

Subject:
Probability Question with Deck of Cards
Category: Science > Math Asked by: tonyvuga List Price: $10.00 
Posted:
24 Feb 2004 09:50 PST
Expires: 25 Mar 2004 09:50 PST Question ID: 310306 
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. 

Subject:
Re: Probability Question with Deck of Cards
Answered By: hailstormga on 24 Feb 2004 14:48 PST Rated: 
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. 
tonyvuga
rated this answer:
Thanks for the in depth answer 

Subject:
Re: Probability Question with Deck of Cards
From: omnivorousga on 24 Feb 2004 10:07 PST 
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, OmnivorousGA 
Subject:
Re: Probability Question with Deck of Cards
From: ahtga on 24 Feb 2004 10:10 PST 
tonyvuga: 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, ahtga Google Answers Researcher 
Subject:
Re: Probability Question with Deck of Cards
From: darrelga on 24 Feb 2004 10:23 PST 
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. 
Subject:
Re: Probability Question with Deck of Cards
From: mathpassionga on 04 Mar 2004 11:42 PST 
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(BA) + P(A')*P(BA') 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(BA) is the probability of B given A. Line 3. Clearly, P(BA)=0 while P(BA')=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!(nk)!).) Therefore, P(A')=1P(A)=42/52. Best. MP 
Subject:
Re: Probability Question with Deck of Cards
From: darrenwga on 11 Mar 2004 21:51 PST 
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. 
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