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This is how I thought about this problem, but I am not completely sure it is right:

There are four Queens so the number of arrangements of the Queens = 4!

There are 52! possible combinations of all the cards.

There are (52-4)! = 48! combinations for all the cards without the Queens.

Then the probability of the four Queens occurring consecutively would = (4!48!)/52!

Is this logic alright?

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up vote 9 down vote accepted

Think the four consecutive queens as if they were a single card. Then we have $49!$ ways to shuffle the remaining $48$ cards with this special card. Each of these shufflings entails $4!=24$ actual deck permutations, for a total of $49!\cdot 4!$ arrangements. Clearly, $\frac{49!\cdot 4!}{52!}$ is the desired probability.

What's going wrong in your answer is that you're not accounting where do the queens occur in the deck.

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The answer seems to be $\frac{4!49!}{52!}.$ There are $48!$ ways to arrange the deck without queens, $4!$ ways to arrange the $4$ queens, and $49$ ways to insert an arrangement of queens into a deck containing no queens such that all four queens occur consecutively. Noting that doing these actions consecutively results in all decks where all four queens occur consecutively and no double counting occurs, we conclude that the number of decks in which all four queens occur consecutively is $4!49!.$

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There are $52$ chairs arranged in a row, to seat $52$ honoured guests, among them $4$ queens (real queens). We choose $4$ of these chairs at random, and put Reserved signs on them. What is the probability these $4$ chairs are consecutive?

The set of chairs that will have the Reserved signs can be chosen in $\binom{52}{4}$ ways, all equally likely.

How many of thse choices are consecutive? The leftmost of these $4$ consecutive chairs can be any one of chairs $1, 2, \dots, 49$, so there are $49$ ways to choose $4$ consecutive chairs. The required probability is therefore equal to $$\frac{49}{\binom{52}{4}}.$$

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