Hi, distributions confuse me.

Hero plays a diamond, one player drops a spade, and the other two players play a diamond. From counting, there are only 6 hearts left and we have 3 of them. What is the equation to solve the probability of one player having all 3 remaining hearts or one player having 2 hearts and the other player having just 1?

Thanks

You gain a fair bit more info from the players' actions so far (who played which heart, when, etc) than you do an exact probability.

Also... it would help to know how many cards are left?

I messed up this thread lol XD. I meant that there are only 6 diamonds left and we have 3 of them. Oops on the confusionoojuns

The 2,1 case is a 3-to-1 favorite since there are 3 ways to hold 1 heart and only 1 way to hold 3 hearts. So P(2,1) = 3/4 and P(3,0) = 1/4.

EDIT: Saw your other post, and now I have no idea what you are asking. You can change hearts to diamonds in my answer. I'm assuming that you're playing partners, that there are 6 unplayed diamonds, and that you and your partner together hold 3 of them, so there are 3 in the other team's hands, and you want to know the probabilities for how they are distributed. It doesn't matter how many cards are left to be played.

EDIT2: If you're not playing partners and you have 3 diamonds in your own hand with 3 diamonds in your 3 opponent's hands, then there are 6 ways for 1,1,1; 6*3=18 ways for 2,1,0 and 3 ways for 3,0,0. So P(2,1,0) = 18/27 = 2/3, P(3,0,0) = 3/27 = 1/9, and P(1,1,1) = 6/27 = 2/9.

Bruce: when the last diamond was played, one player played a spade which means he had no more diamonds. So there are 2 players that could have the remaining 3.

Then my original solution would apply.

...if there was no passing.

On hearts deals where passing has taken place, odd distributions happen much more often than they would on random deals, and I've never come across a good way to try to quantify it (since it depends so much on what players like to pass.)