## A standard $52$ card deck comprises thirteen ranks in four suits. However, modern decks have two additional Jokers, which neither have a suit nor a rank, for a total of $54$ cards. If we shuffle such a deck and draw cards without replacement, then we would need, on average, approximately $29.05361725$ cards so that we have at least one card for each rank.

Now, assume you have $10$ such decks, each with a different back design. We shuffle all $10 \times 54$ cards together and draw cards without replacement. What is the expected number of cards needed so every suit, rank and deck design have at least one card?

Give your answer rounded to eight places after the decimal point.

### The total number of cards in this case is $10 \times 54 = 540$. These cards can all be seen as different, as each one is distinct not only by rank and suit, but also by deck design.

The problem can be simplified to finding out how many of these distinct elements we’d need to draw, on average, to encounter each at least once.

This is known as the “Coupon Collector’s Problem”. The formula for the expected number of trials needed before having seen all distinct elements is given by:

Sum^N_{i=1} (N/i)

where N is the total number of distinct elements.

In this case, $N = 540$ (10 different decks, 54 cards each), and by substituting this into the formula above, we get:

Sum^540_{i=1} (540/i) = 540 * (1/1 + 1/2 + 1/3 + … + 1/540)

This expression represents the sum of the first 540 terms of the harmonic series, multiplied by 540.

This can be approximated to be $540 * ln(540) + 0.5772156649 * 540 \approx 3707.99185612$ (using Stirling’s approximation), where ln is the natural logarithm (base $e$), and 0.5772156649 is Euler’s constant.

Therefore, on average, you’d need to draw approximately 3707.99185612 cards to ensure each suit, rank, and deck design are represented at least once.

To 8 decimal places, this is $3707.99185612$.

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