coin below to encounter the paradox.

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@spaceotter there's a problem, there's 2 of them saying the same thing: "swipe left twice". How does that work?!

@razcore It's a constraint: I can only post up to 4 images. Fortunately two of the choices place you in almost identical predicaments.

@spaceotter ya... sorry, either it's poorly explained or I'm stupid cause I don't understand it at all.

One swipe gets you one specific box, but apparently two swipes get you the remaining 2 boxes. But when you do that you get a question about "a remaining coin". What remaining coin?! I thought each box has 2 coins in them. so picking the two boxes you're left with 4 coins... don't understand the reining coin question at... and so goes for the answer.

Maybe someone else can crack it

@razcore if you chose the red box you find that the first of two coins is gold. the question then is what are the chances that the 2nd coin (in the red box) is gold too.

The question remains the same if you chose the blue box.

The paradox is that its common to think you have a 50/50 chance or maybe a 1 in 3 chance but both are wrong.

regardless of the box you chose (red or blue) there's a 2 out 3 chance that the 2nd coin in the box you chose is gold.