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In a normal pack of playing cards (without jokers), the probabilities of the following events are the same:

(1) Drawing a set of 4 different cards (irrespective of its color and suit) - let us call it as set X

(2) Drawing another set of 4 different cards (irrespective of its color and suit and different from those of X) - let us call it as set Y and

(3) Drawing another set of 4 different cards (irrespective of its color and suit and different from those of X and Y) - let us call it as set Z

As the problem statement is too broad, I try to narrow it down by saying:

The sets of cards are drawn at different times from a full pack of cards and if the cards in each of those sets are not consecutive ones, then find the possible card denominations in each set.

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closed as too broad by Ankoganit, Sconibulus, APrough, Beastly Gerbil, user58 Oct 25 '17 at 19:57

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ They could be the same word. That would easily solve it. $\endgroup$ – Deusovi Oct 25 '17 at 15:08
  • $\begingroup$ @Apep, well, you spotted it. Edited my question. $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 15:54
  • $\begingroup$ @Deusovi Kinda gotcha you noticed, corrected in my edited question. $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 15:54
  • $\begingroup$ @MeaCulpaNay Can a two digit card (10) and an ace - 1/13 be counted as an answer? $\endgroup$ – prog_SAHIL Oct 25 '17 at 17:19
  • $\begingroup$ @prog_SAHIL, well hyphenated( that is two words combined) type words are expected. Though your logic and answer suit as answer. $\endgroup$ – Mea Culpa Nay Oct 26 '17 at 3:59
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This could have so many answers, so I'll submit one that I think's kind of interesting.

The probabilities of drawing a

prime-numbered card (2, 3, 5, 7)

and a

"non-numbered" card (A, J, Q, K)

are the same, both having a chance of

4/13

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  • $\begingroup$ Yes, you got one set( or rather a pair) of words correct! $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 15:51
  • $\begingroup$ +1 I was thinking of a strange-numbered set instead of your first one. This answer leaves less room for debate on the value of a certain card. $\endgroup$ – Apep Oct 25 '17 at 16:13
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This problem is way too broad. For instance, in addition to all of the answers already given, I give you

One-eyed (3)

and

mustachioed king (3)

This question should have more definition in it, or at least some hints in the puzzle to provide a singular answer.

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Answer:

Odd and Even

Explanation:

Aces are frequently considered to be 1 OR 11, however, it is still only one card, giving us A, 3, 5, 7, 9 for Odd cards, and 2, 4, 6, 8, 10 for Even cards. Each card value has 4 instances in a normal pack, so the probabilities of drawing one from either set is the same, provided the cards are replaced once drawn.

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  • $\begingroup$ good attempt, though requiring some assumptions to prove your point! $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 15:43
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Are the words you're looking for

'right-side-up' and 'upside-down'? Depending on how you're holding the deck, you could draw your card upside down. (I don't mean "face-down", but rather "with the symbols on the card flipped".)

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  • $\begingroup$ well, your logic applies well for all, except for Diamonds I am afraid! $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 15:45
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I think the answer might be that your odds of drawing a

red-faced (as in the color of the King, Queen, or Jack's face)card are equal to your chances of drawing a black-faced card since there are 6 of each in every deck of cards.

And I know that

regardless of the suit but this is based on the color of the person's face on the card...but the question is too wide open.

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