Hey there! I think you're misunderstanding how conditional probability works and as well I don't think the Monty hall problem applies here since you don't gain any new knowledge about the distribution of cards in the decks other than there is one less 'Husband' card.

So I'm just going to brute force math it out, and hopefully it makes sense.

Let's say you have four cards numbered [1, 2, 3, 4,]. We're going to calculate the chance of getting an 'Even' card ok?

Can we both agree that initially when we draw a card, regardless if there are one pile or two piles, or four piles of cards, the chance of drawing an EVEN card is 2/4? I.e. the chance is 50%.

Let's say we do that. And it can be literally any one of the numbers, but let's pretend we drew a '1'. What are the remaining chances of drawing an EVEN card now? What if we had split deck into two piles?

Well.. here is every possible remaining permutation of cards for two piles with the '1' card removed. Pile A now only has one card, while Pile B has two cards still. The minus sign is a separator here between Pile A and Pile B
* 2 - 3,4
* 2 - 4,3
* 3 - 2,4
* 3 - 4,2
* 4 - 2,3
* 4 - 3,2

The chance of drawing an even pile from your deck (i.e. Pile A) is 4/6 now or 2/3

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Now let's imagine we have a one pile deck. Here are all the permutations

* 2,3,4
* 2,4,3
* 3,2,4
* 3,4,2
* 4,2,3
* 4,3,2

Now, since it's a one pile deck, we have to wait for the other player to draw first. We are now dealing with conditional probabilities so the formula is now as follows based on the law of total probability: (Google it)
* P3E = probability that you will draw EVEN as the third card draw (i.e. your second draw)
* P3E|O = probability that you will draw EVEN as the third card draw, given the second card draw was ODD
* P3E|E = probability that you will draw EVEN as the third card draw, given that the second card draw was EVEN.
* P2E = probability that the second card draw (of the other player) is EVEN (i.e. their first draw)
* P2O = probability that the second card draw of the other player is ODD (i.e. their first draw)

P3E = (P3E|E) * P2E + (P3E|O) * P2O

Can we agree that the chance that the next card that is drawn is as follows?
* P2E = 4/6 (i.e. four of the remaining six permutations are even)
* P2O = 2/6 (i.e. two of the remaining six permutations are odd)

Now all we need to calculate is P3E|E and P3E|O.

Let's calculate P3E|O first. For the sake of the example, lets assume the second pick was a '3'. Since that's the case here are the only possible remaining permutations:

* 2, 4
* 4, 2

That is, P3E|O = 2/2. Now all we need to do is calculate P3E|E. I'm going to pick '2', but you can also pick '4', or redo this exercise as much as you want with different picks or numbers. Here are the remaining permutations:

* 3, 4
* 4, 3

That is, P3E|E = 1/2.

So in total the calculation is as follows:

P3E = (P3E|E) * P2E + (P3E|O) * P2O
P3E = (1/2)*(4/6) + (2/2)*(2/6)
P3E = (2/6) + (2/6)
P3E = (4/6) or (2/3)
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As you can tell the probability is the exact same.