The trump is picked last by picking one card from the remaining deck. A player has a trump card if it's of the same type as the trump card (ex. picked trump is 2 of clubs, player has 10 of clubs).
What are the odds nobody has a trump versus only one player has a trump and how do you calculate it?
Thanks
On
I would think, the probability for nobody having a trump is just:
39/51 * 38/50 * 37/49 * 36/48 * 35/47 * 34/46
On
i think it is 39/52*38/51*37/50*36/49*35/48*34/47 the reasoning is that probability for the first player to not get trump card is 39(no of non trump cards)/52(total no of cards) the probability for the second player to not get trump card is 38(remaining no of non trump cards)/51(total no of non trump cards) and so on and lastly all the probabilities are multiplied.
$1.$ No body having a trump: $$4*39P_6$$ Logic: Assuming one of the $4$ types a trump, the number ways in which $6$ players can get a card which is not a trump is : $39*38*37*36*35*34$. This is because since the trump is fixed, you have to now select cards from a set of only $52-13=39$ cards. Also number number of ways of selecting trump is $4$.
$2.$ Exactly one player having a trump: $$4*13*39P_5$$ Logic: Assuming one of the $4$ players has got a trump, the number of ways that can happen is $4*13$. So the other players should get non trumps which can be done in $39*38*37*36*35$ ways.