Pre Flop Odds

Hi, Juanrlos922

Welcome to PokerStrategy.com !

Check out the free tool Equilab.

This is Equilab's output for your scenario:

       Equity     Win     Tie
MP2    43.24%  43.03%   0.21% { AKo }
MP3    56.76%  56.55%   0.21% { QQ }

So QQ has an edge -- but only a slight one.

Yes, there are only 6 cards out of 48, but there are 3 streets to get a shot at hitting one of them. Plus which, those straights and flushes do help -- but not much.

Dead Cards: A♦A♣A♥K♠K♣K♥
       Equity     Win     Tie
MP2     4.72%   4.40%   0.31% { AKo }
MP3    95.28%  94.97%   0.31% { QQ }

As you can see, w/o the Aces and Kings, AK has only a 5% shot at winning.

Once you're a bronze member there are articles that cover this.

Best of luck,
--VS

Hi, Juanrlos922,
I thought about this on the way to work this morning, and I think I have a reasonable explanation of WHY it is near 50%

So, as you say, there are 6 cards that give you a pair, and 48 unknowns.
5 Cards will be dealt.

The chance that the firs it not an A or K is 42/48.
On the next it is 41/47 then 42/46 etc.

So the odds of reaching showdown without hitting either is:
42/48 * 41/47 * 40/48 * 39/45 * 38/44

If you do that, the answer is 49.6% chance that AK will not hit.

So as a first approximation, we are right around 50%.
Now all the other scenarios make small adjustments to that.

I mentioned the articles, bronze status etc in my prev post.
To find out more, have a look around the site, but be sure not to miss the free money offers offers.

Best of luck,
--VS

Thanks for the reply.

But you are using the software to demonstrate how the software works. I am still not convinced.

What if I have 4 cards in my stack and 1 out and I draw 3 times?

Then my odds are

1/4 + 1/3 + 1/2 = 108%

Same scenario with 3 cards

1/3 + 1/2 + 1/1 = 183%

Simply adding percentages can not be correct.

Maybe if I understand this I can be a good poker player :-)

Originally posted by VorpalF2F
...

So the odds of reaching showdown without hitting either is:
42/48 * 41/47 * 40/48 * 39/45 * 38/44

If you do that, the answer is 49.6% chance that AK will not hit.

...
--VS

These approaches with adding percentages confuse me.
You multiply 42/48 * 41/47 * 40/48 * 39/45 * 38/44 to receive 49.6% and

6/48*6/47*6/46*6/45*6/44 is supposed to be some 38%

I am having dificulties to understand this.

Hey Juanrlos922,

Adding probabilities indeed doesn't make much sense. That's why we should be multiplying probabilities, like Vorpal did.

His calculation shows the probability of NOT hitting your pair. For the first community card, there are 48 unknown cards remaining. Subtract your 6 pair outs from this, and you have 42 cards remaining that don't give you a pair. On every next community card, you subtract one from both.

So the chance of not hitting your pair is:

42/48 * 41/47 * 40/46 * 39/45 * 38/44 = ~49.7%

Which means the chance of hitting your pair is roughly 1 - 49.7% = 50.3%. Then of course there's a chance that the QQ will actually hit a set or or a straight or that AK will hit the straight or flush, which changes the equity a bit.

Also to continue for the example you gave, having a deck of 4 cards and 1 out, your chances of getting your 1 out are actually this, using the same logic as above:

1 - (3/4 * 2/3 * 1/2) = 75%

Hopefully this isn't too confusing!

Getting a good understanding of how this works exactly certainly can help with becoming a better poker player. But if you are for example just starting out, it might not be overly important to learn all the ins and outs of the math behind it. But it certainly can't hurt either!

Just ask if you have any further questions

Regards,
Tino

Hello Tino,
thanks for your reply. Basically I have 2 options, I can hit a pair or not (holding AK in this scenario)

Originally posted by TinoLaan
So the chance of not hitting your pair is:

42/48 * 41/47 * 40/46 * 39/45 * 38/44 = ~49.7%

Which means the chance of hitting your pair is roughly 1 - 49.7% = 50.3%. ...

Shouldn't we then get roughly 50% multiplying the odds of hitting? No need to calculate to see that

6/48 * 6/47 * 6/46 * 6/45 * 6/44

doesn't come close to the not-hit-example.

I hope I will not confuse everyone else in the end

Hi again,

The problem with your calculation is that you are effectively calculating the chance of hitting a 6 outer on every single card. So basically what you're trying to do is calculate the probability of hitting an A or K on every single community card. And even for that your calculation would be wrong

The calculation Vorpal and I gave you is the easiest way of calculating how often you will hit at least 1 A or K. We do that by calculating the chance that we won't hit our A or K on any of the 5 community cards. Subtract that chance from 100%, and you have your chance of hitting at least 1 A or K.

Does that make sense?

Originally posted by Juanrlos922

Originally posted by VorpalF2F
...

So the odds of reaching showdown without hitting either is:
42/48 * 41/47 * 40/48 * 39/45 * 38/44

If you do that, the answer is 49.6% chance that AK will not hit.

...
--VS

These approaches with adding percentages confuse me.
You multiply 42/48 * 41/47 * 40/48 * 39/45 * 38/44 to receive 49.6% and

6/48*6/47*6/46*6/45*6/44 is supposed to be some 38%

I am having dificulties to understand this.

I'm not adding them. The * means multiply.

The reason that we work with NOT hitting is that it is much easier to calculate.
If you try to do it the other way there is a wee problem:
If you calculate the odds of hitting, on the first card you use 6/48.
But if you hit, then the calculation stops, because you hit.
the calculation
6/48*6/47*6/46*6/45*6/44
tells you what the odds are of hitting one of the 6 outs 6 times in a row.
To do that, you would need to put your card back in the deck each time.

Cheers,
--VS

Thanks guys, that makes perfect sense.

Glad we could help!