# How do I use combination math to calculate at least two spades in the first 3 cards in a stud game

I'm generally good with math, but probability calculation just seems to fight me.

I'm trying to develop a formula to determine the odds of hitting at least 2 spades, given that the player is dealt their first 3 cards.

The probability is about 15.05%. I have confirmed this by running 5 million simulated 3 card hands and comparing how many times 2 or 3 spades hit. This is agrees with the online calculator here: https://www.ohrt.com/odds/index.php?t[]=52&d[]=3&o[]=13&c[]=2&&s=any&p=9

By my reasoning, the formula to arrive at the probability should be:

`combin(13,2)*combin(50,1)/combin(52,3)`

or "from 13 spades, take 2, from the remaining cards, take 1, and divide by the total number of 3 card combinations in a 52 card deck".

But that gives me a result of 17.65%.

I'm trying to develop a general formula for any number of outs or cards dealt(similar to the online calculator), but this was the test case I've been stuck on for a while.

We want to sum the probabilities of drawing three spades in three cards, plus drawing two spades in three cards. The reason why we need to treat the cases separately is because while there is only one arrangement of suits when you draw three spades, there are three arrangements of suits when you only draw two spades.

As such, we have (13/52)(12/51)(11/50)+3(13/52)(12/51)(39/50)=15.0588%.

• Thanks, that makes sense. I'm trying to figure out a way to generalize that to a formula in terms of outs, needed outs, and remaining cards to draw. I was trying to figure it based on numbers of combinations of acceptable cards vs the total number of combinations of cards. For example there are 22100 possible 3 card hands in poker, based on 52C3. So I was trying to figure how to compute 'X' for p=X/N. In my example case, if p=15.06% and N = 22100, then X, must be 3328. So there must be 'X' 3-card combinations that contain 2 or 3 spades. I just can't figure out How to arrive at 3328. – Jon Wolfe Jun 6 at 6:54

Ok, Andrew Chin's answer was not exactly what I was asking for, he did make me rethink my approach, and I was able to come up with a general solution that works for my example and others.

if:

d = number of cards dealt

r = number of required outs

u = number of unknown cards remaining

t = number of outs

then the general formula to compute the probability of getting at least the number of required outs when 'd' cards are dealt is (note the expressions in parenthesis are using combination notation, not fractions. the combin function in excel will evaluate that): For the example case in my question, plugging in the numbers: When yields 15.058% when evaluated. The top part of the fraction sums up all possible 3 card hands with 2 or 3 spades, and the denominator expression are the number of possible 3 card hands in a 52 card deck.