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-   -   chances of flush vs flush (http://archives2.twoplustwo.com/showthread.php?t=334334)

jacknine 09-11-2005 08:05 AM

chances of flush vs flush
 
Im playing a 5-handed holdŽem game. I am dealt 6[img]/images/graemlins/spade.gif[/img] 7[img]/images/graemlins/spade.gif[/img] .
Flop: 8[img]/images/graemlins/spade.gif[/img]K[img]/images/graemlins/spade.gif[/img]T[img]/images/graemlins/spade.gif[/img]

Now, what are the chances of me losing this to a higher flush?
IŽve taken the chance of someone having a higher flush on the flop + someone having 1 higher spade than me and another spade coming off on the turn/river. I came up with approx. 8% chance of losing to a higher flush. Correct?

(Im not taking into account anybody having a str8flush)

AaronBrown 09-17-2005 07:26 PM

Re: chances of flush vs flush
 
With 47 cards out, there are 47*46/2 = 1,081 possible sets of pocket cards. There are four spades out higher than your 8, plus four other spades. These can be arranged into pocket cards in 24 ways. 24/1,081 = 2.2%. With four other players, there's about an 8.8% chance that one of them was dealt two spades that give a higher flush than yours. Many of these hands would have folded preflop in most games.

One of the four spades higher than 9 could be paired with 39 non-spades in the deck, that's 4*39 = 156, or a 14.4% chance than a single other player holds a higher spade. If exactly one other player holds a spade, there are 7 cards that can beat you. The chance of one of them showing up on the turn or river is 1 - (38/45)*(37/44) = 30.0%. Multiplying by the 14.4% chance that gives a 4.2% chance that one player will beat you.

It gets a bit complicated to figure four other players, but it's not a bad approximation to multiply by 4 again and get 16.8%. Add to the 8.8 and you get 25.6%, or about one chance in four that your flush will lose if any player holding a spade stays in to showdown. The actual probability of loss is closer to your 8% estimate, given how many of the hands that beat you are unlikely to have been played.

bruce 09-17-2005 10:48 PM

Re: chances of flush vs flush
 
Pardon my ignorance but how did you arrive at there being 24 combos of someone having a hand with two spades? I thought that since there are 8 spades we could 8*7/2 and
come up with 28 combos. Also if there is a 2.2% chance of a player having spades I assume you're multiplying 2.2 by
4 and coming up with 8.8. But isn't 2.2 the percentage of someone having two spades, not necessarily two higher spades? Thanks for your help. It's been 25 years since I have taken a class in math.

Bruce

AaronBrown 09-17-2005 11:24 PM

Re: chances of flush vs flush
 
I always worry when someone says "pardon my ignorance" because it usually means I've done or said something stupid.

In this case, I think we're both right in both cases, we just reversed our meanings like two people in a door moving side to side and always blocking each other.

Yes, with 8 spades, there are 8*7/2 = 28 ways to combine two of them. But with only 4 higher spades, there are 4*4/2 = 8 ways for someone two have two higher spades plus 4*4 = 16 ways to have a higher one and a lower one. 8 + 16 = 24, so that's how many ways someone can have two spades that result in a higher flush. That's what I was trying to compute; and that's why I can multiply my 2.2% by 4.

KJL 09-17-2005 11:46 PM

Re: chances of flush vs flush
 
How come when figuring out the chance that someone has 2 higher spades the calculation is 4*4/2(wouldn't it be 4*3/2) but when you have a higher and lower one it is not divided by 2. Also why can't you calculate this by doing 4*7=28. This is just not making sense right now and I'm sure I'm doing something stupid but can you explain this to me.

AaronBrown 09-17-2005 11:57 PM

Re: chances of flush vs flush
 
Nothing stupid, it's a tricky subject.

When you pick two items out of one set, you divide by 2, because you can do them in either order. When you pick one item from set one and one item from set two, you don't divide by two because they are distinguishable.

I know is seems odd, but it's true. If you want to pick two cards out of the deck, and you don't care about the order, there are 52*51/2 combinations. But if you pick one red card and one black card, there are 26*26 combinations, no need to divide by 2.

KJL 09-18-2005 12:15 AM

Re: chances of flush vs flush
 
Thanks alot Aaron, I was being stupid tho cause I actually knew that, but for some reason, I didn't realize it was two sets.

bruce 09-18-2005 12:09 PM

Re: chances of flush vs flush
 
Thanks a lot Aaron for your responses. Thanks KJL because
I had the same question.

On a seperate note what's a good book or website for me to learn more about this, particularly in response to holdem?
I'm close to 50 and even though I have four semesters of math in college, I never once took a class in probability?

Thanks again.

Bruce

LetYouDown 09-18-2005 12:11 PM

Re: chances of flush vs flush
 
Might Help - Intro to Probability

bruce 09-18-2005 12:14 PM

Re: chances of flush vs flush
 
Much thanks.

Bruce


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