Are you drawing with or without replacement?

Assuming without replacement, question 1. With six packs of standard playing cards, your odds of drawing one specific card could be expressed as P(Ace of Spades): 6/(52*6), which would be 6/312, which simplifies to 1/52. Any probability for drawing a specific number of cards over several iterations without replacement would necessarily decrease the total number of cards by one, so if you drew an Ace of Spades on the first draw, your odds the second draw would be 5/311.

Simple probability is pretty easy to understand. You should realize, though, that probability is usually expressed as a fraction or a decimal (percentage), rarely as a gambling odds ratio.

2ndWhiteLine's question is valid - and it will affect the calculation... but, with so many cards to draw from and only three cards being drawn, the difference would be relatively small between the two scenarios (with or without replacement).

Q1: with replacement: 1 in 16 chance... It goes like this: the second card has a 1 in 4 chance of matching the first, the third card has a 1 in 4 chance of matching the first two... 1/(4^2)... without replacement it gets much more complicated - but with nearly the same final result.

I will pass on answering the other two - leaving it to people that haven't gone 25 years without calculating such things, as I have.

Q1 should be slightly longer odds than 16:1 - the actual odds are 77/311 times 76/310, which 16.47:1

Q2 - needs a bit more time, but it is 4 calculations taking into account the probability of the first card being each given suit. The answer to Q3 should drop out of that.

I think (for example) the probability of 3 Diamonds is [(D/(D+H+C+S))*((D-1)/(D+H+C+S-1))*((D-2)/(D+H+S+C-2))]. You need to do all 4 calcs to get the total. 1/x that to get the odss to 1

when I say "matching" I mean matching suit - which was the original question

Usually all card probabilities are without replacement.

Depends on whether he's doing beginning probability, which it sounds like he is. Also, there's no "rule of thumb" for assuming a card probability problem is without replacement. Have you ever taken statistics putin?

Yes. Unfortunately.

I also play cards. Never heard of drawing a card and replacing it with the exact same card.

In "word form", Q1 is (TotalSuit - 1) / (TotalDecks -1) * (TotalSuit - 2) / (TotalDecks - 2)
...or...
77/311 * 76/310 = 2926 in 48205 (I'm to lazy to factor further) or PeteU's 1 in 16.47.

Q2 is the sume of the four suits probabilities calculated in a similar way.

Q3 - Yes, we have all explained them to you.

And I have to agree with Putin. As a poker and blackjack player, I've *never* seen anyone do replacement when doing card probabilities. If they want to do those types, they usually do dice probabilities with multiple dice (ala craps) which lend themselves well to the probabilities for replacement.

Thanks guys this is really helpful. Am I right to assume that I work out probability of each suit and add those probabilities together and divide by 4 in order to work out probability of drawing a flush in any suit?

You don't divide by 4.

Nope, just add the together. You do have to get a common denominator so if, say (and these are made up numbers) 1 in 3 of diamonds, 1 in 4 of spades, 1 in 5 of hearts, and 1 in 6 of clubs, you would make them be 20 in 60 of diamonds, 15 in 60 of spades, 12 in 60 of hearts and 10 in 60 of clubs. Then add them together and get 57 in 60 of getting it in that suit. Obviously those numbers were made up because the odds of getting a flush are *way* lower than 95% of the time. :-)

Ask help from your teacher when you have a homework problem you can't figure out.

Ouch, fk. That was just... well... mean. This is maniac. He's a nice lunatic. And we never did give him a proper answer to the two questions by doing the multiplcation and addition on the second or fully factoring the first, so he still has to understand our explanation inorder to completely solve the problem.

I absolutely disagree with limiting yourself to the instructor for help.
My first Crypto class was a nightmare until we started skipping the lecture to teach ourselves.
http://www.youtube.com/user/numericalmethodsguy is the only reason I passed Numerical Methods.
I research textbooks and buy the one I think will teach me the best, not the one the professor chose.
And I am constantly using google and the forums it points to to learn concepts I don't quite grasp.
No one is just giving Maniac the answers. He knows he'll need to understand the question to repeat it on an exam.

All of that said, statistics is not something I'm good at. =)

@fortknox - I'm 30 years too late to ask my teacher :(

@others - I can now work out the chances of being dealt 3 of any suit from the top of six shuffled decks = 16.47:1 thanks. But when I use the same formula to work out the probabbilty of being dealt 3 of same suit from 3 decks the result is 16.97:1 - my instinct says the odds should be the same, are my instincts wrong or have I messed up?

havent read all above, but your instincts are wrong - everytime you remove one of the desired suit, the probability of getting one more decreases faster, the fewer starting cards you had to begin with. just think of your odds of getting three of the same suit, when there is only 4 of each suit to begin with :)

Without doing the multiplication myself, I am pretty sure you are right. The difference between 3 and 6 decks is that in 3 decks is that with a smaller set of cards drawing your first diamond hurts your chances of getting that second diamond more than in a bigger deck.

@Per Olander - thanks for that it illustrates it beautifully - the odds of getting 3 of same suit from 16 cards (4 cards x 4 suits) is 35:1 correct?

that seems right - I rarely work with odds though, and dont see the benefits of it over probabilities :)
if you want to learn more about the mechanisms behind the calculations, read about binomial distributions and hypergeometrical distributions.

@Maniac - it is my default answer to question where it is possible someone is having trouble with homework (or just want the answer), since we do have a lot of younger people. Sorry if I offended.

15:3 or 5:1 times 14:2 or 7:1 = 35:1. Yep, those are correct for 16 cards.

Of course, it's always Outs/(52-VisibleCards) is your odds of hitting your out on the river (VisibleCards is usually 6, but can be 7 if a burned or mucked card flashed). Odds of hitting your flopped draw are:

Outs / (52 - Seen Cards)
+ Outs / (52 - (Seen Cards + 1))

The extra plus one is for the turn card which will be seen before the river.

Agreed with Per Olander here. I hate working with odds. Just use p(x).

^Clearly not a card player...

You are just weighing p(x) vs p(y). And when you play cards, you play the players more than you play the odds. Know the odds but play the man.

That's why you work with implied odds, zultar.

There was a touch of sarcasm there. It would be " know the probabilities but play the man."

@Zultar - It depends on the game. TX NL Hold'em live you play the man as much as the cards, but you still have to be prepared for the show down. Limit Hold-em you purely play the cards. We call it no fold'em hold'em for a reason. And online you*always* play the cards. You have no way of knowing if the guy is responding slow cause his SO is bitching at him/he is distracted or because he is thinking, so you play the cards.

Fair enough.

@D, you know I like the contributions you make to the forum when you are not -you know- berserk. Whether I agree with you or disagree with you, I still read what you write. But when you go off the handle, I feel like you just need one of us -make that a mod since that's what they are paid for- to give you a hug. :)

"to give you a hug"

That seems like a good job for Thucy.

Hip hip hooray for Thucy.

I'm 100% behind Draug on his views of hold'm, which is why I only play NL in person. It has been a couple years, but I used to be pretty decent, and it is all in learning the person more than knowing when to hold and when to fold and when to go all-in.

@fk - Draugnar's Den will officially open March 10th, I think. The inaugural event is most likely going to be a charity event. I'm even thinking of trying to have one charity event per month where the buy in is maybe $25 and $5 of it goes to the charity of the previous charity event winner's choice. You interested in participating?

@FK/Draug

If we ever meet up, I really want to play some Hold 'Em with you two.

I'd love to get the second Saturday afternoon of the month to be a charity tournament and the fourth Friday evening to be a $.25/$.50 ring game game (or $1/$2 if people want to risk more).