PDA

View Full Version : chances of flush vs flush

jacknine
09-11-2005, 08:05 AM
Im playing a 5-handed holdŽem game. I am dealt 6/images/graemlins/spade.gif 7/images/graemlins/spade.gif .

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
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
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
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
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
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
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
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
Might Help - Intro to Probability (http://www.letyoudown.com/intro.pdf)

bruce
09-18-2005, 12:14 PM
Much thanks.

Bruce

MickeyHoldem
09-18-2005, 02:20 PM
None of the following calculations take into consideration cards that may or may not be played or folded, and assume that all hands are played to the river and showndown.

On the flop, the prob. of any 1 player having 2 spades...
p(1) = (8c2)/(45c2)
the prob. of any 2 players having 2 spades...
p(2) = (8c4)/(45c4)
similarly for 3 and 4 players...
p(3) = (8c6)/(45c6)
p(4) = (8c8)/(45c8)

Now there are 4 spades higher (and 4 lower) than yours, so p(1) &amp; p(2) must be adjusted for the times they don't beat you...
q(1) = p(1) * (1 - (4c2)/(8c2))
q(2) = p(2) * (1 - (4c4)/(8c4))

Now by inclusion/exclusion we see that the prob. that no one is ahead of you on the flop (with 4 oppenents) is...
P(f) = 1 - 4*q(1) + 6*q(2) - 4*p(3) + p(4) = .913876
so the prob. that you are losing is 8.6124%

After the turn/river, there are 3 possible situations...

In (a), the situation is the same as the flop calculations above, except for there being 43 cards unknown cards instead of 45. You are ahead...
P(a) = .905883

For (b) the spade that fell could be higher or lower than your 7 (equal chance). If it's higher, there are 3 cards that beat you; lower, 4. So you're ahead here...
P(b) = .5 * (40c8)/(43c8) + .5 * (39c8)/(43c8) = .477311

For (c), things are more complicated again...
(c1) both spades are higher.... occurs (4c2)/(8c2)**
(c2) one higher, one lower.... occurs 4*4/(8c2)
(c3) both spades are lower.... occurs (4c2)/(8c2)
so similar to (b) you're ahead...
P(c) = (4c2)/(8c2) * (41c8)/(43c8) + 16/(8c2) * (40c8)/(43c8) + (4c2)/(8c2) * (39c8)/(43c8) = 0.535167

So finally we can combine these, and find out that you're ahead...
P = (37c2)/(45c2) * P(a) + (8*37)/(45c2) * P(b) + (8c2)/(45c2) * P(c)
= .767260
so the prob. that you will lose is .232740

**note when 2 higher spades fall here, you will only chop when no other player holds one of the remaining 2 higher spades.... this occurs .399342% of the time /images/graemlins/wink.gif

MickeyHoldem
09-18-2005, 02:27 PM
[ QUOTE ]
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.

[/ QUOTE ]
Should be...

4c2 = 6 ways to have 2 higher spades
4*4 = 16 ways to have 1 higher and 1 lower

6 + 16 = 22 ways for a higher flush

AaronBrown
09-18-2005, 02:42 PM
Thank you for the correction.