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This goes back to a problem given to my statistics class, 50 years ago. You are dealt 5 cards from a standard deck of 52. What are the odds of you having the Queen of Spades? The Professor said to work on it and be prepared to answer the following day. I said I could give the correct answer right then, which I did. We took most of the next class discussing/arguing about the answer. I may have won the argument but lost the war (grade). What is the correct answer?

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    It's cool that you still think about it! What was your solution, and how did you arrive at it? (Use "@TheCount" so that I see your response) – 2017-01-05
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    A word of advice, the question asking for **the odds** is slightly different than the question asking for **the probability**. Those are related but different concepts. If you know how to calculate one of these though, you know how to calculate the other. I prefer to think with probabilities personally. – 2017-01-05
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    @JohnRichard So who was right? – 2017-01-10

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There are $52$ cards in the deck, of which (after random shuffling) the first $5$ will be dealt to you. The Queen of Spades is equally likely to be in any of the $52$ possible positions, so the probability it is dealt to you is $5/52$.

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    And so the **odds** of it occurring are... – 2017-01-05
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    $5$ to $47$, of course. – 2017-01-05
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    Why assume the player is dealt the first 5 cards? – 2017-01-05
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    @amWhy why **not** assume that? There is a clear bijection between shuffles of decks which result in the first player getting the first five cards and shuffles of decks which result in the first player getting the first, third, fifth, seventh and ninth cards. Regardless, the way the scenario is worded, makes it sound like those are the only cards being dealt. – 2017-01-05
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    If the dealer **doesn't** deal from the top of the deck, you might want to be suspicious... Of course it doesn't matter which cards you are dealt, as long as that choice is independent of the values of the cards. But it may be easier to focus the mind if they are the first 5 cards rather than some other set. Somehow the probability that the top card is the Queen of Spades is psychologically easier than the probability that the bottom card is the Queen of Spades. – 2017-01-06
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The total number of ordered hands - that is, five-element sequences of cards, where each card is different - is $52\cdot51\cdot50\cdot49\cdot 48$.

Now: of these sequences, how many contain the Queen of Spades?

  • There are $1\cdot 51\cdot 50\cdot 49\cdot 48$-many sequences whose first card is the Queen of Spades.

  • There are $51\cdot 1\cdot 50\cdot 49\cdot 48$-many sequences whose second card is the Queen of Spades. Note that we begin with $51$, not $52$: since we know the second card is the Queen of Spades, this only leaves $51$ possibilities for the first card (it can be anything but the Queen of Spades).

  • Continuing this way, we see: there are $5$ ways for a sequence to contain the Queen of Spades (first, second, third, fourth, or fifth card), and each way corresponds to $51\cdot50\cdot 49\cdot 48$-many distinct sequences.

  • So the total number of sequences containing the Queen of Spades is $51\cdot50\cdot49\cdot48\cdot 5$.

Now we compare these two numbers: the relevant probability is $${\mbox{sequences with QoS}\over\mbox{all sequences}}={51\cdot50\cdot49\cdot48\cdot 5\over 52\cdot51\cdot50\cdot49\cdot 48}={5\over 52}.$$

EDIT: looking at JMoravitz' comment, I see I made a silly mistake: the number I calculated above is the probability that you get the Queen of Spades. But the odds are slightly different: think "the odds are $2$ to $1$". When we speak of odds, we're comparing two probabilities - "Yes You Do" vs. "No You Don't" - rather than calculating one of the probabilities on its own.

In this case, you have a $5\over 52$ probability of having the QoS, and a $47\over 52$ probability of not; so the odds of you having it are $5$ to $47$.

Note that these two things are just different ways of phrasing the same information.

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We seem to be showing various solutions for the probability. Here are two and a half more:

(1) Combinations. $$\frac{{1 \choose 1}{51 \choose 4}}{{52 \choose 5}} = \frac{5}{52} = 0.09615.$$

(2) This is essentially the same as $P(X=1),$ for a hypergeometric random variable $X$ that counts the red balls among five chosen at random without replacement from an urn with 1 red ball and 51 non-red ones. In R statistical software: phyper(1, 1, 51, 5) returns 0.09615385.

(3) A simulation in R of a million repeats of the experiment. I have numbered the 52 cards so that the Queen of Spades is # 1. The result should be accurate to three places.

m = 10^6;  get.qs = numeric(m)
for (i in 1:m) {
  hand = sample(1:52, 5)
  get.qs[i] = sum(hand==1)  }
mean(get.qs)
## 0.096086

An Assignment: Write "The Professor is always right!" 500 times, or execute code.

rep("The professor is always right!", 500)