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The probability that the first card will be something is 1 in 1. The probability that the second card will match the number of the first card is 3 in 51. The probability that the third card will match is 2 in 50. The probability that the fourth card will match is 1 in 49. Multiply all of these together, and you get 6 in 124950, or 3 in 62475, or 1 in 20825, or about 0.00004802.

This solution assumes that the "number" of the card includes the possibility of it being an Ace or a face card, so the stated probability is simply the probability of drawing a four of a kind with a draw of four cards.

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12y ago
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12y ago

The first card can by any suit, since you just same the "same suit."

Whatever the first card is, there are 51 cards left.

Also, whatever that card is, there is one less of that suit.

So we have (12/51)(11/50)(10/49)(9/48)=33/16660 which is about 0.001981

or about once every 500 hands.

The problem is not really any harder if you specify what suit it is. The first fraction become 13/52 and you multiply that by 12/51...etc as above.

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Q: If 4 cards are drawn out of a deck of 52 what is the probability that the cards will be the same number in different suits?
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