I have a deck of 52 cards which I have shuffled in random order.

I remove 10 cards from the top of the deck and discard them without
looking at their values.

What is the probability that the next card on the top of the deck is
the Ace of Clubs?  Please account for the fact that it could have been
in the 10 cards that I removed.
Please give the answer and show the math / formula used with exaplantion.

tonyvu,

For this question, I will assume that this is a normal deck of cards,
with exactly one of every card you would expect to find.  This means
that there is one Ace of Clubs in the entire deck of 52 cards.  So
generally, the odds are 1 in 52 that a card you pick at random will be
the Ace of Clubs, and 51 in 52 that the card you pick at random will
not be the Ace of Clubs.

Now, what you have done is split the cards into two decks:  of the 52
card available, the deck of discarded cards has 10 cards, and the
remaining deck has 42 cards.  Now, if we could look at the cards we
discarded and see that none of them was the Ace of Spades, then it
becomes one card out of just 42, and the odds would reduce to 1 in 42.
 Or, if we saw that the Ace of Spades was one of the cards discarded
from the deck, then we would know that there was zero chance of
picking the Ace of Spades from the remaining deck.  But unfortunately,
we still don't have any additional information...the Ace of Spades
still must be considered as one of the 52 total cards we started with.
 So let's look at the three possible scenarios:

1) The Ace of Spades is one of the 10 discarded cards.  Odds are 10 in 52.
2) The Ace of Spades is the next card in the deck.      Odds are 1 in 52.
3) The Ace of Spades is elsewhere in the deck.          Odds are 41 in 52.

Even though we've shuffled the deck, because we know nothing more
about where the Ace of Spades might be, the odds of the next card
chosen being the Ace of Spades remains 1 in 52.

Thanks for the in depth answer

Unless you have assured that the Ace of Clubs was NOT in the 10 cards
removed from the deck, the probability or odds haven't changed from
those with the full deck.

Best regards,

Omnivorous-GA

tonyvu-ga:

The chance that any card in the deck is the Ace of Clubs (assuming
it's a standard deck of playing cards!) is 1 in 52.

Since you have not looked at the 10 cards that you have removed from
the deck and discarded, that is really no different than taking 10
cards from the top of the deck, and placing them on the bottom. Which
is also no different from extending the shuffling process by one more
step, leaving you with what is still a randomly shuffled deck. So, the
probability that the top card is the Ace of Clubs will still be 1 in
52, since you have not changed the size of the pool from which the
card is drawn. Another way of saying this is: what is the probability
that the 11th card from the top of a randomly shuffled deck is the Ace
of Clubs?

If you need to illustrate the math behind this, perhaps for a homework
assignment or other such academic exercise, simply illustrate that the
number of positive cases (ie. drawing an Ace of Clubs) does not
change, and that the pool size does not change and is not affected
during the draw because you are not looking at the drawn cards until
you reach the 11th card.

Regards,

aht-ga
Google Answers Researcher

I would agree.... the probability that the Ace of Clubs is the next
card is no different from the probability that that particular card
was the first card... or the second... or the third... so the odds
would be one in 52.

Let A = event that the Ace of Spaces is removed from the deck.
Let B = event that the first card of the remaining deck is the 
        Ace of Spaces.

1.    P(B) = P(A and B) + P(A' and B)
2.         = P(A)*P(B|A) + P(A')*P(B|A')
3.         =      0     +  P(A')*(1/42)
4.         =  1/52

Line 1. Either the card was removed or it wasn't.
Line 2. Follows from the definition of conditional probability.  
        P(B|A) is the probability of B given A.
Line 3. Clearly, P(B|A)=0 while P(B|A')=1/42
Line 4. Let C(n,k), read "n choose k", be the number of ways of 
        choosing k cards from n.  Then P(A)=C(51,9)/C(52,10)=10/52.  
        (Recall, C(n,k)=n!/(k!(n-k)!).)
        Therefore, P(A')=1-P(A)=42/52.

Best.
MP

This is the same problem if the question was worded 'What is the
probability that the Ace of Spades is the 11th card in the deck'. A =
1/52.