Probability of drawing an ace from a modified deck

Question

Given the following question,
I have a deck of $52$ cards. I shuffle the deck, and drop $20$ cards randomly into a shredder. You then draw two cards from what remains. What is the probability that they are both aces?

A lot of similar questions have been asked (with good answers). For example, ace in the 10th card.

I know that the answer to my question is $\frac{4}{52}\cdot \frac{3}{51}$, but I am having A LOT OF TROUBLE internalising the answer. Intuitively, why is it that randomly removing cards does not effect the probability of drawing aces?

To be more specific:

1. Shouldn't we condition for the fact that some aces were shredded?
   • I find it very surprising that the answer remains the same after we condition for the fact that $0$ to $4$ aces could be shredded.
2. If we added cards to the deck instead, does the same logic hold?
   • If I were to randomly add $20$ cards from a shuffled deck to an existing deck and shuffle it, is the probability of drawing an ace still $4/52$? Or should I factor in the probability that aces were carried over to the deck? There was a similar question Combining random cards from two decks and calculating probability.

Any help will be much appreciated.

Answer 1

For the second question, the probability the first card drawn is an ace will be:

$\Pr(A) = \Pr(A\mid\text{from deck1})\Pr(\text{from deck1}) + \Pr(A\mid\text{from deck2})\Pr(\text{from deck2})$

$=\dfrac{4}{52}\times\dfrac{52}{72}+\dfrac{4}{52}\times\dfrac{20}{72}=\dfrac{4}{52}$

For the extension to the second question where we ask about the first two cards both being aces...

condition not only on the first card being an ace but also on whether it originated from original deck or from the added cards, further condition on whether the second card came from the original deck or not.

$\dfrac{52}{72}\times\dfrac{4}{52}\times\left(\dfrac{51}{71}\times\dfrac{3}{51} + \dfrac{20}{71}\times\dfrac{4}{52}\right)+\dfrac{20}{72}\times\dfrac{4}{52}\times\left(\dfrac{52}{71}\times\dfrac{4}{52}+\dfrac{19}{71}\times\dfrac{3}{51}\right)$

which amounts to $\approx 0.00509$ probability of both first two cards being aces from your deck with twenty extra cards added from a second deck. This being slightly higher than the $\approx 0.00452$ probability had it been just the one original deck.

Answer 2

Intuitively, why is it that randomly removing cards does not effect the probability of drawing aces?

For the moment, temporarily ignore that cards will be shredded. Consider the following two strategies for selecting the $2$ cards to be examined. You select the $52$ cards one at a time, at random from the deck. In the first strategy, you decide in advance that the first $2$ cards randomly selected will be examined. In the second strategy, you decide in advance that the $51$-st and $52$-nd cards randomly selected will be examined.

Consider the probabilities that one or both of the $2$ cards that are being examined is an Ace. These probabilities are not affected by whether the cards being examined are the first two cards that are randomly selected from the deck, or the last two cards randomly selected from the deck. In both strategies, the $2$ cards being examined are selected at random from the $52$ card deck.

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Now, consider a third strategy. Instead of deciding in advance that the first two cards randomly selected from the deck will be examined, or that the last two cards randomly selected from the deck will be examined, you decide in advance that the $21$-st and $22$-nd cards randomly selected from the deck will be examined. Again, the $2$ cards being examined are selected at random from the $52$ card deck.

Under the assumption that the cards that will be examined are the $21$-st and $22$-nd cards that are randomly selected from the deck, consider the probabilities that one or both of the $2$ cards that are being examined is an Ace. Since these $2$ cards are still being selected at random from the $52$ card deck, the probabilities are the same as they would be if the cards being examined were the first $2$ cards randomly selected from the deck.

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Now, make another change in strategy. As in the previous strategy, you have decided in advance that you will be examining the $21$-st and $22$-nd cards randomly selected from the deck. However, for the first $20$ cards randomly selected from the deck, set these $20$ cards aside. Then, randomly select the $21$-st and $22$-nd card from the deck, but do not yet examine these $2$ cards. Instead, first shred the $20$ cards that were set aside. Then, examine the $21$-st and $22$-nd cards that were randomly selected from the deck.

Has this shredding affected any of the probabilities? No, because shredding the first $20$ cards that were randomly selected does not change the spots on the $21$-st and $22$-nd cards that were randomly selected.

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Now, make one last change in strategy. Instead of :

• Setting aside the first $20$ cards randomly selected from the deck,

• Randomly selecting the $21$-st and $22$-nd cards from the deck,

• Then shredding the first $20$ cards,

Alter the strategy as follows:

Shred each of the first randomly selected $20$ cards, as the card is being selected. Has this affected any of the probabilities? No, because the only difference between this strategy and the last one, is that here, the $20$ cards are shredded immediately, instead of being set aside and then shredded. Here, the $21$-st and $22$-nd cards selected are still selected at random.