quick odds calculation question

[Goodnews]

If i had two overcards and a flush draw on the flop, how would i go about calculating my odds of hitting one of my outs by the river?

is it 15/47 + 15/46 = 65% ? Or am I calculating these odds incorrectly?

My friend argue that the real percentage is around 50% (through some method through permutations), and I'm not sure how they got this number, as you can probably see, an 15% difference is something to be concerned about.

He mentioned something about HPFAP, but i was quite drunk when we were discussing this.

[TStoneMBD]

if you need to do on the fly calculations, the 4/2 rule is your best bet.

for instance, if you have 8 outs, the odds of connecting in the next 2 cards is about 8x4=32%. if you have 8 outs on the turn, your odds of connecting on the river is about 8x2=16%.

once you get up to around 14 outs though the system doesnt work.

[Balla2hot4u]

i want to fully understand why these out percentages are what they are.. thanks
-Adam

[TStoneMBD]

quite honestly, i have no clue, but they work out pretty well.

[Balla2hot4u]

[ QUOTE ]
if you need to do on the fly calculations, the 4/2 rule is your best bet.

for instance, if you have 8 outs, the odds of connecting in the next 2 cards is about 8x4=32%. if you have 8 outs on the turn, your odds of connecting on the river is about 8x2=16%.

once you get up to around 14 outs though the system doesnt work.

[/ QUOTE ]

ok so why is it 8 X 4 on the turn and 8 X 2 on the river? is it because your outs are automatically multiplied times 2 ( because the math of the amount of outs u have divided by the cards remaining almost always multiplies it by 2?) then times the amount of cards remaining to come.. so u have 8 outs on the flop to hit a straight right.. 8X2(x2 again for the amount of cards still to come) = 34% then now on the river u have 8 outs still but only one card to come so its just 8X2= 16%? im just a little confused and would to clarify to myself thanks

[TStoneMBD]

it should be pretty selfexplanitory why its Xx4 and Xx2 since youre getting half the cards you divide by 2. again, i dont know why this system works, all i know is that it works.

[jimmyd2]

Clonie Gowen wrote a poker tip on computing % on Fulltiltpoker.com. She used the 4X2.
# of outs X number of cards remaining X 2. It is not exact but can help you while in play. The exact site is:
http://www.fulltiltpoker.com/prolessons/prolessons1.html

[Cobra]

The easiest way to figure the probability of getting an out on the turn or the river is 1 - the probability of not getting it on the turn and river. This gives you the probability of one or more of your outs hitting.

In this 15 out example:

= 1 - (32/47*31/46) = 54.11%

Cobra

[jba]

[ QUOTE ]
The easiest way to figure the probability of getting an out on the turn or the river is 1 - the probability of not getting it on the turn and river. This gives you the probability of one or more of your outs hitting.

In this 15 out example:

= 1 - (32/47*31/46) = 54.11%

Cobra

[/ QUOTE ]

this is it. OP's original post is off because it ignores the short circuit case.

The 4x rule is a rough approximation of the function Cobra points out, which is formally:
f(x)=(1-((47-x)/47)*(46-x)/46))*100;

note that this is a quadratic function (x squared is in the house) so the function f(x)=4x can only possibly be an approximation, because it's a linear function. It just so happens to closely approximate the real function for values of x between 0 and 10. Past ten it starts to breaks down - note that with 25 outs you win 100% of the time, which clearly isn't right.

The 2x is actually pretty damn close. the real function here is:

f(x) = (1-(46-x/x))*100

note that there's no x squared in the hizouse, so we have a linear function, so it can be closer. in fact:

(1-(46-x/x))*100 = (46-(46-x))/46*100 = 100x/46 = 2.17x

close enough to 2x.


check out this link (http://www.walterzorn.com/grapher/grapher_e.htm), you can graph the functions and see how they relate.

fix up the x/y coordinates to reasonable numbers (0-50x and 0-100y is good) and paste these equations in:
(1-(((47-x)/47)*((46-x)/46)))*100;
4x;
(1-((46-x)/46))*100;
2x;

peace

[CallYNotRaise06]

[ QUOTE ]
The easiest way to figure the probability of getting an out on the turn or the river is 1 - the probability of not getting it on the turn and river. This gives you the probability of one or more of your outs hitting.

In this 15 out example:

= 1 - (32/47*31/46) = 54.11%

Cobra


--------------------------------------------------------------------------------



this is it. OP's original post is off because it ignores the short circuit case.

The 4x rule is a rough approximation of the function Cobra points out, which is formally:
f(x)=(1-((47-x)/47)*(46-x)/46))*100;

note that this is a quadratic function (x squared is in the house) so the function f(x)=4x can only possibly be an approximation, because it's a linear function. It just so happens to closely approximate the real function for values of x between 0 and 10. Past ten it starts to breaks down - note that with 25 outs you win 100% of the time, which clearly isn't right.

