Question

How do I approach this problem: (probability)

If I shuffle a deck of 52 cards, then split it into to decks of 26, what is the probability that all 13 cards from one suit (spades, hearts etc) will be in one of the stacks?

Answer 1

Is this just a counting problem? Something as simple as identifying the chance that each card from a suit will be picked?

My intuition is that there is a 1/2 chance for each card to land in one of the piles..say pile A. so (1/2)^13? Sort of a conditional probability?

Answer 2

the number of ways to split the deck is \(\binom{52}{26}\) so that is your denominator

Answer 3

then we can compute the probability that out of the 26 chosen, say you have 13 spades

Answer 4

13 out of 13 spades, only one way to do that, i.e. \(\binom{13}{13}=1\) and then for the remaining cards you have \(\binom{39}{13}\) ways to get those

Answer 5

so the probability you get all the spades is
\[\frac{\binom{39}{13}}{\binom{52}{26}}\]

Answer 6

@Zarkon how am i doing so far?

Answer 7

hmmm, I understand what you did for the denominator, and for the (13 choose 13), but why is the numerator (30 choose 13)?

Answer 8

you have 26 cards
13 are spades, no choices there
but you have 13 other cards in your pile, out of the remaining 39 cards that are not spades
that is why it is \(\binom{39}{13}\) in the numerator

Answer 9

ahhhhhh ok ok. I was thinking in terms of the divided piles still. That accounts for the whole deck. So, there is only 1 way for a suit to be chosen. (13 choose 13). Since the suit doesnt matter, just that at least one suit is all in one pile, would I multiply this by 4?

Answer 10

yes

Answer 11

Oh, thank goodness. Finally got something right haha. Also, Since your both here at the moment. Thank you for helping me last night Satellite and Zarkon. I spent hours on that other problem :D

Answer 12

that other one was a pain in the tuchas

Answer 13

still i am wondering about it, because something disturbs me about the 1 and 20 at the end

Answer 14

I think zarkon was right, I found a paper about a lottery question that was very similar, and the general term matched with zarkon's work. Link inc in a moment if I can find it.

Answer 15

but i would go with @Zarkon answer for sure

Answer 16

you have a ways to go to get the answer

Answer 17

On this problem?

Answer 18

alright, im gonna think through this one if it kills me. Ill post what I come up with soon.

Answer 19

Okay, so, I found a very similar problem that asks the probability of making a full house with a 5 card poker hand. The solution is \[\frac{ 13*12*\left(\begin{matrix}4 \\ 2\end{matrix}\right) * \left(\begin{matrix}4 \\ 3\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 5\end{matrix}\right) }\]

13 is the number of types (2,3,A, K etc) for the first card, and 12 is the # of types for the 2nd. (4 choose 2) and (4 choose 3) represent choosing 2 and 3 of same-type-different-suit (to make the full house). And the denominator is all possible 5 card hands.

Answer 20

Now, relating this to my problem is the hard part. I dont have to worry about card type. I only have to worry about suits. As we discussed, there are (13 choose 13) ways to make a full suit of cards. There are (52 choose 26) ways to make the two stacks of cards. (30 choose 13) covers the remaining cards in the first stack. This gave us: \[\frac{ 4*1*\left(\begin{matrix}30 \\ 13\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 26\end{matrix}\right) } = .000000966\]

When you say I have a ways to go @Zarkon , I am not sure what this is missing. Do I need to do separate cases for when more than one full suit appears?

Answer 21

The calculation should be
\[4\times \frac{\left(\begin{matrix}39 \\ 13\end{matrix}\right)}{\left(\begin{matrix}52 \\ 26\end{matrix}\right)}\]

Answer 22

Thanks for taking a look at this problem!

39 is the remaining cards from which to choose the other 13 in the stack. That makes sense to me. Its really that simple? I don't need to have separate cases for more than 1 full suit or anything like that? Zarkon got me worried that it was much more complex than I thought. Thank you for the response!

Answer 23

You're welcome :)

Answer 24

Do you have a result for the calculation?

Answer 25

.000065514

Answer 26

so about a .0065% chance of getting a full suit just by cutting the deck. Seems a reasonable answer as it seems a quite unlikely occurrence.

Answer 27

i think the 30 was a mistake, should be 39

Answer 28

I agree, but it was good that you mentioned it, definitely something that hadn't occured to me, including all those permutations. I worked on this some more this morning, and I cant find anything else thats left out. I think we've beaten this one to death, but left it open for you and Zarkon to look over hehe. I certainly appreciate the help everyone!

Answer 29

just curious, but do you think the answer
\[\frac{4\times\binom{39}{13}}{\binom{52}{26}}\] is incorrect?

Answer 30

nope, thats what Im going with.

Answer 31

what if ...


you first choose spades...then with the \({39 \choose 13}\) you actually get all hearts

...

Answer 32

Would that matter? I was trying to think about how this might effect the problem, but, if it says "at least 1" full suit, I feel like the cases in which the half-deck contains 2 full suits is included in that.

Answer 33

My understanding is that if it had said 1 and only 1 full suit, I would have to subtract the cases in which more than 1 occurred, but since its at least 1, shouldn't more than 1 not matter?

Is it maybe because by getting a full suit of say clubs, eliminates the chances of any clubs being in the other 39 cards?

Answer 34

you are double+ counting. you are also not counting enough

Answer 35

Hmm. For not counting enough, should i multiplying by 2 to account for both half decks? For double counting do you mean I should be subtracting out intersections where 2 3 and 4 full suits appear because having spades and hearts for example is the same as having hearts and spades?

Answer 36

Well, 3 full suits would necessarily imply 4. So I think I need to subtract the permutations that occur in the two half decks. Like (HD CS) is equivalent to (CS HD) ?

Answer 37

Well ive got another few days to work this out, I'm ggonna get it