On "Let's Make a Deal," you pick Door #1. Monty opens Door #2--no prize. Do you stay

No, no, no, no, no.

The answeers to Jordan Drachman’s first question is NOT 2 in 3. I’ll explain:

50% odds the first card is an ace. 100% odds that the second card is an ace. 50% odds that we will pick each card. Therefore we have a 50% of 50% plus 100% of 50% odds of picking an ace. That gives 75% (.5 * .5 + 1.0 * .5 if you prefer) odds of getting an ace out of the hat. This equates to 3 out of 4 times. Not 1/2, not 2/3 but 3/4.

I wonder if Cecil will read this…


Nice calculation, acb, but you answered the wrong question. You asked, “What are the odds of drawing an ace?” and answered that one correctly (I think). The original question, however, was, “If you draw an ace, what are the odds the original card in the hat was an ace?”


Sorry to all about the missed point regarding the cards, but the question for which the problem tries to establish probability is not “What are the odds of drawing an Ace?” but “What are the odds that the original card was an Ace?”

As there are only two possibilities for the original card (ace or king), the odds that the original card was an ace is one out of two (1/2 or 50%).

All the subsequent activity, adding the ace, drawing, etc, are irrelevant to the odds regarding the original card’s identity.

You missed the point again. I will acknowledge that the wording of the question was a little imprecise. The question does read “What are the odds that the original card was an ace?” But given the additional information that was given, the implied question was “Given that you have drawn an ace, what the the odds that the original card was an ace?” The answer to the former (technically) is 1/2. The answer to the latter is 2/3 by the logic that has already been discussed.