How many possible 2 card draw combinations are there in 2 decks of cards?

[Accipiter22]

say I'm playing a card game with 2 52 card decks...plus 2 jokers in each deck, so that's 108 cards altogether...what are the odds of any 2 cards being drawn at a time?

 

[RaynorWolfcastle]

Are jokers wild?

Assuming that you just want to draw two identical cards, it's simply
1/103

edit:simplified the original eq by supposing that permutations were fine.

Basically regardless of what card you're given, there's a 1/103 chance of getting an identical one

 

[2Xtreme21]

... What are the odds that you draw 2 cards? 1:1

 

[Stiganator]

QFT!

 

[her209]

I'd say 100% chance of drawing 2 cards from a deck of 108 cards.

 

[Special K]

1/nCr(108,2) = 1/5778 = 0.000173

 

[UncleWai]

Originally posted by: Special K
1/nCr(108,2) = 1/5778 = 0.000173

No..... the initial card can be anything... plus this is just wrong.

 

[Accipiter22]

not the same card, just any two random cards. Jokers are wil.

 

[Special K]

Originally posted by: UncleWai

Originally posted by: Special K
1/nCr(108,2) = 1/5778 = 0.000173

No..... the initial card can be anything... plus this is just wrong.

Well there's 108 choose 2 different pairs of cards in the deck, so why wouldn't that be the probability of any particular pair of cards being drawn?

 

[UncleWai]

Originally posted by: Special K

Originally posted by: UncleWai

Originally posted by: Special K
1/nCr(108,2) = 1/5778 = 0.000173

No..... the initial card can be anything... plus this is just wrong.

Well there's 108 choose 2 different pairs of cards in the deck, so why wouldn't that be the probability of any particular pair of cards being drawn?

2/108 * 1/107 is your answer right. What you are saying in your calculation is that the initial drawn card has to be one of the two specified card. What the OP is asking is drawn any same 2 particular cards. Think of this as two processes, you draw an initial card first, then the second draw has to be identical to the initial card. Your probability becomes 108/108 (drawn any particular card)* 5/107(match the initial card).
The probability is more than that with 4 wild jokers in the deck, I still haven't thought about it.

 

[rikadik]

This depends on some things we don't know:

Are the jokers 'wild'?
Do all the jokers count as the 'same' card, or are there two types of joker (e.g. a black and a red joker)?

 

[Schfifty Five]

What exactly are you asking here?

 

[yosuke188]

I'm

 

[Yossarian]

Originally posted by: Accipiter22
what are the odds of any 2 cards being drawn at a time?

this is meaningless, please clarify if you want a real answer

 

[esun]

I think the question is: given any two cards (e.g. J diamonds, 10 hearts), what is the probability to pick those two out of a deck on the first try without replacement?

In which case, (2 / 108)(1 / 107), assuming each card is unique.

 

[engineereeyore]

Originally posted by: esun
I think the question is: given any two cards (e.g. J diamonds, 10 hearts), what is the probability to pick those two out of a deck on the first try without replacement?

In which case, (2 / 108)(1 / 107), assuming each card is unique.

Wouldn't it be (2/108) * (2/107) since there are two of each card? Shouldn't it depend on if the cards needed are the same? So it could be either (2/108)*(1/107) or (2/108)*(2/107), depending on whether the needed cards are the same (first case) or different (second case).

Feel free to correct me though.

EDIT: Plus, if you have 4 jokers, 2 from each deck, wouldn't that change things also?

 

[mzkhadir]

so why are we doing your homework for you ?

 

[KoolAidKid]

I am assuming that you are asking what are the odds of getting a pair. If so:

p(pair) = [p(pair | a joker is drawn as the first card)*p(a joker is drawn as the first card)] + [p(pair | a joker is not drawn as the first card)*p(a joker is not drawn as the first card)]

= [1 * 4/108] + [11/107 * 104/108] = .037 + .099 = .136