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"In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"

Are my calculations correct?

  • $4*51$ possible events with an ace on the first draw
  • $52*4$ with an ace on the second draw
  • $4*3$ with an ace on both draws

Thus we have $4(51+52+3) = 4(106)$ events with at least one ace.

$(4*3)/(4*106) = 3/106$ is the probability of two aces when you get at least one ace.

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  • $\begingroup$ The idea is fine, the count is not right. $\endgroup$ – André Nicolas Aug 9 '15 at 15:19
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Not quite. If you have $4 \cdot 51 = 204$ events with an ace on the first draw, then you have only $48 \cdot 4 = 192$ with an ace on the second draw. Furthermore, the $204$ already contains two-ace draws. (Alternatively, you can count $4 \cdot 48 = 192$ events with only an ace on the first draw, and then add in the $4 \cdot 3 = 12$ events with aces on both draws.) So the total count is $204+192 = 396$. Then $(4 \cdot 3)/396 = 1/33$ is the desired probability.

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  • $\begingroup$ Great. I didn't realize that to have one ace meant that not only does the deck have 1 less card but also the cards can't be another ace. $\endgroup$ – John Aug 9 '15 at 15:42
  • $\begingroup$ Yup. Incidentally, as André Nicolas points out, the easier way to count two-card hands that have at least one ace is to count up the hands that have no aces $\binom{48}{2}$ and subtract that from the total number of two-card hands $\binom{52}{2}$. $\endgroup$ – Brian Tung Aug 9 '15 at 19:38
  • $\begingroup$ A small doubt, if we are counting case of 2 Ace in 4 x 51 why add 4 x 48 again? Because choosing a card from remaining 51 card would cover even other 3 ace's right? $\endgroup$ – JeeZ Apr 14 '17 at 20:30
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The idea is fine, but the count is not quite right.

The probability of at least one Ace is $1-\frac{\binom{48}{2}}{\binom{52}{2}}$. The probability of $2$ Aces is $\frac{\binom{4}{2}}{\binom{52}{2}}$. For the conditional probability, divide.

Remark: A way to count that is closer to yours is to note that we can select, in order, $2$ non-Aces is $(48)(47)$ ways. So the number of ways to choose, in order, a hand with at least one Ace is $(52)(51)-(48)(47)$. Then a calculation like yours gives conditional probability $\frac{(4)(3)}{(52)(51)-(48)(47)}$.

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You have that

$$P( 2As | 1As ) = \frac{P( 2As \text{ and } 1As )}{P( 1As) }$$

But $2As \text{ and } 1As = 2As$, so you just need to find $ P(2As) $ and $P(1As)$

$$P(2As) = \frac{4 \choose 2}{ 52 \choose 2} $$

$$P(1As) = 1- P(0As) = 1- \frac{{2\choose 48} }{ 52 \choose 2} $$

So

$$P( 2As | 1As) = \frac{4\choose 2}{ { 52 \choose 2} - {2\choose 48} } = \frac{6}{ \frac{52\times 51}{2} - \frac{48\times 47}{2}} = \frac{6}{198} = \frac{1}{33} $$

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  • $\begingroup$ Where you wrote $2 \choose 48$, did you mean to write $48 \choose 2$? $\endgroup$ – Mathemanic Jun 18 '16 at 6:26

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