I'm trying to learn probability and statistics but I can't really get my head around this one. I realize after drawing the first card there will only be 51 cards in the deck but I'm having trouble calculating the chance that the second one is an Ace if I don't know what the first card is?
Assuming that the initial card was an Ace, I could say the probability of the second being an Ace is 3/51... but how do I approach this with an unknown card?
$\endgroup$4 Answers
$\begingroup$There are $52\times 51$ outcomes when you draw 2 cards one after another without returning the first drawn card to the deck.
There are $48\times 4$ outcomes when the first card is not Ace and the second card is Ace. There are $4\times 3$ outcomes when both cards are Ace. The total outcomes are $48\times 4 + 4 \times 3 = 51\times 4$.
The probability is $\frac{51\times 4}{52\times 51}=\frac{51}{52}\times\frac{4}{51}=\frac{4}{52}=\frac{1}{13}$.
$\endgroup$ $\begingroup$For any given position $n$ in the deck (here $n=2$, meaning the second card,) if the deck is shuffled then the probability that the card in position $n$ is an ace is $4/52 = 1/13$, because $4$ out of the $52$ cards in the deck are aces.
$\endgroup$ $\begingroup$Break the answer into two scenarios:
- If the first card is an ace and the second card is also an ace
The probability, in this case, is $$\dfrac{4}{52}\dfrac{3}{51}.$$
If the first card is not an ace but the second card is an ace
The probability, in this case, is $$\dfrac{48}{52}\dfrac{4}{51}.$$
Since these two scenarios are mutually exclusive, the final probability is the sum of these two probabilities: $$\dfrac{4}{52}\dfrac{3}{51} + \dfrac{48}{52}\dfrac{4}{51}.$$
$\endgroup$ 1 $\begingroup$Hint: If the first one you draw is an Ace then the probability for the second one to be an Ace is $\frac3{51}$;
however if the first one drawn is not an Ace then the probability is $\frac4{51}$.
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