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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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    $\begingroup$ 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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