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Here's the situation:

You have a deck of 52 cards that is already shuffled. You pick the first card, and the first card is not an Ace. Is the probability of drawing an Ace as second card the same if

  1. The second card is immediately drawn from the deck
  2. The remaining deck is first shuffled without adding the first card back in, and then the second card is drawn.

For me, it is obvious that probability of drawing an Ace in the second case is 4/51, but I'm not entirely sure that the probability is the same in the first case.

Also, what would be the probability of the first case if we have drawn n cards and none of these are aces?

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    $\begingroup$ yes, its the same. $\endgroup$ – supinf Sep 15 '15 at 11:54
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    $\begingroup$ 4/(52-n). $ $ $\endgroup$ – Did Sep 15 '15 at 11:56
  • $\begingroup$ As an aside, this is also the reason why it is pointless to doubly randomise a choice (e.g. tossing a coin to see who goes first for the real coin toss!) $\endgroup$ – Oscar Bravo Sep 15 '15 at 14:04
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    $\begingroup$ Another aside... if you and a friend each draw a card, and then your friend reveals he didn't draw an ace... your odds of getting an ace are better if you replace your card with a new one from the 50 cards remaining. $\endgroup$ – kbelder Sep 15 '15 at 23:33
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If it's not intuitively clear, you can check your calculation using contitional probability:

  • $X$... the event "the second card is an ace"
  • $Y$... the event "the first card is not an ace"

Then, $P(X\land Y) = \frac{48\cdot 4}{52\cdot 51}$ and $P(Y) = \frac{48}{52}$

Then $$P(X|Y) = \frac{P(X\land Y)}{P(Y)} = \frac{\frac{4\cdot 48}{52\cdot 51}}{\frac{48}{52}} = \frac{4}{51}$$

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Given you know the first card of your shuffled deck, the remaining $51!$ ways of arranging the remaining $51$ cards are equally likely. For each of the $52$ possible values of the first card you have $51!$ equally likely arrangements of the remaining $51$ cards. This is consistent with there being $52\cdot 51!=52!$ arrangements of the whole deck to begin with.

You can conclude that, after drawing the first card, the remaining $51$ cards are equally likely to be in any shuffled order, and so shuffling them changes essentially nothing. This argument should lead to the conclusion that the probability is the same in either case.

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Yes, the probability of drawing an Ace is the same if you shuffled the rest of the deck or didn't. You are correct that it's 4/51. If you drew n cards and none were Aces, then the probability of drawing an Ace next would be 4/(52-n).

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In both cases, there are only 51 cards left, thus $Pr = \dfrac{4}{51}$

And if $n$ non-aces have been drawn, $Pr = \dfrac{4}{52-n}$

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