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# PreFlop odds calculation

• Bronze
Joined: 14.09.2007
Howdy community,
i don't know what i'm doing wrong, i hope you can help me here.
I know that the probability to be dealt one ace and any other specific card (Ax) is about 15%. So if i'm on a full table (10 players) then the probability that there was at least one ace dealt should be 15% * n (where n is the number of players). So, if you have 10 players. then the outcome would be 150% (which is of course not possible). So where did i make the mistake? How do you calculate it the right way? Of course you can't assume that there will be an ace in the GAME that percentage, because a lot of aces are thrown away (when with bad kicker). Nevertheless, would be kind if somebody helps me.
Thanks!
Greetings
• 10 replies
• Bronze
Joined: 06.02.2007
pretty basic statistics calculation.

you calculate it P(atleast someone has an ace) = 1- P(no-one has an ace)
• Bronze
Joined: 21.09.2007
the sheer fact that you have an ace lowers the 14.4% chance:

Given fact: you have Ace X

50 cards in deck

Chance of player p hitting ace X is

(3/50 * 47/49) * 2 = 0.115%

That's the mistake.

The chance of absolutely no ace is a douts calculation. Chance for no ace is 85%, so chance of no ace at all is 0.85 * 0.85 * 0.85 * 0.85 * 0.85 * 0.85 * 0.85 * 0.85 * 0.85 * 0.85.
• Bronze
Joined: 14.09.2007
umph, i am a little confused now.
It's probably better if someone could give me the formula for calculating the probability. That would be kind.

1. Let's say I have K3o. What is the probability that any other player holds at least a A or/and a K? Let's say if there a X players.

2. I have Q3o. What is the probability that any other player holds at least an Ace or a King or a combination of both? Let's say if there a X players.

3. Again I have Q3o and the Flop comes Q-4-5. What is the probability that any other player holds at least a Queen?

It would be very nice if somebody could compile a general formula for these cases. I'm searching the internet for hours now, but wasn't successful, yet. I hope somebody could take the time. Thanks very, very much in advance!
Greetings
cobby
• Bronze
Joined: 26.02.2007
Hi Cobby,
Did you have a look at the strategy articles bronze section? There is a good one on outs, odds, equity, that might help you.
• Bronze
Joined: 03.07.2007
There's no general formula. It depends on the number of opponents and your position. Precise calculation are quite complex and not really useful (practically u're not able to do them during the game) . Whats really important is that the more players in the game and the earlier your position is the bigger hand u should have to avoid domination. For example with 10 player when u're in the early position you should play only AK, AQ, QQ, KK and AA any other hands will likely be dominated in this position. This is a practical knowledge that you will get with experience rather than through calculation that are simply to difficult to do them while playing.
• Bronze
Joined: 26.03.2007
Originally posted by cobby
umph, i am a little confused now.
It's probably better if someone could give me the formula for calculating the probability. That would be kind.

1. Let's say I have K3o. What is the probability that any other player holds at least a A or/and a K? Let's say if there a X players.

2. I have Q3o. What is the probability that any other player holds at least an Ace or a King or a combination of both? Let's say if there a X players.

3. Again I have Q3o and the Flop comes Q-4-5. What is the probability that any other player holds at least a Queen?

It would be very nice if somebody could compile a general formula for these cases. I'm searching the internet for hours now, but wasn't successful, yet. I hope somebody could take the time. Thanks very, very much in advance!
Greetings
cobby
There is not a single general formula, but there *is* a common methodology in calculating these kinds of things. I agree with Rubisilesia that experience will teach you most of these, but IMO calculation can be a valuable addition to get a feel for certain probabilities beforehand, even though you can't perform these calculations during play.

1. The chance that at least one other player holds at least an A or a K is 1 (or 100%) *minus* the chance that *noone* holds either A or K.

What is the chance that your first opponent doesn't hold A or K? You have Kx, so there are 4 Aces and 3 Kings left in a deck of 50 (52 minus your 2) cards. Your first opponent needs one of the remaining 43 (50 minus 7) non-AK cards (chance of 43/50) and then another one of the then-remaining 42 non-AK cards from the remaining 49-card deck (chance of 42/49), for a total chance of (43/50)*(42/49) = 0.74 or 74%

Given the above, what is the chance that the second opponent *also* doesn't hold A or K? There are still 4 Aces and 3 Kings in the deck, but the deck in total now only contains 48 cards (you hold 2 cards, and your first opponent also holds two cards). That makes the chance of the second opponent also not holding an A or a K: (41/48)*(40/47) = 73%

You can repeat this for any additional opponent.

Assuming two opponents in total, the chance that *both are not holding A or K is 74% * 73% = 54%. Therefore the chance that there is *at least* one more A or K in play is 100 - 54 = 46%

Try and calculate the probability yourself for 3, 4, and 5 opponents, then see if you can calculate 2. and 3. yourself. If not, let me know, and I'll do those as well.
• Bronze
Joined: 25.03.2005
Nice post Classical. Trying to get a feeling by calculating these things is definatly a good thing to do.
• Bronze
Joined: 14.09.2007
hi classical.
I think you're wrong. Also i don't know which mathematical approach you used, i come to another resultby using the binomial coefficient.

If you hold KQo. And you want no other A or K in the game (which means that none of you opponents holds a A or K, then you can do it like this:

C(7,0) * C(43,4)
--------------------
C(50,4)

Here you are assuming that you play against two opponents (thus you have 4 cards) and that none of these 4 cards should be a K or an A (which are the 7 cards left, represented by "C(7,0)"), but any other cards (of the remaining 43, represented by "C(43,4)").
So the result would be about 54%, not 46%.
Greetings
• Bronze
Joined: 15.06.2007
Yea, I always make those kind of calculations during my hands when I multitable
• Bronze
Joined: 26.03.2007
Hey Cobby,

We both agree! Probability of *no* A or K with 2 opps is indeed 54%, and therefore probability that there is at least an A or a K is 100 - 54 = 46%.

Your calculation with binomial coefficients is correct and equivalent to mine.

@SonicXT: wow, you must be a math wizard. No way I can do these calculations at the table. I do do them beforehand though, sometimes.