We can look at this as cards on a table too.
1. 52 cards are arranged in a line on the table.
2. Player wins if they can turn up the Ace of Hearts.
3. Player points to card #27.
4. There is a 1-in-52 chance that the player's chosen card is the Ace of Hearts.
5. Monty turns 50 cards face up - all except card #27 and (for some mysterious reason) card #13.
6. Monty has eliminated 50 possibilities. There are now only two cards on the table. One is the card the player has chosen (card #27) and the other (card #13) is unknown (to the player).
7. As before, there is still a 1-in-52 chance that the player - having chosen card #27 - had correctly chosen the Ace of Hearts.
8. But, of 50 original possibilities, there are now only two.
9. The chances that card #13 is the Ace of Hearts is 1 in 2.
If you can find any logical flaw in the above steps - as per the rules you stated, feel free to point out the error.
This is all fine up to a point, but step 9 needs further examination/explanation to determine if there's a "logical flaw". So, let's unpack that.
Yes, it is true that once the number of unrevealed cards has been reduced from 52 to 2, the chances that either one of the 2 remaining unrevealed cards is the ace of hearts would be 1/2,
if no other information is available to us, other than "There are two unrevealed cards on the table and one of them must be the Ace of Hearts". But in DaveC's scenario (quoted), the player
does have more information than that.
To see the difference, consider the following variant on the above:
0. The player is out of the room and cannot yet see what is going on.
1. 52 cards are arranged in a line on the table.
2. Player wins if they can turn up the Ace of Hearts.
3. A random device selects one of the 52 cards at random. It is card #27. This card is designated as "the player's chosen card" (even though the player has no choice in the matter).
4.
There is a 1-in-52 chance that the player's chosen card is the Ace of Hearts.
5. Monty turns 50 cards face up - all except card #27 and (for some mysterious reason) card #13.
6. Monty has eliminated 50 possibilities. There are now only two cards on the table. One is the card the player has chosen (card #27) and the other (card #13) is unknown (to the player).
6a. The player is then shown into the room and sees two cards unrevealed on the table, while 50 other cards are revealed as cards that are not the Ace of Hearts.
7. There was
a 1-in-52 chance that the player's chosen card - card #27 - was the Ace of Hearts at the time that card was selected at random.
8
. But, of 52 original possibilities, there are now only two.
Does this change things
for the player, compared to the initial scenario? I think it does.
Now, the interesting thing here is what happens in each scenario when we add step 10:
10. The player is then given the option to "stay" with card #27 or to "switch" to card #13.
In the original scenario (at the top, above), the player is fully aware of everything that happened. That is, the player witnessed steps 1 through 6. In the variant scenario, however, the player is not aware of what happened because he wasn't in the room prior to step 6a.
Assuming the player in each case
does not know the "rules" of the game, but has to infer them from what he has seen, what is the best strategy for the player to "win" by turning up the Ace of Hearts?
In the
variant game:
The player, upon entering the room, can see 50 cards that are not the Ace of Hearts face-up on the table and two cards face-down. He can deduce (assuming a fair deck) that one of the two face-down cards must be the Ace of Hearts. Now here's the important bit: he has
no reason to suppose that card #13 is any more likely than card #27 to be the Ace of Hearts. So, it's a 50-50 guess. "Stay" or "Switch" therefore makes no difference to him. He can only guess at which card the Ace of Hearts might be. So, in this case, the probability that the card #13 is the Ace of Hearts,
based on the information the player has, is 1/2. There is no advantage of choosing "stay" over "switch", as far as the
player can tell. (
We know better, because
we have advantage of knowing exactly how the game played out, but the player is not in our position.)
But what about the
original game:
In
that game, the player is aware of exactly how it came to be that 50 cards were revealed and 2 cards were left unrevealed. He is aware that Monty
knew which of the 52 cards was the Ace of Hearts from the start, and that Monty
deliberately turned over 50 cards that Monty
knew were not the Ace of Hearts.
The player can (and should) reason as follows:
1. The probability that I chose the correct card (#27) in the first place was 1/52.
2. Therefore, the probability that the Ace of Hearts was one of the 51 cards I
didn't choose was 51/52 at the time I chose card #27.
3. After Monty eliminated 50 of those 51 cards by revealing that they
aren't the Ace of Hearts, I now know that there's a 51/52 chance that card #13 is the Ace of Hearts, while the probability that card #27 is the Ace of Hearts remains unchanged at 1/52.
4. Therefore, the sensible thing to do is to "switch" to card #13, because I'll have a 51 times better chance of winning if I switch, compared to if I "stay".
