> 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
There are three possible outcomes of the experiment: reset, success, and fail. If you pick a silver coin first we “reset” and discard the outcome. If you pick a gold coin first, then a silver, it’s a fail. If you pick a gold coin first, then another gold, it’s a success.
The question asks for P(success) / (P(success) + P(fail)). We can calculate this as follows:
33% – picked the silver-only box and picked a silver coin, reset
33% – picked the gold-only box, next coin is gold, success
16.5% – picked the silver/gold box and picked the gold coin first, next coin is silver, fail
16.5% – picked the silver/gold box and picked the silver coin first, reset
P(success) = 33%
P(fail) = 16.5%
P(success) / (P(success) + P(fail)) = 33% / (33% + 16.5%) = 66%.
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