Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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($O^c$)

share|cite|improve this question
The same way as you do the Monte Hall problem: when your initial choice was made, the probability that you are correct is $\frac1{52}$. This probability does not change with all of the additional information you're given; ergo, the probability that the other card is the Ace of Spades is $\frac{51}{52}$. – Steven Stadnicki Sep 22 '13 at 19:44
The way I was shown the monty hall problem was that we used tables. So, Im not quite sure how to do something other than using a table and it may be noticeable that using a table would be time consuming? Would you care to show how to do this problem without using a table? – John Petryk Sep 22 '13 at 19:47
What do you mean by academically. You are using the counting principle, which though not advanced, is still a rigorous principle and completely accepted. – PyRulez Sep 22 '13 at 20:24
sorry i meant formal. cheers! – John Petryk Sep 22 '13 at 20:38

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).

share|cite|improve this answer
W- win by switching and O- win by not switching someone used this formula i dont seem to understand it P(W|O)P(O) + P(W|$O^c$)P($O^c$) – John Petryk Sep 22 '13 at 20:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.