Flawed Author-Cardplayer Articles

[dragon14]

On page 31 of Middle Limit Holdem Poker Ciaffone and Brier state "Whatever your first card may be, there are only 3 cards out of 51 that match it's rank. That is 1 out of 17, or 16 to 1 against making the pair".

Thus I'd give Ciaffone the benefit of the doubt in this case.

[maurile]

[ QUOTE ]
5.878% chance the SB has a pair. If he does,
5.954% chance the BB does to, If the SB does not,
5.920% chance the BB has a pair

5.878 + 0.350 + 5.920 = 12.148% chance one or both will have a pair.

[/ QUOTE ]
You don't add them!!!

If you flip two coins, what's the chance that one or both will land on heads? 50% + 50% = 100%? (Hint: No.)

[binions]

[ QUOTE ]
[ QUOTE ]
5.878% chance the SB has a pair. If he does,
5.954% chance the BB does to, If the SB does not,
5.920% chance the BB has a pair

5.878 + 0.350 + 5.920 = 12.148% chance one or both will have a pair.

[/ QUOTE ]
You don't add them!!!

If you flip two coins, what's the chance that one or both will land on heads? 50% + 50% = 100%? (Hint: No.)

[/ QUOTE ]

Sure you do, bud.

This isn't flipping coins. I just went thru all of the dependent calculations from the perspective of the button.

There is a 5.878% chance the SB will have a pair. I assume you accept this.

Of those times when the SB has a pair, there is an additional 0.35% chance that both blinds have pairs. The rest of the time, the BB won't have a pair simultaneously with the SB.

So, from the button's perspective there is a 5.878% + 0.35% chance that either the SB or the SB & BB have pairs. You add these two chances, no?

And if the SB doesn't have a pair, there is a 5.92% chance the BB has a pair.

If you are suggesting that you need to multiply the BB's 5.92% by the 94.122% chance the SB doesn't have a pair, that's double counting.

Because the 5.92% was calculated for the specific situation when the SB doesn't have a pair.

[AngryCola]

As much as I hate to agree with binions [img]/images/graemlins/grin.gif[/img], you're wrong here maurile.

[maurile]

No you don't.

I'll use your numbers:
[ QUOTE ]
5.878% chance the SB has a pair. If he does,
5.954% chance the BB does to, If the SB does not,
5.920% chance the BB has a pair

[/ QUOTE ]

The chance that at least one of them has a pair is equal to one minus the chance that neither of them has a pair. The chance that neither of them has a pair is equal to the product of (a) the chance that the SB does not have a pair, and (b) the chance that the BB does not have a pair (given that the SB does not).

Chance that the SB does not have a pair = 1 - 5.878% = 94.122%.
Chance that the BB does not have a pair (given that the SB does not) = 1 - 5.920% = 94.08%
Product of the two = 94.122% * 94.08% = 88.55%
One minus the prouct = 1 - 88.55% = 11.45% (Not 12.148%)

If you want to add things together (instead of multiplying and subtracting from one), you have to add (a) the chance that the SB has a pair, and (b) the chance that both (1) the SB does not have a pair, and (2) the BB does have a pair.

Chance that the SB has a pair = 5.878%
Chance that the SB does not have a pair = 1 - 5.878% = 94.122%
Chance that the BB has a pair (given that the SB does not) = 5.92%
Chance that both (1) the SB does not have a pair, and (2) the BB does = 94.122% * 5.92% = 5.572%

Sum of (a) and (b) = 5.878% + 5.572% = 11.45%

[maurile]

[ QUOTE ]
There is a 5.878% chance the SB will have a pair. I assume you accept this.
.
Of those times when the SB has a pair, there is an additional 0.35% chance that both blinds have pairs. The rest of the time, the BB won't have a pair simultaneously with the SB.
.
And if the SB doesn't have a pair, there is a 5.92% chance the BB has a pair.

