PDA

View Full Version : Sorry I must be an idiot

drew
05-29-2003, 01:47 AM
I am reading "Theory of Poker." On p38 Sklansky says, "For example, with 3 spades on your first three cards and no other cards seen, you will make a spade flush in seven cards 18 percent of the time." Could someone show me this calculation please? I just cannot do this for some reason. /forums/images/icons/confused.gif Thanks

Cyrus
05-29-2003, 02:51 AM
"With 3 spades on your first three cards [in seven-card stud] and no other cards seen, you will make a spade flush in seven cards 18 percent of the time. Could someone show the calculation for this?"

There are 10 more spades remaining.

C(10,4)=210 combinations that give you 4 more spades, for a 7-card hand of all spades.

C(10,3)=120 , so (120*39 non-spade cards)=4680 combinations that have 3 spades and 1 non-spade card, which you give you a 6-card Flush.

C(10,2)=45 , so (45*741)= 33345 combinations that contain 2 spades along with 2 non-spade cards. That's because there are C(39,2)=741 combinations of 2 cards that the 39 non-spade cards can produce.

All in all, 210+4680+33345= 38235 total combinations that give you a Flush of at least 5 spades.

Dividing that number with all the possible combinations of 4 cards that can be made from the remaining 49 cards that are unseen, gives 38235 / C(49,4) = 38235/211816 = 0.180510443026022585640367111077539 or 18%

Short Answer : Sklansky said so.

drew
05-29-2003, 07:31 AM
Thanks. I knew it had to be a math error. I was forgettting to multiply in the combinations of non-spade cards. Very frustrating! Do people actually calculate binomial coefficients at the table? Is there a shortcut to estimating these percentages? Or, are they just roughly memorized?

Cyrus
05-29-2003, 04:54 PM
"Do people actually calculate binomial coefficients at the table? Is there a shortcut to estimating these percentages? Or, are they just roughly memorized?"

Some odds are so often encountered that they get to be memorised, eg the odds of flopping a set when holding a pair. Other odds have to be worked at the table. Besides Sklansky, no one does complex calculations involving combinatorials at the table. (Malmuth has a nail-size computer with a heads-up display in his glasses.)

The usual method, as far as I know, is shotgunning the outs. Suppose you wanna work out the odds of flopping a set when holding a pair : You have 2 outs and 3 cards to come on the flop, so you have 3 'chances' to hit any of your 2 outs. That means 2*3=6 'chances'. The odds against hitting your outs are (50-6) to 6 = 46 to 6 = 7 point something to 1. That's quick and dirty -- and slightly beautifying the odds (which are actually 8.3:1 against).

chuckl8r
05-29-2003, 06:57 PM
In 'Killer Poker', by Jon Vorhaus, he gives a method of figuring the chances.

(Number of outs)times(Number of draws)times 2%

That gives an 8.5 - 1 figure for the flop-a-set scenario

(2)X(3)X2% = 12% 100/12 = 8.5

That seems to give an awfully high figure for the flush, though, doesn't it?

BamaGambler
05-30-2003, 02:33 PM
I think this shortcut only works when you need one more card to complete your hand. Therefore, it can't be applied to the 7 card stud example b/c you need 2 more spades. It could be applied to an example where you had 4 spades after 4th street (or 5th or 6th).

dogsballs
06-02-2003, 09:46 AM
These shotgun methods become increasingly more inaccurate as you increase your outs, with the real chances being lower than the quick and dirty methods calculate.

You may also have figured it where you're only looking at the probability of catching one more spade, not the two that you need.

dogs