Common card math puzzle

the answer appears to be the number of ways to partition the set {1, 2, 3, ..., 52} into partitions all of which are size 2 or larger

Here's why:
The order of Axel's deck doesn't matter, since we are only interested in matches, so assume it's {1, 2, 3, 4, ..., 52}
Bob's deck can be expressed as a set of chains - for example, if the first card is 6, the sixth card is 22, and the 22nd card is 1, then {1, 6, 22} is a chain.
It is possible for all 52 cards to be a single chain.
If any chain of length 1 exists, then that card is in the same position in both decks.

I am under the impression that there is no "closed form" solution to count the number of ways to partition a set of size, say, 52, much less to count the number of ways where there are no partitions of size 1.
However, a quick check of the Online Encyclopedia of Integer Sequences says that the number of partitions is the 53rd term of the recursive sequence:
s_n = n s_n-1 + (-1)ⁿ, where s₁ = 0
The probability would be s₅₃ / 52!

Sounds like a derangement problem.

In that case the answer would be:

1 - 1/e = ~ 63.2 %

1 - (51/52)^52 = 63.6%

That would be a good estimate of the answer, but not an exact form. Certainly the answer is an integer divided by an integer, thus the answer must be rational, unlike your estimate.

Much like Ace2, a pretty good estimate, but not the exact answer. There is an effect of removal you must consider.

The sum of (-1)^(j+1)/j! from j=1 to j=52 which is ~63.2120558828558%.

My initial “estimate” of 1 - 1/e comes out to ~63.2120558828558%.

sum(i=1,52,(-1)^(i+1)/i!)

As ThatDonGuy pointed, we can assume cards are numbered and deck A is ordered A1 = 1, A2 = 2 .. A52 = 52.
Then the probability is the number of orders for deck B for which no card Bi = i divided by the total number of possible orderings, 52!
After a quick search that number is called subfactorial, !52.
So p = !52/52!
!52 = 52! - 52!/1! + 52!/2! - 52!/3! + 52!/4! ...
or !52 = 52! * sum[for n=0 to 52]( -1^n / n!)
Interesting result is that seems to tend to "1/e"
lim !n/n! = 1/e
so !52/52! = 0.3678794411

Subfactorials.

I did some googling and it's an interesting concept.
https://en.wikipedia.org/wiki/Derangement

In particular that page gives and example of considering giving hats to people and derives the function using a form of recursion. It also gives the answer, which I won't repeat here (near reference (4) ).

So the hint is to see if you can look at the hats problem and derive a recursion.

The following solution is just rewording what others have said. However, I hope the elaboration will be helpful to some.

It is easier to see the logic if we assume the decks are each consist of cards numbered 1 to 52. The first deck is in order, from 1 to 52. The second deck is randomly shuffled. The answer will be the ratio of the number of ways the second deck can be ordered where there is at least one card that matches to the total number of ways to order 52 cards.

Step 1: To start, consider the number of ways the second deck can be ordered where the first card is number 1. The answer is the number of ways to order the other 51 cards, which is 51! = 1551118753287382280224243016469303211063259720016986112000000000000.

Any card can match the first deck, so we must do this for all 52 cards. This gives us 52*51! = 52! combinations where at least one card matches.

Step 2: However, step 1 double counts every situation where two cards match. For example, if the first two cards were 1 and 2, we would have counted the 50! ways to arrange the other cards twice, once for 1 as the first card and a second time for 2 as the second card. The number of ways to choose 2 cards out of 52 is combin(52,2) = 1326. For each combination of two cards, there are 50! = 30414093201713378043612608166064768844377641568960512000000000000 ways to order the other cards. Thus, for step 2 we need to subtract out combin(52,2)*50! = (52*51/2!)*50! = 52!/2! combinations.

Step 3: Next, consider the situation where the first three cards in the random deck are 1, 2, and 3 in order. There are 49! ways to order the other 49 cards. We would have counted them three times in the initial step of counting for at least one matching card. Then we would have subtracted out all combin(3,2)=3 ways to choose 2 out of these three cards in the second step. So this situation would have been counted 3-3=0 times, so we need to add them back in. There are combin(52,3) such situations of choosing at least 3 cards that match. So we need to add back in combin(52,3)*49! = 52*51*50*49!/3! = 52!/3! combinations back in.

Step 4: Next, consider the situation where the first four cards in the random deck are 1, 2, 3, and 4 in order. There are 48! ways to order the other 48 cards. We would have counted them four times in the initial step of counting for at least one matching card. Then we would have subtracted out all combin(4,2)=6 ways to choose two out of these four cards in step 2. Then we would have added all combin(4,3)=4 ways to choose 3 out of these four cards. So we are at 4-6+4=2 ways each such situation would have been counted. So we need to subtract one of those ways out, so that each situation will be counted once. There are combin(52,4)*48! = 52*51*50*49*48!/4! = 52!/4! such situations that need to be added back in.

We will keep doing this, alternating adding and subtracting to correct for double counting.

In the end, the number of situations where at least one card in the random deck matches the ordered deck =
combin(52,1)*51! - combin(52,2)*50! + combin(52,3)*49! - combin(52,4)*48! ... - combin(52,52)*1! =
52!/1! - 52!/2! + 52!/3! - 52!/4! ... - 52!/52! = x = 333239808909468890675694068318655265019682314241643033726180828783.

There are 52! = y = 527177615496365219422618541545122659969212453861982208000000000000 total ways to order 52 cards.

Thus, the answer is x/y = 0.6321205588285576784044762298385391325541888689682321654921631983025382745851401611680269165474009520.

The probability that there are no matches is 1-(x/y) = 0.3678794411714423215955237701614608674458111310317678345078368016974617254148598388319730834525990480.

If this number looks familiar, it should. 1/e = 0.3678794411714423215955237701614608674458111310317678345078368016974614957448998033571472743459196437.

So, the answer can be VERY closely estimated, to 69 decimal places, as 1-(1/e).

Acknowledgements

Mathematical calculations were done in Pari/GP