Card probabilities at short and full tables

cobby

Joined: 14.09.2007

22.11.2007, 17:38 0

Hi,
i'm not sure if i do the right math, so i wanted to get it checked by somebody who's more experienced in that.
1.) Let's assume that i hold KJs (just an example) under following circumstances:

-10 players at the table
- i'm in 7th position (there were 6 folds in front of me and 3 players still to come)

2.) Now let's assume that i'm playing under the same circumstances, but only at a 5 handed table, where i'm sitting in 2 position (there was 1 fold and again 3 players still to come)

Now i'll come to the problem:
Let's further assume that i only want to play the hand if i think that one of the remaining players doesn't hold two specific cards.
So, now to the question: Is the probability that one of the remaining three players holds that specific two cards the same, or is it different? In other words - if i have two cases which only differ in the number of players sitting at the table - does the "overall number of players at the table" affect the probability that somebody holds a certain hand/card (e.g Ax), ALTHOUGH in both cases there're the SAME NUMBER of players left to act (in our cases 3 players).

I'll try to put it in another way, because it's very hard to explain what i mean:
Is e.g the seventh position at a ten handed table the same as the second position at a five handed table (in both cases there three players left to act) if at the ten handed table the first 4 players fold?
Or is there still a difference when i'm considering playing a hand? Because the probability of somebody holding a specifc card combination is higher at the ten handed table than at the five handed table BEFORE the action begins?

I hope you understand my problem and that you can give me some help.
If you don't understand my question post it and i will try to include an example.
Thanks very much in advance!
Greetings
cobby

3 replies

_Anonymous_

22.11.2007, 18:12 0

I can't do the math but I say it's the same probability because of the fact that the cards are not revealed.

But if you know for example that a player with this specific hand would act differently than with all other cards then the probabilities are different in my eyes.

For example if the players only play if they hold any ace and fold ALL other hands then you can say that there wasn't a single ace in one of the seat ahead of you and you can calculate with 52 - 2 (you own) - 12 (your opponent) = 38 cards for a fullring table and 52 - 2 - 2 = 48 cards at a short handed table if you want to know how likely an ace is behind of you.

cobby

Joined: 14.09.2007

Posts: 73

22.11.2007, 18:53 0

Yes that's similar to what i was asking, i guess.
If a few players fold ahead of me, than they were likely to have bad cards, which makes it more likely that the remaining players have better cards.
And if only one player folded ahead of me, it is likely that the remaining players hold worse cards than in the example above, because the "bad" cards (which the players in the example above folded) are still in the deck, which could now have hit one of the remaining players.

Am I right? On the other side players get generally more loose when the number of players shrink, which forces me to play looser although to stay alive (at least in tournaments) so the hand quality may be the same?!?

Looking forward to more ideas and thoughts.
Greetings

_Anonymous_

24.11.2007, 03:20 0

Well looking at it purly mathematical, the chances are still the same.

But as you said, the more players folding is a sign of a lot of "bad" cards beeing folded, leaving better ones for the rest.

When the number of players shrink the value of one single high card increases, so you should be playing more hands than at a full 10 seated table. Not sure if I would call that looser. But taking it to the extreme in Heads up its usually the player which is slightly more active that wins.

[Closed] Card probabilities at short and full tables