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I was thinking about the possibility of storing information in the ordering of a pack of playing cards. Since there are $52$ distinct cards, there are $52!$ different ways to order them. This gives $\log(52!)=225.58$ bits of information (using the base $2$ logarithm).

I then wondered about shuffling together $2$ packs of cards. If they have different coloured backs, then this gives $104$ distinct cards, and $\log(104!)=551.48$ bits of information. If they have the same coloured backs then this gives $2$ each of $52$ distinct cards, and $\log(104!/2^{52})=499.48$ bits of information. It seemed intuitively clear that having the same coloured backs would give less information (fewer distinct arrangements).

What is confusing me is that when I considered much larger numbers of packs of cards, having distinct coloured backs did not seem to add as much information as I would have expected. For example, I had expected that $52$ packs of cards, each in one of $52$ different colours, would be able to store far more information than $52$ identically coloured packs of cards.

For $52$ different colours, there are $\log((52\times52)!)=26,933.95$ bits of information.

For $52$ packs the same colour, there are $\log((52\times52)!/52^{52})=26,637.53$ bits of information.

Am I calculating the total number of distinct arrangements incorrectly, or is it just my intuitive feeling that distinct colours should add far more information that is incorrect?

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In writing out the question, I realised my mistake. I needed to divide by $n!^{52}$, not $n^{52}$. The omission did not cause a problem for $n=2$ because $2!=2$, but for all greater $n$ this overestimates the number of arrangements of $n$ packs of identically coloured cards.

For $n=2$, there are two Ace of Clubs cards, and there are $2$ ways of ordering them. Since both these arrangements are identical, the total number of distinct arrangements needs to be divided by $2$. This applies for each of the $52$ types of card, so overall the total number of distinct arrangements needs to be divided by $2^{52}$.

For $n=3$, there are three Ace of Clubs cards, and there are $3!=6$ ways of ordering them. Since all these arrangements are identical, the total number of distinct arrangements needs to be divided by 6. My mistake was dividing by 3, as if there were only 3 ways of arranging 3 items.

With the corrected approach, there is a much more significant difference between 52 distinctly coloured packs and 52 identically coloured packs.

For $52$ different colours, there are $\log((52\times52)!)=26,933.95$ bits of information.

For $52$ packs the same colour, there are $\log((52\times52)!/52!^{52})=15,203.74$ bits of information.

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