Probability of last person getting two hearts?

Question:

There are 10 people. Each is dealt 5 cards from a 52 card deck. What is the probability that the first 9 people have exactly one heart, and the last person has exactly two hearts?

Here is the solution I am trying to understand:

For notation, the event that the $${i}$$th person has exactly one heart is $$A_i$$.

The solution I am reading states that you must compute $$P(A_1) \times P(A_2|A_1) \times P(A_3|A_1,A_2) \times \cdots \times P(A_{10}|A_1,A_2...A_9)$$ which would be

$$\frac{13*\binom{39}{4}}{\binom{52}{5}} * \frac{12*\binom{35}{4}}{\binom{47}{5}} * \frac{11*\binom{31}{4}}{\binom{42}{5}} ... \frac{\binom{4}{2}*\binom{3}{3}}{\binom{7}{5}}$$

Why are conditional probabilities needed to solve this problem? And doesn't this solution only hold for a specific ordering of people?

Isn't there a way to do this purely with combinatorics?

• Do you mean to be multiplying the probabilities instead of adding them? Commented Dec 9, 2019 at 2:39
• Does the solution use $5$s in the denominator (e.g. $\binom{52}{5}$)?
– Em.
Commented Dec 9, 2019 at 3:42
• Oops, yes I corrected the questions with the correct denominators.
– max
Commented Dec 9, 2019 at 3:54

Conditional probabilities are needed because the fact that the first person gets exactly one heart tells us what deck the second person is drawing from. It has $$12$$ hearts and $$35$$ non-hearts, which is (a bit) richer in hearts than the original deck.

As expressed, the solution is specific to the ordering of people. The probability is the same if you ask the same question with any specific person getting the two hearts. The specific terms will change, but the result will be the same. The easiest way to see it is to imagine dealing all the cards before anybody looks at them. If you want the chance that everybody except the seventh gets one heart and the seventh gets two, just look at the hands in a different order where the last person's hand is looked at seventh.

This solution is purely combinatorics. Why do you think it is not? We have computed the number of successes and divided by the number of total possibilities.

In essence, for that approach, you need to know what happened in the previous draws. Imagine literally dealing the hands. Each successive hand has fewer hearts in the deck to draw from, for example.

If you don't want to think about conditional probabilities, try to consider the following.

Instead of dealing the cards, let's assign when each card will be drawn, or the waiting time until a card is drawn.

Imagine the list $$\{1,\dotsc,52\}$$ representing a particular waiting time (e.g. $$5$$ meaning the cards is the $$5$$th draw. We associate each consecutive group of $$5$$ with the $$k$$th hand (e.g. we associate $$1$$ through $$5$$ with the first hand). Then we can recast the problem as one of assigning waiting times until each heart is drawn.

There are $$\binom{52}{13}$$ ways to assign the waiting times for the $$13$$ hearts (e.g. $$\{1,\dots, 13\}$$ represents that all hearts are drawn in the first $$13$$ draws. Well, in each of the first $$9$$ groups of $$5$$, we just want one arrival (e.g. in draws $$6$$ through $$10$$, we just choose one of those waiting times). In the last group of $$5$$, we want $$2$$ arrivals. Then we have \begin{align*} \prod_{i=0}^{8} \frac{\binom{13-i}{1}\binom{39-4i}{4}}{\binom{52-5i}{5}}\cdot \frac{\binom{4}{2}\binom{3}{3}}{\binom{7}{5}}\cdot\frac{\binom{2}{2}\binom{0}{0}}{\binom{2}{2}} &= \frac{\binom{5}{1}^9\binom{5}{2}\binom{2}{2}}{\binom{52}{13}} \\ &=\frac{390625}{12700271192} \\ &\approx 0.000030757. \end{align*}

First, I would like to say that I agree with Ross Millikan that the book solution is essentially combinatorial. But if you like a combinatorial approach, here is another way of looking at the problem.

The number of possible distributions of the deck into ten five-card hands is $$\binom{52}{\underbrace{5 \; 5 \;5 \;5 \;5 \;5 \;5 \;5 \;5}_{10} \;2} = \frac{52!}{5!^{10} \;2!}$$ (a multinomial coefficient), all of which we assume are equally likely.

In order for the first nine players to have one heart and the tenth player two hearts, the hearts can be distributed in $$\binom{13}{\underbrace{ 1 \; 1 \; 1 \; 1 \; 1 \; 1 \; 1 \; 1 \; 1}_{9} \; 2 \; 2} = \frac{13!}{2!^2}$$ ways, and the remaining non-hearts can be distributed in $$\binom{39}{\underbrace{4 \; 4 \; 4 \; 4 \; 4 \; 4 \; 4 \; 4 \; 4}_9 \; 3} = \frac{39!}{4!^9 \;3!}$$ ways. So the probability of the hearts being distributed as described is $$\frac{13!}{2!^2} \cdot \frac{39!}{4!^9 \;3!} \bigg/ \frac{52!}{5!^{10} \; 2!}$$

I'll first try to explain why you need to use the concept of conditional probabilities to solve this problem. Suppose we have ten people, and we arbitrarily assign each one a number, and deal the cards out to the people in the order which they were numbered. Then, when considering the $$k$$th person, we want to find the probability that the $$k$$th person ($$k < 10$$) gets a single heart, given the event that $$k-1$$ hearts have already been given out. Basically, the cards that have already been given out to the first $$k-1$$ people impact what the $$k$$th person gets, which is why we need to use the notion of conditional probability to take this into account. And, as is specified in the solution you posted, we do this by conditioning on the $$k-1$$ events indicating that each of the $$k-1$$ people got a single heart.

In a sense you are right that this answer holds for only one ordering, the ordering in which the people are dealt their cards. However, this is the only ordering which matters since the people themselves are indistinguishable apart from the order in which they were dealt their cards.

As far as I can tell, there isn't a purely combinatoric way to solve this problem.