cameron
|
Best Answer
*Edit* I forgot, there's a simpler formula that does the same thing for certain calculations.
N_a * N_b * ...(for # different types needed) * (#total picks Permute(#total picks) )
divided by
D * D-1 *...(for # total picks)
For the sake of demonstration, suppose you're drawing 3, and you need 3 types of cards to hit your hand. Suppose one type has 8 remaining cards in the deck, another has 5, and the other has 3. Suppose the deck has 47 cards remaining. The math would look like this:
[(8 * 5 * 3) / (47*46*45)] * 3p3 = 0.0074
Now suppose we were drawing 4 cards, and we needed 2 of the card with 8 in the deck, instead of just one. There would be a slight change in the formula:
[(8*7 * 5 * 3) / (47*46*45*44)] * 4p4
/ 2
= 0.00235
The division by 2 is because, needing two of one type of card isn't the same as needing one each of two types of cards. We're dealing with permutations. With two types of cards, we'd have more permutations. Consider: (8/52)*(7/51) and (7/52)*(8/51). But with two of one type of card, there is no (7/52), it wouldn't make sense. It so happens that there are exactly half as many permutations, so we divide the whole thing by 2.
Here's a simpler example where the formula is again different. Find the probability of getting 4 aces when drawing 5 cards. In this example, there's only 1 type of card needed, so order doesn't matter at all, therefore there's no permutation used. There will be a combination used, because there are more cards drawn than needed.
[4! / (52*51*50*49) ] * (5 c 4) = 1.847 E-5
You don't have to memorize the variations of the formula, just understand the simple logic of the formula and you'll know how to tailor it to each situation.
Below is my original answer. It's a one-size-fits-all formula, you'll never have to tweak it for a specific problem, it can solve all the above examples plus more. However, it is more complicated, so it's perhaps better to use it only for more complicated problems.
{ (x C a) * [(z-x) C (n-a)] / (z C n_total) } * { (d C a_s)^r / (n_total C a_ss) } = probability of getting exactly A of X object in N draws, for dependent events
z = total number of objects to choose from (in hold'em this will be 52)
x = number of objects of the type of thing you need (for instance, there are 13 spades)
a = how many of those x objects do you need (for instance, you need 5 spades for a spade flush)
n = how many picks to try to get the A objects (for instance, if you see the river you get 7 cards)
There will be more than one value for X, A and N if you need more than one type of card. You would have two separate numerators, and multiply them together before dividing by (z C n_total). N_total would be the sum of the individual n-values.
In most cases, d=1 and r=1. But some problems require the calculation to be divided into what I call "rounds". For instance, someone recently asked the probability of getting dealt blackjack 8 times in a row. For that there are 8 rounds (r = 8), because order somewhat matters: getting dealt double-aces 4 times and double-tens 4 times doesn't count as blackjack, you need the aces and tens to happen together in eight two-card clusters. The d-value is the number of draws per round. In this example, d=2 because a blackjack hand consists of two cards.
Furthermore, some problems require "sub-rounds". This is when not all rounds are equivalent (or put another way, when order matters on a micro-level as well). This kind of problem requires an additional (d C a)^r to be multiplied into the numerator before dividing by (n C a). (However, my formula might have a flaw for problems with sub-rounds, I'm looking into it.)
It's a powerful formula, but undoubtedly I've made it confusing. So I'll provide some demonstrations of how it can be used:
"An urn has 9 balls that are identical except that 5 are white and 4 are red. A sample of 6 is selected randomly without replacement.
What is the probability that exactly 4 are white...?"
(5 c 4) * [(9-5) c (6-4)] / (9 c 6) = 0.357
Bingo: 75 numbers to draw from, 5x5 card, 1 free space in the middle
Calculate: the probability of marking all 24 spaces in 47 picks or less.
(24 c 24) * [(75-24) c (47-24)] / (75 c 47) = 6.3 E-7
Blackjack: calculate the probability that the dealer gets 8 blackjacks in a row, using a 4-deck shoe.
(16 c 8) * [(208-16) c (8-8)] * (64 c 8) * [(208-64) c (8-8)] = (16 c 8) * (64 c 8)
(16 c 8)*(64 c 8) / (208 c 16) = 1.7548 E-10
(1.7548 E-10) * (2 c (8/8))^8 / (16 c 8) = 3.490556 E -12
Now try this harder formula on the problems I solved using the easier formula (4 aces in 5 cards, etc.). If you get the same answers I did, then you have the basic idea of this formula.
It would be better to have a program that crunches this formula for you, so all you have to do is input the variables and it outputs the answer. I already have such a program, so if you wanted I could email the code for a TI calculator, or attach my java version of the program.
 Posted 654 day ago
|