Probability - 52 cards, 13 hands, each hand with 4 different suits?

Question

Consider a standard deck of 52 cards. The deck will be dealt into 13 hands, 4 cards per hand.
What is the probability that the 4 cards in each hand are from different suits?

I figure for the first hand the odds would be: (52/52)*(39/51)*(26/50)*(13/49)

But what about subsequent hands, and the final odds?

Answer 1

You've got the right idea, now just extend to further hands. Assuming that the first hand is four different suits, in the second hand the odds are (48/48)*(36/47)*(24/46)*(12*45).

Continuing in this fashion and multiplying out, the final result is
1 / 61,204,166,001

Answer 2

Jeffareid's answer is over theory and incorrect. He starts of comfortable with this series: fifty two/fifty two x 12/fifty one x 39/50 x 26/40 9 x 13/40 8 And is right that there are 4 such sequences, counting on while one conceives of drawing the redundant card. yet something is strange and incorrect. There are 4 such sequences, so the threat is the above circumstances 4. although, even that is over theory. it is not considerable what the redundant fit card is, and it is not considerable while that is chosen. there is not any evaluate including 4 such sequences based on the region of the redundant card. that is corresponding to multiplying the numerator via 4, which basically cancels the consequences of calculating 4 distinctive possibilities for the redundant card. So a less complicated series is: fifty two/fifty two * 39/fifty one * 26/50 * 13/40 9 * 40 8/40 8 = And, as he states, any first card will do. So an excellent greater convenient answer is: 39/fifty one * 26/50 * 13/40 9 = .1055

Answer 3

Did you examine the reverse situation:

What is the probability that at least one hand has two cards of the same suit?

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Edit: bad idea. Previous poster is right and his number checks out. I verified with Excel. I also checked that the probability of the 13th hand of being of different suits is 1 (it has to be).