Calculate simple odds for Texas Holdem hand

[Garet Jax]

Let's say I start with AK, what are the odds I will get another A or K on any of the common cards? Assume I am playing heads up.

My answer is 6/48 + 6/47 + 6/46 + 6/45 + 6/44, but it is way too high (comes to ~65%). What am I missing?

I have looked on the calculators, but they never explain what they do to reach their number.

 

[jdub1107]

I may not be correct, but I don't think you can assume the other person doesn't have either

 

[KoolAidKid]

Probability of getting an ace or a king = 1 - probability of not getting an ace or king =

1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 0.487

I could be off on this, though, I need a nap.

 

[SLU MD]

thats very close to being correct. The reason your odds of winning the hand are about 50% vs an underpair in the other guys hand is b/c he has 2 cards he can catch to win regardless of an A or K falling. Also, straights and flushes are other possiblities that lower odds.

so say, AK vs 99

You have 6 outs, and he has 2, even if you catch and A or K. all this taken in, odds are about (give or take a few percent) 50% both ways.

SLU M.D.

 

[Garet Jax]

Originally posted by: jdub1107
I may not be correct, but I don't think you can assume the other person doesn't have either

This is true so there is always some error factor, but generally calculations are done so as to ignore that possibility.

 

[DBL]

Originally posted by: jdub1107
I may not be correct, but I don't think you can assume the other person doesn't have either

True. You can't assume anything other than the fact that you know the two cards n your hand. It doesn't matter what the other person has if you don't know.

 

[DBL]

Originally posted by: Garet Jax

Originally posted by: jdub1107
I may not be correct, but I don't think you can assume the other person doesn't have either

This is true so there is always some error factor, but generally calculations are done so as to ignore that possibility.

There is no error factor.

 

[amoeba]

Originally posted by: Garet Jax
Let's say I start with AK, what are the odds I will get another A or K on any of the common cards? Assume I am playing heads up.

My answer is 6/48 + 6/47 + 6/46 + 6/45 + 6/44, but it is way too high (comes to ~65%). What am I missing?

I have looked on the calculators, but they never explain what they do to reach their number.

you can't add the probabilities.

I'll give you an example.

Whats the probability of me getting a head at least once when I flip a coin twice?

If I do it your way, it would be 1/2 + 1/2 = 1.

which is obviously wrong as it can come up tails twice.

the correct way is 1- (1/2 * 1/2) = 3/4

in other words the calculation of whether you will get an A or a K should be

1 - (42/48 * 41/47 * 40/46.... 38/44) as another poster mentioned.

But that is not your odds of winning as the underpair can hit a set.

so it really should be (1-(42/48 * 41/47 * 40/46.... 38/44)) ( 46/48*44/47*....42/44)

but then AK can perhaps make flush or straight

so you have to add more variables to equation.

One can easily see that this is not feasible for a calculator algorithm

Most calculator algorithms out there do their calculations by statistical modelling and not actual odds calculation.

simulate say 10 million hands and show you which one comes out on top.

 

[chuckywang]

Originally posted by: KoolAidKid
Probability of getting an ace or a king = 1 - probability of not getting an ace or king =

1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 0.487

I could be off on this, though, I need a nap.

You are correct. This is the easiest way to calculate the probability. Good job.

 

[Garet Jax]

Originally posted by: KoolAidKid
Probability of getting an ace or a king = 1 - probability of not getting an ace or king =

1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 0.487

I could be off on this, though, I need a nap.

OK. I think you are close (you started with 50 because you didn't take his 2 cards into consideration). But this is something I have never understood...

I tried (6/48)*(6/47)*(6/46)*(6/45)*(6/44) and it equals 0.00003....

This to me says there are:
6 chances out of 48 on the first card
6 chances out of 47 on the second card
6 chances out of 46 on the third card
6 chances out of 45 on the fourth card
6 chances out of 44 on the fifth card

but the chances are miniscule.

Why does 1 - ((42/48)*(41/47)*(40/46)*(39/45)*(38/44)) work when the above doesn't?

 

[amoeba]

Originally posted by: Garet Jax

Originally posted by: KoolAidKid
Probability of getting an ace or a king = 1 - probability of not getting an ace or king =

1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 0.487

I could be off on this, though, I need a nap.

OK. I think you are close (you started with 50 because you didn't take his 2 cards into consideration). But this is something I have never understood...

I tried (6/48)*(6/47)*(6/46)*(6/45)*(6/44) and it equals 0.00003....

This to me says there are:
6 chances out of 48 on the first card
6 chances out of 47 on the second card
6 chances out of 46 on the third card
6 chances out of 45 on the fourth card
6 chances out of 44 on the fifth card

but the chances are miniscule.

Why does 1 - ((42/48)*(41/47)*(40/46)*(39/45)*(38/44)) work when the above doesn't?

read my post.

and I meant to start at 44/50 not 42/48.

