Originally posted by eldragonfly
Reword the problem then, this is nonsensical.
"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 o ...[text shortened]... 't make the case for holding up a gold sided card, so your "demonstration" is incorrect.
Completely wrong. This is a simple application of Bayes Theorem, as follows:
P(A g B) = P(A U B) / P(B)
where P(A g B) = probability of A given B
P(A U B) = probability of A and B happening together
P(B) = probability of B
In this case:
A = the other side is silver
B = the silver side is face up
Counting the sides, there are 6 possible outcomes in this game:
gold/gold with gold(1) face up
gold/gold with gold(2) face up
gold/silver with gold face up
gold/silver with silver face up
silver/silver with silver(1) face up
silver/silver with silver(2) face up
To calculate P(A U B), we simply count the number of outcomes where silver is face up (B) and the other side is silver (A). The outcomes that satisfy these conditions are:
silver/silver with silver(1) face up
silver/silver with silver(2) face up
So there are 2. Similarly, to calculate P(B) we count the number of times silver is face up (B). The outcomes that satisfy this condition are:
gold/silver with silver face up
silver/silver with silver(1) face up
silver/silver with silver(2) face up
So there are 3. Therefore, the probability that the other side is silver (A), given that the side face up is silver (B), is:
P(A g B) = P(A U B) / P(B) = 2/3
Since the dealer is only offering even money, or 1:1, the bet is not a fair one. If you still don't believe me, try playing this game and keeping track of the result, or setting up an Excel spreadsheet to run the game for you.