Should I account for opponents hand (when flopping a flush draw)?
If I'm trying to calculate the odds for flush.
And i have two hearts. If I'm alone there is 50 cards left in the deck and 11 of them are hearts. So it's easy to calculate the odds.
But in a game of heads up there will be 48 cards in the deck and 2 unknown in opponents hand.
Are the odds still 11/50 or 11/48 or something else.
Forget about hand-ranges. Just two random cards in opponents hand.
re: Should I account for opponents hand (when flopping a flush draw)?
The denominator in these calculations in the number of unknown cards. The cards in your opponent's hand are still unknown, even though he has them.
You're not going to adjust your calc because you can't get dealt the bottom card in the deck, right? Or the 2nd to the bottom, etc?
Thanks very much, No. It felt reasonable that you would not account for it. But i felt that i had to be sure(doh).
Your example were on point and shows the point without a doubt. Thank you very much!
So if I'd flop a flush draw having 9 outs to hit my flush, do I have to assume that I'm always taking this perfect scenario, that is no one flush card I need was dealt on opponents' hands?
ZOMG but what if one of your flush cards is dealt to the 2 bottom cards of the deck! You can't get dealt those either. So it's the same as if it's in your opponent's hand. There is only one card you can get dealt on the turn.
You don't assume anything. The 9 out calculation is how often you will actually hit, on average, including the times some of those cards are dead. If we KNOW all 9 are available, the chance is greater than the average, because you have reduced the unknowns in the denominator.
You said this in OP "Forget about hand-ranges. Just two random cards in opponents hand. "
Therefore the posters above are correct.
In reality, you can draw some small inference on the flop about the suits in an opponents hand, but that's not what you asked.
What are the odds one of your flush cards was in another hand? Two cards? Three? ... All? None? In each scenario what are the odds that the flush card will come on the turn? What is the result when all scenarios are combined?
Yep, thanks for the answers.
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 an average of 1 in 4 cards are a flush card, should I assume that instead of nine outs to make my flush, I have less than this since odds are that some of my outs were dealt to other players?
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)?
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.
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?
You sir, should be a math teacher, you explained it perfectly so that my brain can understand. When you compared both methods it clicked in my head.
Since this question comes up quite often, I will sticky the thread. Thanks for the questions and the great answers.
This question arose in my home game last night. It is a very special home game. My middle school daughter and I are the only two players.
To make a long story short, she was not totally convinced of the efficacy of the "unseen cards" method of calculating the probability of hitting a flush draw even though I gave all the good and valid arguments appearing in this thread.
Like a good used car salesman, I asked her what it would take to convince her of the validity of that derivation method. She responded that she'd have to see that the calculations of doing it the "long way" result in the same answer as doing it the "shortcut way".
Here is what we did for the "long way" (she agreed that showing the two methods give the same result in a heads-up pot is sufficient to convince her that they give the same result for any number of players).
Hero (my daughter) holds two hearts (HH). Flop comes with exactly two hearts (HHO) where H denotes a heart and O denotes an other non-heart.
Villain (me) can hold 0 hearts, 1 heart, or 2 hearts. Let's consider these cases one by one.
Since Hero's hand and the Flop contain exactly 4 hearts and 1 other card, Villain's hand essentially emanates from a 47-card stub of which 9 are hearts and 38 are others.
In this case, what is the prob of Hero making her flush? In this case, the turn and river are dealt from a 45-card stub of which 9 are hearts and 36 are others.
In this case, what is the prob of Hero making her flush? In this case, the turn and river are dealt from a 45-card stub of which 8 are hearts and 37 are others.
In this case, what is the prob of Hero making her flush? In this case, the turn and river are dealt from a 45-card stub of which 7 are hearts and 38 are others.
Putting it all Together for the "Long Way"
Using the "unseen cards" method, Hero sees 4 hearts and 1 other card, meaning that there are 47 unseen cards of which 9 are hearts and 38 are other non-hearts.
Seeing the results of all these calculations finally convinced my daughter that the "unseen cards" method is indeed a valid method of computing poker probabilities (under the assumption that all other cards are "random" unknown cards).
But what if it's a full 9-handed table? Surely some of the others folded some hearts. How do you account for that? Huh?
At least you didn't mention the burn cards!
In our friendly family home game, we don't use burn cards.
Good idea, everybody has a better chance to make their hands!