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In class, while illustrating the topic of conditional probability, my professor presented the following card example:

You have 3 cards that have been randomly shuffled: card1, card2, and card3. One is an ace and the other two are non-aces. We are interested in the location of the ace. Thus, the sample space is S = {card1, card2, card3}.That is, the ace can either be card1, card2, or card3. We assume that each outcome is equally likely(ie classical probability formulation). Let event
A1 = "card1 is the ace", thus P(A1) = 1/3. Let event B = "turn over card3 and it is not an ace".

My professor says that once B occurs it becomes the sample space since it becomes the full set of possibilities given that it actually occurred. He says that A1 = {1} and B = {1,2}, so now the probability of the ace being the first card is $P(A1|B) = \frac{|A1\cap B|}{|B|} = \frac{1}{2}$.

The answer to this problem made me uneasy because this problem looks very similar to the Monty Hall problem. Applying the "Monty Hall problem" reasoning to this problem would give the probability of the ace being card 1 as 1/3, since the probability of the ace being
card 2 (ie "switching your pick") would be 2/3.

Is the answer to the card example really 1/2? Are these problems the same? If not, what makes them different?

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When card 3 is turned over, was it chosen at random, or was it chosen by someone who knew it was not an ace? Strangely enough, this information matters. –  Ryan Sep 2 '13 at 0:55
It was chosen at random. Why are those two scenarios different? Doesn't the unaware chooser have the same "information" regardless of whether the person who turns over card 3 knows which card is the ace? –  user92638 Sep 2 '13 at 2:09
Suppose that there are 100,000 cards, and only one of them was an Ace. If 999,998 cards are removed at random and no ace is revealed, do you switch or stay? –  Ryan Sep 2 '13 at 2:18

1 Answer 1

up vote 6 down vote accepted

First, I would encourage you to actually simulate each situation a few times (say 20), note the outcomes and estimate the probabilities. The simulation process will make you appreciate how important the protocol is.

The Monty Hall problem should be stated carefully. If Monty knows which door hides the ace, and Monty always makes sure to open one of the remaining doors at random, then it's the classical Monty Hall, and you should switch choices. If Monty knows which door hides the ace, but has a strange preference for the number 3 and thus will always open door number 3 when it does not hide the ace, then it's a different situation. If Monty does not know where the ace is, and always opens door number three, which just happens not to hide the ace (but Monty did not know that), then it's yet another situation, giving indeed the probability $1/2$ above.

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One way to look at your problem is to see that there were initially three places for the Ace. If it is place 3 you don't get beyond the first stage. If it is in place $1$ or $2$ these remain equally likely. In classic Monty Hall there are two possibilities once I have picked - I have chosen the Ace (1/3) or I have not (2/3). Whichever of these is true, Monty can (deliberately) choose a card which is not an Ace. The difference is that the professor might have picked the Ace, Monty never would have. –  Mark Bennet Sep 2 '13 at 2:33

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