[/ QUOTE ]
The second and third paragraphs are the parts you're not allowed to add.

The second paragraph because it's double-counting. (If the SB has a pair, it doesn't matter whether the BB also has one. There's no bonus points for them both having a pair, as opposed to just the SB.)

The third paragraph because it's not discounted by the probability that the SB has no pair (which is the only time the BB's paired-ness matters).

You can see this easily if you change the numbers around. Suppose we're using a special deck such that the following is true:

90% chance the SB has a pair. If he does,
50% chance the BB does too, If the SB does not,
60% chance the BB has a pair

What's the chance that at least one of them has a pair? If you do it your way, you end up with an answer greater than 150%.

[maurile]

Here is one last attempt at an explanation.

A = the chance that the SB is paired.
B = the chance that the BB is paired.

You're adding A + A*B + B.

You should be adding A*(not-B) + B*(not-A) + A*B
. . . or, more simply, A + B*(not-A)

[TimTimSalabim]

Thanks, I didn't realize there was already a "classic" [img]/images/graemlins/grin.gif[/img].

[binions]

[ QUOTE ]
[ QUOTE ]
There is a 5.878% chance the SB will have a pair. I assume you accept this.
.
Of those times when the SB has a pair, there is an additional 0.35% chance that both blinds have pairs. The rest of the time, the BB won't have a pair simultaneously with the SB.
.
And if the SB doesn't have a pair, there is a 5.92% chance the BB has a pair.

[/ QUOTE ]
The second and third paragraphs are the parts you're not allowed to add.

The second paragraph because it's double-counting. (If the SB has a pair, it doesn't matter whether the BB also has one. There's no bonus points for them both having a pair, as opposed to just the SB.)

The third paragraph because it's not discounted by the probability that the SB has no pair (which is the only time the BB's paired-ness matters).

You can see this easily if you change the numbers around. Suppose we're using a special deck such that the following is true:

90% chance the SB has a pair. If he does,
50% chance the BB does too, If the SB does not,
60% chance the BB has a pair

What's the chance that at least one of them has a pair? If you do it your way, you end up with an answer greater than 150%.

[/ QUOTE ]

Gotcha.

Like figuring out whether you will hit at least 1 of your flush cards with 2 to come (1-(38/47 * 37/46)

Thanks.

[MicroBob]

correct....if you were playing some form of hold-em that put 10 cards on the board you wouldn't be guaranteed of getting a 5th diamond if you had a flush-draw with 7 cards left to come.


Same for the pocket-pairs....imagine a huge hold-em game with 21 players at the table....if you're UTG and there are 20 players left to act after you it is not a guarantee that 1 of them has a pocket-pair...even if each individuals' chances of having a pair were 5% (5% X 20 = 100%).


Will go back and read Maurlie's calculations as I really need to do a better job understanding this stuff and my above 'simpleton' is pretty much the whole of what I know of such things.



Also - I agree with Rudeback and D.S. here.

These guys are the EXPERTS and are being paid to write articles in a fairly high-profile magazine on their supposed area of expertise.

If Brier disagree's with Ed on the A4s and wants to ignore the free-card aspect that's one thing.
Same goes if Shykovsky wants me to fold my AA on a flop of J52 because there was a raise and re-raise from guys that he thinks he can almost 100% put on a set (which is kind of typical logic for him).

But messing up the math (or the 'terminology' per-se as was kind of the case in saying 17-to-1 instead of 1 out of 17) when you and your editor have every chance to proof-read before going to publication is pretty inexcuseable.

It's a high-profile, widely read magazine and it wouldn't hurt them to get some stuff right every once in awhile.


That said, I'm also glad there are the masses out there who think they are actually learning something by reading this garbage.

I wouldn't mind sitting-in with a bevy of Shykovsky desciples ANY DAY of the week.

So, to that end I say, "Keep up the Great Work Card-Player!!! I'm with ya all the way!!"