Drawing card question, any guess?

I think in the limiting case, as P -> 1/e, the Q of your Spades and Club drawing game will be close to P², that is, √Q -> 1/e, but only slightly less. Let me think about some more why that should be. I've done some Monte Carlos on this, and results so far seem to agree with this. I'll run a much longer trial overnight to see what comes out.

Edit: Okay, seeing that Subfactorial(13)/13! is very nearly the same as 1/e = 0.367879..., Monte Carlo trials can be done with just 13 (or 26) cards as posed by this question. After 10 million trials of each, the percentages of failure to match is as follows:

With 13 cards......0.367688

With 26 cards......0.360783²

S(13)/13! or 1/e....0.367879

If instead of using 13 cards, we used 1000 cards, after 100,000 trials, the difference is even smaller:

With 1000 cards...0.367898

With 2000 cards...0.367423��

S(1000)/1000!.......0.367879

Edit: Let me tabulate exact calculations of number of derangements in the case of 3 & 6 cards, 4 & 8 cards, and 5 & 10 cards (standard and double derangements)

3 cards: 2 out of 6 = 0.333333

6 cards: 80 out of 720 = 0.333333²

4 cards: 9 out of 24 = 0.375000

8 cards: 4752 out of 40320 = 0.343303²

5 cards: 44 out of 120 = 0.366667

10 cards: 440192 out of 3628000 = 0.348289²

The pattern is that both approach 1/e = 0.367879, but the square root of the double derangement approaches it more slowly than the standard derangement.

Edit 2: It has proven really tricky to edit this answer correctly, took too many tries.

Edit 3: See Edit 2. I used 1000 & 2000 cards, running 100,000 trials each. Also, I used software routine to tabulate all possible permutations, including the 10 card "double derangement" scenario, rejecting the permutations that resulted in a match. I did not use a formula, because then otherwise I could have worked out the numbers exactly for the 26 card "double derangement" scenario, as we can easily work out for the 13 card straight derangement scenario. A detail of utmost importance is that Spades and Clubs are distinguishable when tabulating the permutations. Otherwise, we'd end up with a kind of a Bose-Einstein statistics kind of thing, where the total number is reduced by a factor of 2^n, which would throw off odds in an actual game play. For example, with a 4 card "double derangement", there are 4! permutations, but if suits were indistinguishable, then there are only 6.

Obviously, if the Spades and Clubs were dealt into 2 separate piles (you know, the separate but equal thing), and 1 card was drawn from each in parallel succession, then P² = Q exactly. But it gets tremendously more complicated if the suits were mixed up between those equal sized 2 piles.

Edit 4: In all the different simulations and direct computations I've done, for large n Q does seem to trend upwards towards 1/e², and doesn't get past it. I've had to run some pretty large trials to convince myself that it doesn't end up with a slightly lower value asymptotically. If it's important to you, maybe I can run a super large trial?

Edit 5: Okay, why don't you hold on while I run some large trials for n = 1000 or greater.

Edit 6: gianlino, using that terrifying-looking formula you've just posted, I plugged in 1000 in place of 13, and ended up with 0.367787 (which is the sq root). Which is pretty close to the results I've been getting at attempted 1M trial runs. So we do agree that (1/e²)(1 - 1/(2n)) is the trend, so now what is the question?

Edit 7: Okay, I understand the question now. I haven't quite figured that one out yet.

I will try to give a simplified answer as follows

The derangement ( no of ways miscalls) of 13 cards of spades 13!(1-1/1!+1/2!-1/3!+......(-1)^13/13!)out of a total of 13! ways the calls can be given. Hence P = (1-1/1!+1/2!-1/3!+.........-1/13!). Thus P= first 14 terms of 1/e series. Now the 15 th term is +ve and > 16 th term for 1/e series . This means 1/e is slightly greater than P. Now if we say P= 1/e then Q=P^2 is 1/e^2 but actually slightly less than 1/e^2 because the probabilities are fractions. However , to answer the question I feel Q has to be always equal to P^2 as the 2 fold derangement problem is actually two independent derangement problems combined which by the multiplication rule of probabilities should give Q=P^2. But you can always work out the full number of terms in either case and confirm the result which I believe would be the same.

@Giliano you are right. Now if you have identified that the first probability classical case or P= first fourteen terms of 1/e series and second probability is first 27 terms of 1/e^2 series we can easily see from the first five terms of each of these series P = 0.375 or P^2=0.141 where as Q = 0.073 for first five terms. So for a finite series out of an infinite series as is the case here Q is definitely less than P^2 higher terms being inconsequential under finite conditions as they impact little on the out come.

I'm just thinking out loud here. For a given pair, there is 50% chance that one

of the cards will be assigned one of the first 13 spots and the other card assigned

one of the second group of 13 spots. This would be business as usual. But there

would also be a 50% chance that both cards are either among the first 13 spots

or the second group of 13 spots. In this case the likelihood that one of the cards

would be in the correct spot would increase, and hence have the effect of

decreasing Q. So although the precise calculation of Q would be complicated,

I would expect that Q < P^2. Perhaps I've misinterpreted.

Edit: On second thought, if the two cards were both in either the first group

or second group of 13 then there would be no representation of that denomination

in the other group. So although the probability that one of the cards is in the

right spot for one group of 13 is doubled, it is reduced to nil for the other, the net

result being a wash. Can we then extend that result when we consider the

distribution of all the pairs between the two groups of 13? Probably, since the

ordering of the cards is random. I'm looking forward to Scythian's computer results.

Edit #2: I'm just trying to get my head around this. Is then the 'redundancy' of

double hits for one or more pairs the primary reason why Q < P^2 for a double

derangement?

Are you talking about the derangement problem?

This is to do with getting a set of numbers, cards, letters etc in the right order, or getting them all in the wrong place. Like putting letters in envelopes at random.

It has nothing to do with taking out all the spades, clubs or whatever.

The probability of getting every card, out of a set of n cards, in the wrong place is 1/e as n gets large.

Meaning that it is more likely than not that at least one card will be in the right place. (Pr = 1 - 1/e)

For n = 52 the Pr is certainly very close to 1/e.

they practice. you never drew a random card. they gave you one without you even noticing.