Let's ask what the answer is for the spade suit alone.
There are 2^{13} = 8192 up-card configurations, and I just counted the card values for each of them. In **632** cases the total was a multiple of 13.

**p** = 0.0771484375 = **1/12.96702532**.

Fairly close to 1/13. I went back and got the count for the 12 other remainders. It was **630** in every other case. There was an excess of **2 **cases for exact multiples. That is, 632 + 12x630 = 8192 cases. So now I am curious about more suits in the deck.

I added hearts to the spades and counted remainders for those 2^{26} cases. It turned out that the zero remainder cases outnumbered the others by **4**. But there were astronomically more cases, making that small excess really insignificant. So for a deck comprising hearts and spades we get

p =** 1/12.9999907.**

The only question remaining for adding diamonds and clubs is whether the excess numbers for zero remainder would be **6** and **8**, or** 8** and **16**. The numbers were getting too large, so I tried a deck with only seven cards to a suit and checked for divisibility by seven. I could enumerate 3 and 4 suit decks. It turns out the excess number for an **n**-suited deck is **2**^{n}. Here are the distributions for a deck with seven cards/suit, having one, two, three and four suits:

2^{7} = 128 = 20 + 6x18 p = 20/128 = **1/6.4...**

2^{14} = 16384 = 2344 + 6x2340; p = 2344/16384 = **1/6.9897...**

2^{21} = 2097172 = 299600 + 6x299592; p = 299600/2097172 = **1/6.999906...**

2^{28} = 268435456 = 38347936 +6x38347920; p = 38347936/268435456 = **1/6.99999749...**

This validates the analysis. Note that the red numbers differ by 2, 4, 8 and 16 for 1, 2, 3 and 4 suits.

So back to the 13-card per suit deck. The calculation procedure is to take the number of up-card configurations, in our case that's **2**^{52}, which is of the order of 10^{15}, divide that number by 13, which is the number of possible remainders, and find a nearby integer **n** such that **n** + 12(**n**-16) = 2^{52}.

Then the probability of divisibility by 13 is **p** = **n**/2^{52}. This is approximately

**p = 1/12.99999999999999...**

where there are 14 '9's (1 less than the exponent of 10 for the number of configurations.).