basic probability question counter intuitive?

[arden street]

well it's counter intuitive for me at least [img]/images/graemlins/confused.gif[/img] and not so basic! [img]/images/graemlins/tongue.gif[/img]

so i'm trying to figure out the % chance of hitting an Ace or a King on the flop when you hold AK and are heads up and you're assuming your opponent doesn't hold an A or a K.

there are 3 remaining Aces and 3 remaining Kings available and i thought the equation was 6/48 + 6/47 + 6/46 = .383?

but that's wrong apparently?

could someone please give me the correct formula?

and if anyone wanted to explain why the 'intuitive' formula is wrong that would be appreciated too!

cheers

[jay_shark]

One approach is to use the inclusion/exclusion .

P(AuBuC) = P(A) + P(B) +P(C) - P(AB) - P(AC) - P(BC) + P(ABC).

Where P(AuBuC) = the probability of event A or event B or event C happening .

P(AB)= the probability of event A and event B happening together .

So Let us define what these letters mean in this context .

Let A denote the event that the first card is an ace or a king .
B is the event that the second card is an ace or a king .
C is defined similarly .

P(AuBuC) = 6/48+ 6/48 + 6/48 - 3*[2*3/48*3/47 + 2*3/48*2/47]
+ [3/48*2/47*3/46*3*2 + 3/48*2/47*1/46*2]

= 0.336262719

The tricky part is to figure out the probabilities of the joint events happening .

i.e, P(AB) = the probability the first card is an ace (3/48) , times the probability the second card is a K ( 3/47) . However the reverse may be true as well and we multiply by 2 . Moreover , the probability the first card is an ace followed by a second ace is 3/48*2/47 . The same is true for kings and so we multiply that by 2 .

Solution 2: This is the easiest approach .

The probability the flop does not contain an A or a K is (48-6)C3/48c3 . Therefore the probability the flop contains an A or a K is the complement of that or 1 - 42c3/48c3 = 0.336262719 .

[BruceZ]

[ QUOTE ]
well it's counter intuitive for me at least [img]/images/graemlins/confused.gif[/img] and not so basic! [img]/images/graemlins/tongue.gif[/img]

so i'm trying to figure out the % chance of hitting an Ace or a King on the flop when you hold AK and are heads up and you're assuming your opponent doesn't hold an A or a K.

there are 3 remaining Aces and 3 remaining Kings available and i thought the equation was 6/48 + 6/47 + 6/46 = .383?

but that's wrong apparently?

could someone please give me the correct formula?

and if anyone wanted to explain why the 'intuitive' formula is wrong that would be appreciated too!

cheers

[/ QUOTE ]

To do it your way, we must do it like this:

P(flopping A or K) =
P(1st card is A or K) +
<font color="red">P(1st card is NOT A or K)*</font>P(2nd card is A or K when 1st is not) +
<font color="red">P(1st and 2nd cards are NOT A or K)*</font>P(3rd card is A or K when 1st and 2nd are not)

The probability of getting an A or a K on just the first 2 flop cards, which we will call P2, is:

P2 = 6/48 + <font color="red">(1 - 6/48)*</font>(6/47) =~ 23.67%

and then for all 3 flop cards it is:

P2 + <font color="red">(1 - P2)*</font>6/46 =~ 33.6%.

We must multiply the 6/47 by (1 - 6/48) which is the probability that we miss on the first card since the 6/47 only applies to the times that we miss on the first card. Otherwise we would be double counting the times that we hit on both cards since these are already accounted for by the 6/48. Similarly, the 6/46 only applies to the times that we miss on both the first and second cards, which has probability 1 - P2.

Note that instead of (1 - P2) above, we could have written 42/48 * 41/47 since this is also the probability of no A or K on the 1st and 2nd cards.

A much easier way to do this problem is to take 1 minus the probability of NOT getting an A or K on any of the 3 cards. This is:

1 - (42/48 * 41/47 * 40/46) =~ 33.6%.

This post answers the same type of question for completing a flush draw.

[R Gibert]

Here is a formula you could use for this:

P = A + A'B + A'B'C

The apostrophe (') denotes a math object called a "conjugate." In math conjugate means different things in different contexts. In this context it means simply:

X' = 1 - X

which is not too exciting, but it turns out to be a convenient shorthand for an all too common operation. More usually it is denoted by a horizontal bar written over whatever X is, but I don't know how to type that easily so I have resort to using an apostrophe.

Here is the 2nd version of "my" formula expanded out:

P = A + (1 - A)*B + (1 - A)*(1 - B)* C

As you can see, the first version is easier to remember.

Now we're ready for those 3 fractions you summed up, lets call them:

A = 6/48
B = 6/47
C = 6/46

and plug them into the first formula:

33.6% = 6/48 + (42/48)*6/47 + (42/48)*(41/47)*6/46

which is the correct answer calculated in a fairly economical and easy to remember way that corresponds roughly to the way you calculated it.

