Originally posted by PBE6
This one comes from the Old West, apparently.
A man with a hat shows you 3 cards, one completely gold, one completely silver, and one gold on one side and silver on the other. He puts them in his hat, and picks one at random. He then shows you one side of the card he picked, which happens to be silver. Now he says "I'll bet you even money that the other side of this card is silver too...whaddaya say, partner?"
Is this bet a fair one?
If my way of thinking is redefining the problem, let let me revisit the problem as stated.
The issue here seems to be some vagueness in the details of how it works in the wording. But the problem follows a well-known pattern, and it makes sense to assume that the problem is identical.
However, let me look at what questions are not answered which may affect the problem at hand.
According to the problem, the selection of the actual card is random, meaning the dealer does not have pre-knowledge of which card will be selected, and any of the three cards have an equal chance of being selected, even the one that is gold-gold, even though later on it is stated the one which was selected had a silver side, because that is what you see.
The main vagueness which occurs in my mind is whether or not the dealer knows which card it is before making the offer.
My assumption here is that he does not, and the analysis I gave works off that assumption.
However, what if I am wrong on this count? Then the dealer has full knowledge of whether the bet he makes is good or bad, and this ceases to be a probability problem at all. And the answer most likely is that the bet is a bad one.
So it is reasonable to assume the dealer has not seen the other side of the selected either for purposes of analyzing the problem.
That also means that it could just as easily been a gold side showing. In fact, given the setup, Silver and Gold had equal chances of coming up.
Now, the fact that it is a silver side showing provides us with information we did not have with the setup alone.
Before the card was drawn and placed in front of us, we knew one of six possible events would happen. One of the three cards would be drawn, and one of the two sides on that card turned face up. The card and side selection is assumed to be fair in the absence of information to the contrary, which means any of the 6 possibilities are equally likely.
Now the drawing occurs, and we see a silver side. This eliminates the three gold sides from consideration. It could have just as easily been a gold side selected, and we would then be eliminating the silver sides.
The important thing is that any of the 3 silver sides are equally likely to have been the silver side picked.
We are then offer even money on the other side being silver (i.e. the same color).
To examine the chances of this, we can examine how many of the possible sides match their opposite side.
And the answer is that TWO of the silver sides are on the silver/silver card, while only ONE is on th silver/gold card, which means the dealer has a 2-in-3 chance of being correct.
If you were to repeat this process until silver turned up 30 times, and then tally the results, you would expect to see the silver /silver card 20 times, and the silver/gold only 10 times. The other 30 times, it would be a gold side showing up.
You would lose twice as often as you win.
Or think of it this way.
When the Silver/Silver card is picked, you'll ALWAYS see a silver side.
When the Silver/Gold card is picked, you'll only see silver HALF the time. (The other half of the time, you'll see the gold side).
Since either card is equally likely to be picked, then half the time the latter is chosen, it will come up on the gold side. This means of the cases in which a silver side is showing, the Silver/Gold card is half as likely to have been chosen as the Silver/Silver.