Quote:
Originally Posted by kb5zcr
I have also wondered about this. If the game is 9 handed and I have (had) 8 opponents and each opponent was dealt two cards, this makes 16 cards that will never come on the board.
If this were true, you could also ignore the twenty cards at the bottom of the deck, as they will absolutely never come on the board either.
"In a nine handed game, 18 cards are dealt ( two per player), on average, 4.5 of these 18 cards will be of the same suit, if I have 2 of them, then shouldn't I assume that 2.5 of the same suit aren't left available to hit the board (on average)?"
Sure you can do that. But you then don't count those unseen opponents cards when working out the chance of hitting.
So the chance of the next card being a heart normally calculated is 11 / 50
(11 = the number of hearts in the deck)
(50 = the number of unseen cards)
So the chance of the next card being a heart calculating it your way is 8.5 / 34
(8.5 = the number of hearts you estimate are in the stub on average)
(34 = the number of cards in the stub)
These two come out with the same answer (not exactly the same because your 2.5 number is not exactly correct). This should show you that you don't have to do those interim calculations... just use the total 'unseen' cards method and it saves a lot of effort.
Alternatively think of it this way... you deal the hands out and get two hearts in your hand. Then lets say one of the other players just simply swaps his hand with the bottom two cards on the deck. Would that affect the probabilities of hearts on the flop? Obviously not. So why do you need to consider what's in the other players hand at all?