 

[KoolAidKid]

Originally posted by: Garet Jax

Originally posted by: KoolAidKid
Probability of getting an ace or a king = 1 - probability of not getting an ace or king =

1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 0.487

I could be off on this, though, I need a nap.

OK. I think you are close (you started with 50 because you didn't take his 2 cards into consideration). But this is something I have never understood...

I tried (6/48)*(6/47)*(6/46)*(6/45)*(6/44) and it equals 0.00003....

Why does 1 - ((42/48)*(41/47)*(40/46)*(39/45)*(38/44)) work when the above doesn't?

Because these are conditional probabilities, e.g., P(no A or K on turn | no A or K on flop) = 41/47. In many probability calculations of this kind where you are trying to figure the probability of one or more events occurring it is much more straightforward to calculate the probability that none of the events occur.

 

[Garet Jax]

Originally posted by: amoeba
read my post.

and I meant to start at 44/50 not 42/48.

Thanks for the reply. Make sense, but I am still a little confused.

Throwing texas hold em out the window for now:
The chances of getting an Ace on one of two card pulls = 1 - (48/52 * 47/51) = ~15%

The changes of not getting an ace on one of two card pulls = 1 - (4/52 * 4/51) = ~0.6% (should be 85%)

I know you can solve it by (48/52 * 47/51) = ~85%, but I don't understand it works one way and not the other?

What am I missing? Thanks.

 

[cchen]

You can get a much better estimation by writing a monte carlo simulation, if you're really interested

 

[amoeba]

Originally posted by: Garet Jax

Originally posted by: amoeba
read my post.

and I meant to start at 44/50 not 42/48.

Thanks for the reply. Make sense, but I am still a little confused.

Throwing texas hold em out the window for now:
The chances of getting an Ace on one of two card pulls = 1 - (48/52 * 47/51) = ~15%

The changes of not getting an ace on one of two card pulls = 1 - (4/52 * 4/51) = ~0.6% (should be 85%)

I know you can solve it by (48/52 * 47/51) = ~85%, but I don't understand it works one way and not the other?

What am I missing? Thanks.

the chances of not getting an ace on one of two card pulls does not equal 1-(4/52*4/51).

do you see why?

 

[Garet Jax]

Originally posted by: amoeba

the chances of not getting an ace on one of two card pulls does not equal 1-(4/52*4/51).

do you see why?

Not really, it seems to me that there are 4/52 chances to get the ace on the first pull and 4/51 chances on the second pull. I just don't know how to combine them into one percentage.

It also makes sense that there are 48/52 ways not to get an ace on the first pull and 47/51 ways not to get an ace on the second pull. Multiplying them together and subtracting them from 1 works here but doesn't work above and this is what confuses me.

 

[amoeba]

by the way, Demorgan's law can be worked in to this.

The probability of me pulling an ace on the first card OR me pulling an ace on the second card

is equal to 1 - the probability of me not pulling an ace on the first card AND me not pulling an ace on the second card.

in other words.

A + B = !(!A * !B )

 

[amoeba]

Originally posted by: Garet Jax

Originally posted by: amoeba

the chances of not getting an ace on one of two card pulls does not equal 1-(4/52*4/51).

do you see why?

Not really, it seems to me that there are 4/52 chances to get the ace on the first pull and 4/51 chances on the second pull. I just don't know how to combine them into one percentage.

It also makes sense that there are 48/52 ways not to get an ace on the first pull and 47/51 ways not to get an ace on the second pull. Multiplying them together and subtracting them from 1 works here but doesn't work above and this is what confuses me.

First of all your calculation in the post above me is off.

1- (4/52*4/51) = .99

what this calculates is the probability that you won't pull an ace on the first pull and say, pull a K on the second pull.

if it was 1 - (4/52*3/51), that would be the probability that you don't pull an ace on both pulls.

I suspect you are getting your Or, Ands, not and their probability mathematical equivalents mixed up.

Just do this. whenever you see Or, add.

whenever you see and, multiply

whenever you see not x, do 1 - x

 

[KoolAidKid]

Originally posted by: Garet Jax

Throwing texas hold em out the window for now:
The chances of getting an Ace on one of two card pulls = 1 - (48/52 * 47/51) = ~15%

The changes of not getting an ace on one of two card pulls = 1 - (4/52 * 4/51) = ~0.6% (should be 85%)

I know you can solve it by (48/52 * 47/51) = ~85%, but I don't understand it works one way and not the other?

What am I missing? Thanks.

As I said, calculating the probability of drawing an ace directly is much less straightforward than calcualting the probability of not drawing an ace.

If you draw 2 cards, one after the other, there are 4 possible outcomes:

XX
AX
XA
AA

X = not an Ace, A = Ace

The probabilities of these events are as follows:

(48/52)*(47/51) = 0.85
(4/52)*(48/51) = 0.0724
(48/52)*(4/51) = 0.0724
(4/52)*(3/51) = 0.00452

The probability of drawing at least one ace in 2 draws = 0.0724 + 0.0724 + 0.00452