what is the probability of getting consecutive pairs of cards in a deck after shuffling the deck?

Answer 1

The number of possible orderings of a deck of cards is 52!, or 52 times every positive integer less than 52.

If you have consecutive pairs, then there is a first card of the four. This may be any card. The second card must match the first, and there are three left. The third card may be any card. If it doesn't match the first two, the next card has 3 options. If it does match the first two, there is one option left (and you'll have four in a row.) We don't care what order any other cards are in.

So, let's split up the cases a bit more, into the case where it's one of the 48 remaining cards, and not matching.

We have 52*3*48*3*(48!) orientations of the cards where we have two consecutive pairs that are not of the same type.

We have 52*3*2*1*(48!) orientations where there are four in a row.

Note that neither of these cases exclude more pairs (or quads), but you only asked for one of them. If you want, you can extend it to more.

You asked for the odds.

(52*3*3*48+52*3*2)(48!)/(52!) will give the probability that you have such an orientation, which is about .003505. which is really pretty low.

Answer 2

I assume you mean off the top of the deck, and not anywhere in the deck - that would be slightly longer.

Card 1: can be any of the 52 cards
Card 2: can be any of the 3 matching cards to card 1
Card 3: can be any of the 48 remaining cards not matching card 1
Card 4: can be any of the 3 cards matching card 3

Total choices: 52*3*48*3
Total possibilities:52*51*50*49
Chance of top 4 cards being 2 consecutive pairs: 3*3*48/(51*50*49)
= 432/124950 = 72/20825 about 0.0035

Answer 3

If you have a full 52 card deck and draw 1 card then it leaves 51 cards, with 3 matching the one you drew. So, drawing a second card that matches the first is 3 out of 51.

Oops, you said consecutive pairs and not consecutive cards making a pair, I misread it... here's how to do that...
After that, you'll have a 48/50 chance of drawing a different card than the first two (I assume you don't want 4 of a kind in a row to count as 2 pairs). After that, you'll have 3/49 chance of getting the fourth card to match the third. So, you have 3/51 chance of getting the first pair and a 48/50 to not get 3 of a kind, and out of that you have 3/49 chance to get the second pair... multiply those (3*48*3) and (51*50*49) and get 432 out of 124950 I think.

Your first drawing can be 52 of 52 cards, so that first drawing can be ignored since it's a 100% chance.

Answer 4

The number of ways to choose 4 cards from the deck is given by the binomial coefficient
C(52,4) = 52!/[4!48!] = 49*50*51*52 / (2*3*4) = 270,725.

To compute the number of pairs, we note that there are 13 possible values (2,3,4,...,Q,K,A), 4 cards of each, so the number of pairs is 13*C(4,2) = 78.

There are 78*77=6,006 possible pairs of pairs (repetition is not allowed, but the pairs can be chosen in either order). Double this again, because the cards within a pair can be chosen in either order, so you have 12,012 sequences of 4 cards consisting of two consecutive pairs.

Therefore, the probability is 12,012 / 270,725 = 0.044.

Answer 5

This seems to be on the threshold of a binomial regulation with parameter p = 3/fifty two. In different words putting the enjoying cards decrease back interior the deck or no longer would not seem to electrify the consequence very plenty.

Answer 6

slim and none, i would tend to think

Answer 7

lowish!