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What is the probability in 4 handed Euchre, with each person getting 6 cards, and my having Jack of clubs, Jack of spades and ten of clubs, that any of the three other hands will have 3 or 4 clubs against me?

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1 in 6.

Since you're asking about the the possibility of four clubs in another hand, I'll go ahead and state the unstated assumption: you mean that you don't have any clubs other than the Jack and Ten in your hand.

9,T,J,Q,K,A in four suits: 24 cards
Six cards dealt each player: 24 cards

To figure the odds, counts the number of ways to deal what you see and the number of ways to deal the target arrangement. Then divide the latter by former.

Deal what you see

Deal yourself JC, JS, TC and 3 of the other 17 non-clubs.
Number of ways to do so: C(17,3) = 1,330
Deal the other three hands: C(18,6)*C(12,6)*C(6,6) = 17,153,136

So the number of deals that result in you getting JC,JS,TC and no other clubs: 11,664,132,480

Deal the same, but with 3 or 4 clubs in a single other hand

The number of ways to deal JC,JS,TC and no other clubs to you and three or four other clubs to one of the other players:

Deal your hand: 1,330

Then either

  • deal three clubs exactly to one player: C(4,3)*C(14,3) = 1,456

or

  • deal four clubs to one player: C(4,4)*C(14,2) = 91

For a total of 1,547 ways to deal 3 or 4 clubs to that player.

The deal the other hands: C(12,6)*C(6,6) = 924

= 1,901,139,240

Figure the odds

The odds are then 1,901,139,240/11,664,132,480

This comes out to about 16.3% or 1 in 6

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