Way back in the day ca. 1969 or so, my uncle proposed a bet that I wouldn't get a randomly shuffled deck of cards to not have at least one pair.  His bet was something like Pay $1000 : to $1 bet for each random shuffle---no stacking of the deck allowed.

The other day, I found a deck of cards while reorganizing and I shuffled it, and remembered the challenge.  I went through it and to my surprise, there was not one pair.  Upon shuffling it a number of times, there appeared at least one pair.  Since then, I have been thinking about how to solve the probability without the use of a calculator, or computer.  These would have not been available to my uncle back in the day.  

Some of my thoughts:  

There are 52! possible combinations of card order in a deck.  Figure out the possible combinations where there are no pairs, and divide from 52!   Or, figure out the possible combinations of at least one pair to get the inverse of the above.  

The first card is always drawn and does not matter what it is.  So you have the first draw cancel from each side of the proportion. On the second draw, you have a 3 in 51 chance of drawing a like value.  On the third draw you have . . . . Now this is where I get a bit lost.  You have a chance to draw a like value of your first card which reduces the chance of drawing it later (or does this matter), and a chance to draw a like value of your second card that is 3 (minus something?) in 50 chance.  As it goes down the line, it seems to me to get more complicated.   Doing this by hand would be impractical at best.  

How did my uncle likely calculate this? I can see that you could get a sequence where the n! numbers factor out, but I'm not sure on the other parts.  

There are exponentially more combinations than just 52.

Take 9 numbers only for example.

It could be

123456789
132456789
142356789
152346789
162345789
172345689
182345679
192345678
213456789
312456789
412356789
512346789
612345789
712345689
812345679
912345678
192837465
184795326

And so on...

Now add 4 of each number, 2 jokers, and a good shuffle.

It's a crazy amount of ways the cards could land

The exclamation point indicates factorial.  That is, you are multiplying 52 x 51 x 50 x . . . x 2 x 1.

52! is about 8.0358 x 10 ^ 67.  This is an impractical thing to put on paper, hence my question.  

ETA:  Jokers were omitted in the deck for the purposes of the bet.

Also... with some quick math.

The probability of the second card having the same value as the first is 3/51. And of the 48/51 times where the first two cards are unequal, 3/50 of them will have the second card equal the third. But following this, it becomes very, very complex: if you get no pairs among the first three cards, the probability of the fourth card matching the third depends on whether the first and third cards matched. If any matched, its 2/49. if not it becomes 3/49

Exactly why my I'm wondering how my uncle figured out the odds.  It seems way too complicated without using a computer program to do this, yet he seem to have done so.

Maybe he was a heavy drinker...

If so it'd be way easier to calculate. Lol

As a quick guess, if randomly shuffled, you are looking at the chance that any card is the same as the previous card.  Since there are 4 of any specific card (jack, ace, etc), then there are 3 more cards that can match the first card.  Therefore, the odds of any two consecutive cards being a pair are 3/51.  The odds of the cards NOT being a pair are then 48/51.

Going sequentially through the deck, there are 51 potential pairs you are looking at, so the odds that none are pairs are  (48/51)^51.  The calculator app on my computer says that amounts to about a 4.5% chance of no pairs in a randomly shuffled deck.

Mike

Most of the work in these kinds of questions is correctly describing the 'problem.'

ETA:

And the odds of NO pairs is 1-[the odds if at least one pair].