What I did is added adjustments to your calc to account for the "overlap" that was inflating your result. Your formula assumes that the chance of getting a desired card on the 1st card of the flop does not overlap with the chance of getting a desired card on the 2nd card of the flop, etc.

The adjustments are the A' and A'B' factors that screen out the overlapping cases.

Naturally this formula can be extended to any number of cards and not just 3.

Here is another way that does not try to correct your version and takes a different tack:

P = 1 - A'B'C'

Here the A'B'C' term is simply the probability of not getting any of the desired cards in any of the 3 cards of the flop and subtracting that from 1 for the result.

Now if we plug in our values for A, B and C, we get:

33.6% = 1 - (42/48)*(41/47)*(40/46)

This is easier to remember and more economical to do than the 1st method.

BTW, it could also be written as

P = (A'B'C')'

but I don't see this as a more readable improvement.

[R Gibert]

[ QUOTE ]
[ QUOTE ]
well it's counter intuitive for me at least [img]/images/graemlins/confused.gif[/img] and not so basic! [img]/images/graemlins/tongue.gif[/img]

so i'm trying to figure out the % chance of hitting an Ace or a King on the flop when you hold AK and are heads up and you're assuming your opponent doesn't hold an A or a K.

there are 3 remaining Aces and 3 remaining Kings available and i thought the equation was 6/48 + 6/47 + 6/46 = .383?

but that's wrong apparently?

could someone please give me the correct formula?

and if anyone wanted to explain why the 'intuitive' formula is wrong that would be appreciated too!

cheers

[/ QUOTE ]

To do it your way, we must do it like this:

P(flopping A or K) =
P(1st card is A or K) +
<font color="red">P(1st card is NOT A or K)*</font>P(2nd card is A or K when 1st is not) +
<font color="red">P(1st and 2nd cards are NOT A or K)*</font>P(3rd card is A or K when 1st and 2nd are not)

The probability of getting an A or a K on just the first 2 flop cards, which we will call P2, is:

P2 = 6/48 + <font color="red">(1 - 6/48)*</font>(6/47) =~ 23.67%

and then for all 3 flop cards it is:

P2 + <font color="red">(1 - P2)*</font>6/46 =~ 33.6%.

[/ QUOTE ]
I get 23.67 + (1 - 23.67)*6/46 = 20.7 [img]/images/graemlins/confused.gif[/img]

[ QUOTE ]
We must multiply the 6/47 by (1 - 6/48) which is the probability that we miss on the first card since the 6/47 only applies to the times that we miss on the first card. Otherwise we would be double counting the times that we hit on both cards since these are already accounted for by the 6/48. Similarly, the 6/46 only applies to the times that we miss on both the first and second cards, which has probability 1 - P2.

A much easier way to do this problem is to take 1 minus the probability of NOT getting an A or K on any of the 3 cards. This is:

1 - (42/48 * 41/47 * 40/46) =~ 33.6%.

This post answers the same type of question for completing a flush draw.

[/ QUOTE ]

[Pokerfarian]

[ QUOTE ]
[ QUOTE ]
P2 + <font color="red">(1 - P2)*</font>6/46 =~ 33.6%.

[/ QUOTE ]
I get 23.67 + (1 - 23.67)*6/46 = 20.7 [img]/images/graemlins/confused.gif[/img]


[/ QUOTE ]
You need 100 instead of 1 because you've put everything else in %s (or put everything in decimal/fractions &amp; then convert the answer to a %)

[R Gibert]

Oops! [img]/images/graemlins/blush.gif[/img]

Nevermind.

[R Gibert]

[ QUOTE ]

The probability of getting an A or a K on just the first 2 flop cards, which we will call P2, is:

P2 = 6/48 + (1 - 6/48)*(6/47) =~ 23.67%

and then for all 3 flop cards it is:

P2 + (1 - P2)*6/46 =~ 33.6%.


[/ QUOTE ]
This is how what you're doing looks like in Python:

reduce(lambda x,y: x + (1 - x)*y, [6.0/48, 6.0/47, 6.0/46])

which appears to use 6 mul/div operations, but one of them is a multiply by zero, so it is actually only 5. What you did looks like 6 mul/div operations too, but 6/48 does not need to be computed twice. This puts it on a par with:

33.6 = 1 - 42/48*41/47*40/46

The number of add/sub operations is the same too.

Interesting.

[R Gibert]

Note that

P = A + A'B + A'B'C

= A + A'(B + B'C)

The latter is the method given by BruceZ where one multiplication less gets performed.

[R Gibert]

See

http://forumserver.twoplustwo.com/showth...D=#Post12527998

It seems we are essentially doing the same thing.