# Probability in a Card Deck using Combinations

Consider the following experiment. Three cards are drawn from a standard deck of 52 cards, one after the other. Before the second card is drawn, the first card is put back in the deck and the deck is re-shuffled. Before the third card is drawn, the second card is put back in the deck and the deck is re-shuffled.

1. Find the probability that all three cards will be in the suit of hearts.

2. Find the probability of not getting all three cards in the suit of hearts.

3. Find the probability of not getting any cards in the suit of hearts.

So far I solved (1) by using

$$\frac{\binom{13}2}{\binom{52}3}=\frac{\frac{13!}{(13-3)!3!}}{\frac{52!}{(52-3)!3!}}=\frac{2860}{22100}=0.129\approx12.9\%$$

The possible number of three hearts in a card deck is $2860$, and the possible number to have three hearts in a card deck is $22100$. Using these values yield $0.129$ as probability to pick three cards where all are suit of hearts.

The second point confuses me a bit. Not having all three cards could mean one, two, or maybe zero cards in the suit of hearts. I thought about using the counting principle and multiple $$\frac{\binom{13}0\binom{13}1\binom{13}2}{\binom{52}3}\;,$$ but I am not sure if this is right. If somebody is able to point into a direction that would be great.

The third part of this exercise I assumed to be $$\frac{\binom{13}0}{\binom{52}3}\;,$$ since none of the cards ($0$) is in the suit of hearts.

I appreciate your time and efforts. Thanks

-Daniel

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In the calculation in (1) you have a typo in the first numerator $-$ you clearly meant to have $\binom{13}3$, not $\binom{13}2$ $-$ and an arithmetic error: $\binom{13}3=286$, not $2860$. However, you’ve made a more fundamental error at the start: with the errors corrected, your calculation would give you the probability of getting three hearts if the cards were drawn without replacement, but here they are drawn with replacement. Thus, at each draw you are choosing one card from $52$ and have a probability of $\frac14$ of drawing a heart. The probability of drawing three hearts is therefore $$\left(\frac14\right)^3=\frac1{4^3}=\frac1{64}=0.015625\;.$$

If you wish, you can instead count the successful outcomes amongst the equally likely outcomes: there are $13^3$ possible sequences of three hearts, and there are $52^3$ possible sequences of three cards, so the probability of success is $\frac{13^3}{52^3}$, which works out to the same result as before.

Your numerator in (2) makes no sense: it’s the number of hearts times the number of pairs of hearts times the number of triplets of hearts. If you were to calculate this probability by brute force, you’d want the probability of getting no hearts plus the probability of getting exactly one heart plus the probability of getting exactly two hearts. However, this is the hard way to do the problem. The event whose probability is wanted in (2) is just the complement of the event in (1), so its probability is the probability that (1) does not occur. The probability that (1) does occur is $\frac1{64}$, so the probability that (1) does not occur must be ... what?

In (3) you’ve gone off the rails altogether: this would not be right even if you were drawing without replacement. Think a moment: $\binom{13}0=1$, and there is certainly more than one possible three-card hand with no hearts! In fact to get no hearts you must draw a non-heart each times. The probability of drawing a non-heart the first time is $\frac{39}{52}=\frac34$. The probability of drawing a non-heart the second time is also $\frac34$, as is the probability of drawing a non-heart the third time. Now how should you combine these numbers? Note that this problem is structurally exactly like (1).

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I'm a bit confused with the card drawing process. From your statement, it seems that the three cards are all drawn from a full deck following a uniform distribution. So the answer should be

(1)$$P(\text{all three are hearts}) = \left(\cfrac{13}{52}\right)^3 = \cfrac{1}{64}$$ (2)$$P(\text{not all three are hearts}) = 1-P(\text{all three are hearts}) = \cfrac{63}{64}$$ (3)$$P(\text{none of three is heart}) = \left(\cfrac{39}{52}\right)^3 = \cfrac{27}{64}$$

If there is no putting back operation in drawing process, then the answer should be

(1)$$P(\text{all three are hearts}) = \cfrac{{13\choose3}}{52\choose3}$$ (2)$$P(\text{not all three are hearts}) = 1-P(\text{all three are hearts})=1- \cfrac{{13\choose3}}{52\choose3}$$ (3)$$P(\text{none of three is heart}) = \cfrac{39\choose3}{52\choose3}$$

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