> There are three boxes:
> – a box containing two gold coins,
> – a box containing two silver coins,
> – a box containing one gold coin and one silver coin.
>
> Choose a box at random. From this box, withdraw one coin at random. If that happens to be a gold coin, then what is the probability that the next coin drawn from the same box is also a gold coin?
My thinking is this… Taking a box at random would be 33% for each box. Because you got one gold coin it cannot be the box with TWO silver coins, therefore the box must be either the gold and silver coin or the box with two gold coins. Each of which is equally likely so the chance of a second gold coin is 50%
I understand that this is a veridical paradox and that the answer is counter intuitive. But apparently the real answer is 66% !! I’m having a terrible time understanding how or why. Can anyone explain this like I was 5?
In: Mathematics
Label the possibilities: box 1 has coin 1 (gold) and coin 2 (gold), box 2 has coin 3 (gold) and coin 4 (silver), box 3 has coin 5 (silver) and coin 6 (silver). There are now six equally likely draws:
* Box 1, coin 1: gold
* Box 1, coin 2: gold
* Box 2, coin 3: gold
* Box 2, coin 4: silver
* Box 3, coin 5: silver
* Box 3, coin 6: silver
There are three ways you can get a gold coin, and two of those three ways are from box 1.
The problem with considering that it’s just box 1 or box 2 when drawing a gold coin is that these two boxes are not equally likely. This is because all of the draws from box 1 will be considered, while half of the draws from box 2 are discarded because this is conditional on drawing a gold coin.
Another way to look at it: let’s say you have two boxes. One has 99 gold coins and 1 silver coin. The other has 99 silver coins and 1 gold coin. If you draw a gold coin, is it equally likely that it came from either of the boxes?
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