During a certain game show, contestants are presented with a standard 52-card deck. If the contestant picks the Ace of Spades they win the grand prize. The contestant is asked to select a card and put it aside without looking at it. The game show host then goes through the deck and discards 50 out of the remaining 51 cards that are not the Ace of Spades with the game show host or with the contestant. The contestant is offered the choice of cards. Which one do they choose?
How do you solve this kind of Problem? I know the probability has to be 0.980 with winning by switching. I just did that in my head by $\frac{51}{52}$. I dont know how to prove this in a formal manner
Ok.. This is the formula used, which I have trouble understanding
W- win by switching O - win by not switching P(W|O)P(O) + P(W|$O^c$)P($O^c$)
There is a 1/52 chance that the card is the one picked initially. The card can only be either the one you picked, or the alternative, therefore the chance of it being the 'other one' is 1 - 1/52 = 51/52.
It is precisely the Monty Hall problem with more 'doors', and the same resolution (switch).