What is the probability that from a well shuffled, 52 card deck that is dealt to 4 different people (13 cards each), the 5, 4, 3 and 2 of hearts end up in different hands?
The 5 will be in one of the hands.
Probability the 4 ends up in a different hand: ~75%
Probability the 3 ends up in a different hand than the 5 and 4: ~50%
Probability the 2 ends up in the 4th hand: ~25%
0.750.50.25 = 0.09375 or about 9.3%
Does that seem about right?
I get the same answer by a different method - I think the general solution for n cards distributed in n hands is
n!/(n^n), so
n=1 gives 100%
n=2 gives 50%
n=3 gives 22.2%
n=4 gives 9.3%
My lunch break is finishing up, so I can’t complete my analysis, but I think it’s more complex than that. Once you put the first card in hand 1, then there are 51 cards left, then 50, then 49. This is assuming it stays a 1 out of 52 chance for subsequent cards.
It is more complex than that but only a bit. Deal out all the cards face down. Someone else turns up the 2H. What is the probability the 3H is not in the same hand? Not quite 3/4 but a bit more 39/51. There 39 slots in the other three hands and 12 in the one holding the first card, so 39/51. The other person finds the 3H and turns it up. Assuming things are still good and it’s in a different hand, then the 4H has a probability of 26/50 of being in one of the other two hands, and the 5H has a probability of 13/49 being in the last had so
39/51 * 26/50 * 13/49 = 10.5%.
The last answer is correct (~10.55%).
But there’s a more compact way of expressing it:
C(52,4) = 270725 is the total number of ways to pick 4 cards out of the entire deck of 52. In this case, the 2, 3, 4, and 5 of hearts.
C(13,1)=13 is the number of positions each card can show up in each hand, and there are 4 hands, so:
C(13,1)^4 / C(52,4)
And wrapping up the calculation:
C(13,1)^4 / C(52,4)
= 13^4 / 270725
= ~0.1055
Hmm. My thought process was to ignore all the cards but the 2, 3, 4 and 5. I figured there were 4^4 different ways that they could end up distributed among the four hands, and 4! of those cases were “successes” (they had one of those cards in each hand). But that gets me 9.3%.
Ignoring all the other cards is where that went a bit awry. You can’t just limit the analysis to 4 cards without taking the other 48 into account in at least some way.
For example, the degenerate case is a deck of only 4 cards. Now there’s a 100% chance they all get distributed evenly between 4 hands.
But even at the next level up, if there are 8 cards (the 4 we want, and 4 ‘blanks’), there are 8! = 40320 ways of dealing these cards. Of these there are 9216 ways where each of the 4 cards ends up in a different hand, i.e. 22.9%.
How many cards are in the deck makes a difference.
More interestingly, the probability dropoff slows markedly. It doesn’t really fall below 10% until you get close to a 100 card deck. And even at 1000 cards with 4 ‘hands’, the probability is still ~9.4%.
ETA: should note, you do get close to 9.3%, but you need an infinite number of cards for those 4 players. Your formula is basically the limiting behavior at infinite n.
The probability of you getting, e.g., both the 2 and the 3 will be less than the probability of North getting the 2 and East getting the 3.
I was unclear. I was considering every possible way that 4 cards could be put in 4 hands. One possibility is that hand a has card 1, 2, 3, 4. Another possibility is that hand a has card 1, hand b has cards 2 and 3, hand c has no cards, and hand d has card 4. I thought that amounted to ignoring all the cards but cards 1 through 4, but apparently I’m mistaken.
A simpler example would be the two cards, two hands scenario. You could have hand a have card 1 (which means that hand b has card 2), hand a have card 2 (which means hand b has card 1), hand a have both cards, or band b have both cards. 4 possibilities, of which 2 are successes. For three cards and three hands, there are 27 possibilities of which 6 are successes.
But again, I’m apparently wrong, so I need to think about it more.
Indeed so, counting the number of ways those cards can be distributed depends directly on how many cards are in the deck. Ignoring the other cards was mistaken. The case of a 4 card deck is the direct example, since there’s no way for a single player to get more than 1 of the cards.
But as I missed the edit window before the responses, I’ll elaborate a bit more. If there are (4n) cards and each of the 4 players gets a hand of (n) cards, the general formula is:
C(n,1)^4 / C(4n,4)
Doing some algebra, this comes out to:
4!n^4 / [(4n)(4n-1)(4n-2)(4n-3)]
The limiting behavior, as we take n to infinity, is 4! / 4^4.
So you can safely ignore the other cards as you did, but only if there are infinitely many other cards in the deck to choose from.
Ah. Thanks. So I came up with a valid solution to the wrong question. Story of my life…