Probability of Ace of Spades in Monty Hall Like Scenario I select a card at random from a standard deck of playing cards.
The probability of this card being the Ace of Spades is 1/52.
From the remaining cards, someone turns over all but one, and none of these is the Ace of spades. This person is aware of which card is which and has deliberately avoided turning over the Ace of Spades.
Why is the probability that the remaining card is the Ace of Spades 51/52?
Specifically, how can I understand this in terms of the basic formula for probability?
$\frac{\text{no. successful outcomes}}{\text{no. possible outcomes}}$
Even though I know that since the original card has a 1/52 chance and all the possible outcomes must sum to a probability of 1, this is not convincing me at the moment, and I want to understand the result in terms of what "is" happening with the cards, rather than what "isn't".
Please not that this last point makes this question different to the existing similar question here: Cards Probability like Monty Hall Problem
 A: First, let's just note that the formula
$$\frac{\text{no. successful outcomes}}{\text{no. possible outcomes}}$$
applies only to situations in which you can identify a finite number of equally likely outcomes, and is not "the basic formula for probability".
Now let's apply it to the cards.
The ace of spades is equally likely to be in any of the $52$ places where a card is located. Of these $52$ places, $51$ are where the "remaining cards" are.
Therefore the chance is $\frac{51}{52}$ that the ace of spades is among the remaining cards.
The procedure of turning over cards is something that a person who knows where the ace is can always do, regardless of whether the ace is in the $51$ remaining cards or is the card you originally chose. The ability to do this procedure tells us nothing about the probability of whether the ace is in the remaining cards.
What turning over the cards does is to tell us, if the ace is one of the remaining cards, which of the remaining cards it is.
As far as the probability of winning is concerned,
it's really no different than having the opportunity to discard your original card, pick up the remaining $51$ cards in a pile, and if you can find the ace of spades in that pile yourself you win.
Think of it as just a bit of showmanship.
