# Probability of drawing an Ace: before and after

this is my first question on the website.

When I took probability, one of the first questions in my textbook was this:

What is the probability of drawing an ace in a standard 52-card deck?

That one was pretty easy, thankfully. The next one threw me for a loop...

Take the top card away from the deck. What is the probability of drawing an ace?

And the answer was the same as the first one! I can't remember how my professor explained it, but I feel the intuition at least. All one card removals are equally likely: it's almost like still having the original 52 cards. But I'm still bothered. "The question is how to understand this."

Hope someone can help.

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Draw this probability tree.

            First        Second
1     A                 1/221
-    /
17  /
A
1   / \   16/17
-  /   \
13 /     not A             16/221
/
*
\
12 \       A               16/221
--  \     /  4/51
13   \   /
not A
\   47/51
\
not A           564/663


Now add up the branches with an A on the second and get

$${16\over 221} + {1\over 221} = {17\over 221} = {1\over 13}.$$

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This is a lot more complicated that it needs to be. The second event is independent of the first event; there's no more reason to make the second even conditional on the first in a tree than there would be to make it conditional on the third. – joriki Jun 28 '11 at 2:00
That the second event is independent of the first is something you must establish. It is not automatically true even if it seems "intuitively true." – ncmathsadist Jun 28 '11 at 2:02
How would one establish this? – mathmath8128 Jun 28 '11 at 2:13
You show that $P(S\cap F) = P(S)P(F)$, or that $P(S|F) = P(S).$ Neither of these is true. The tree tells that. It is true that the probability of an ace on any given draw, with no knowledge of the others is 1/13. – ncmathsadist Jun 28 '11 at 2:42
I think the reason it seems counter intuitive is because it's hard to imagine discarding the card and not looking at it, not learning its value, and thus making them dependent events. The probability of the second card being an ace is 1 in 13 until you find out whether or not the first card is an ace. – D. Patrick Jun 28 '11 at 4:13

One way to think about it is that you are drawing the second card of a standard 52 card deck. Shouldn't that one have the same probability to be an ace as the first one?

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Yeah. This is what I was thinking. You're just placing the first card aside; in a sense, it's still there (in a probabilistic way). You're just looking at the next one. – mathmath8128 Jun 28 '11 at 1:54

I never heard of this before but it is certainly true, even for different number of aces and total number of cards (I computed the conditional probability to make sure.) Then I came up with the following explanation: Take the top card and put the top card on the bottom of the deck, which is equivalent to removing it. Then nothing has changed in terms of probability and so we get the same probability. Learn something new everyday.

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That's a nice way to frame the problem, Steve. – amWhy Jun 28 '11 at 2:11

All permutations of the cards are (in principle) equally likely. There are exactly as many permutations that have an Ace as the first card as there are permutations that have an Ace as the second card, or the seventeenth.

An interesting fact: If you are playing a bunch of bridge hands, all permutations are not equally likely, because shuffling between hands is inevitably of poor quality. When tournament organizers started to use computer programs to generate "random" hands, players complained that they were getting too many "weird" hands. They were right, they had gotten accustomed to human-shuffled poorly randomized hands.

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You haven't formulated a question. (Rather, you already provided the correct answer to the question that you did formulate.) I'll assume that your implicit question is how to understand that the answer to the second question is again $1$ in $13$.

The probability of the first card being an ace is the same as the second card being an ace. In fact, in a sense it's the definition of the deck being shuffled that these probabilities are the same. The only difference between the first and the second question is that you draw the first and the second card, respectively. This is a bit easier to think about than regarding the deck minus the first card as a new population of $51$ cards.

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Yes, I'm sorry. The question is how to understand this. :-) – mathmath8128 Jun 28 '11 at 1:53
Why the downvote? – joriki Jun 28 '11 at 5:10

Taking the top card from a well shuffled deck is the same as picking a random card from the deck. Taking the second card from a well shuffled deck is the same as choosing a random card. Removing the top card from the deck and selecting the second card is the same as choosing the second card and is thus the same as selecting a random card.

If you were to take the first card off of the top of the deck and then look at it, it'd be different story though.

No matter what card you choose from the deck it has a 1 in 13 chance of being an ace (whether it's the first or the second card).

However, if you "take the top card away from the deck" and you look at it in the process, then you no longer have a single independent event. You're not just moving the first card out of the way so you so you can flip the second card over. The odds of the second card being an ace still haven't changed, however, the odds of flipping an ace with the second card change now that it's a new independent trial with either a 4 in 51 chance or a 3 in 51 chance of being an ace.

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Look at the extreme question:

Take the top 51 cards away from the deck. What is the probability of drawing an ace?

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Powerful hint: Taking any card, say ace of clubs or 9 of spades, has to be equally likely. And SOME card has to come out in the second place. This can be done using bijections, to show some cardinals (I.e. amount of decks with an "x" card on the second place) are equal. And thus using good cases divided total cases you are done.

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