Benjamin H. answered • 11/30/24
Math and Science Made Simple! Experienced, Knowledgeable Tutor
I'll be heading into the problem with this assumption: Neither the trump card nor the other two cards in your hand are Q,K, or A of spades.
We can also simplify the problem by considering the other 7 player's hands as one big pool. We can do this because we are not considering each player's hand as unique, so the order doesn't really matter, all we care about is IF one of them has one of the cards. So, that means there are 21 unknown cards in play and 4 known cards (that are not Q,K, or A of spades).
Probability can be written as the number of possible outcomes we want divided by the total number of possible outcomes or one minus the number of outcomes we don't want divided by the total number of outcomes.
Removing the 4 known cards gives us 48 cards left in the deck. The total number of possibilities for the cards in their hand can then be represented as 48C21 (48 choose 21, or the total number of combinations of 21 cards from the deck of 48).
We then need to find the number of possibilities we want. Now this can be tricky because we can say there are cases where there is just a King, a King and a Queen, an Ace and a King (all spades), etc. This would be exhaustive to solve, so our best option is finding the number of possibilities that have NONE of the cards we want as this would only be one calculation. We can then find our target probability like this:
1-(probability that none of the players have a Queen, King, or Ace of spades)
The number of combinations of 21 cards that can be made without those three would just be 45C21. We just consider the deck as if those cards weren't in it, so 48 - 3 = 45. We can find our final answer like this:
1 - (45C21 / 48C21) = 83%
That means it's pretty likely that at least one of them has at least one of those cards