The 2x is actually pretty damn close. the real function here is:

f(x) = (1-(46-x/x))*100

note that there's no x squared in the hizouse, so we have a linear function, so it can be closer. in fact:

(1-(46-x/x))*100 = (46-(46-x))/46*100 = 100x/46 = 2.17x

close enough to 2x.


check out this link, you can graph the functions and see how they relate.

fix up the x/y coordinates to reasonable numbers (0-50x and 0-100y is good) and paste these equations in:
(1-(((47-x)/47)*((46-x)/46)))*100;
4x;
(1-((46-x)/46))*100;
2x;

peace


[/ QUOTE ]

just a little highschool algebra

[SpeakEasy]

"Outs" are your estimate of the number of cards that will give you the best hand. Flush draw and two overcards = 15 outs (may be less, depending upon whether any of these cards will give opponent the better hand).

On the Flop:
1-8 outs --> Outs x 4
9-12 outs --> (Outs x 4) - 1
13+ outs --> (Outs x 4) - 4

On the Turn:
(Outs x 2) + 2

These calculations will tell you the % chance of hitting one of your "outs."

Example:
I hold AQo, board is K J 9

I estimate my outs on the flop to be any ace or ten -- 7 outs. Formula: 7x4=28. 28% chance of seeing one of my cards on the turn or river. Translated into pot odds, this is roughly 4 to 1 odds of hitting one of my outs. If the pot + bets is giving me better than 4 to 1 odds, I call on the flop (or raise, if the situation dictates).

For turning the percentages into pot odds, just remember some basic benchmarks:
50% chance requires 2 to 1 pot odds.
33% chance requires 3 to 1 pot odds.
25% chance requires 4 to 1 pot odds.
20% chance requires 5 to 1 pot odds.
16% chance requires about 6 to 1 pot odds.
10% chance requires 10 to 1 pot odds.

With practice, this becomes very easy to calculate on the fly. Go forth and calculate.

[ianlippert]

[ QUOTE ]
it should be pretty selfexplanitory why its Xx4 and Xx2 since youre getting half the cards you divide by 2. again, i dont know why this system works, all i know is that it works.

[/ QUOTE ]

It because you are changing your outs into a percentage. The reson the 2x works is because there are almost 50 cards in the deck. If you wanted to change 8/50 into a percentage you multiply top and bottom by 2 to get 16/100 or 16%. The 4x is because you are getting a second card your percentage is doubled. Once again, this system gives you an estimate and gets very sketchy around 15. I remember one play I had 18 outs and calculated it as 72%, when really it was more like 66%. Still good enough to call the guy all in though.

[TStoneMBD]

you math people are awesome. i have no idea what you said, but i dont know what id do without you guys.

[gambool]

[ QUOTE ]

50% chance requires 2 to 1 pot odds.
33% chance requires 3 to 1 pot odds.
25% chance requires 4 to 1 pot odds.
20% chance requires 5 to 1 pot odds.
16% chance requires about 6 to 1 pot odds.
10% chance requires 10 to 1 pot odds.


[/ QUOTE ]
Aren't you a point out on each of these? e.g. for a 50% chance you need even money?

[Nicmavsfan28]

please explain as you would a small child how you got those numbers :32/47 and 31/46 (was it?) anyway...with 15 outs why would it not be 30/47. Im really confused. /images/graemlins/frown.gif

[Cobra]

You have 47 unseen cards and you are trying to figure the probability of getting one or more of your 15 outs. To do this you take 1 and subtract the probability of getting none of your outs on the turn and river.

Now with 15 good cards and 47 unseen this leaves (47-15) = 32 bad cards. So the first term of missing your outs is 32/47. Now for the river you still have 15 good cards 46 unseen cards and (46-15) = 31 bad cards. So the second term of missing your outs on the river is 31/46. Hopefully that helps.

Cobra

[boogster]

Everyone is giving you the outs AGAINST hitting your hand. I calculate my odds based on the chances that I WILL hit my hand. So, here's what I do:

If you have 9 outs after the flop, then you have a 9x4=36% chance of hitting one of your outs. Therefore, you need 64%:36% or 2 to 1 pot odds to play till the river. Or if you are playing to just one card, then you have 9x2=18% or 82%:18% or roughly 4 to 1 pot odds to play for one card. Remember to count your own bet when calculating the pot odds. I think this is the easiest way to calculate pot odds versus the number of outs that you have to make your hand.

So, if you're drawing to a four flush after the flop (9 outs) and there's $30 in the pot and you're the last to act, do you call a $5 bet or not? Of course you do, because you're getting roughly 7:1 pot odds and your chances of hitting that flush on the turn is 4:1. If you are calling $10, then you don't have the odds to call ($30:$10 pot odds versus 4:1 outs). Of course, this is assuming that you're playing limit hold'em. In no limit, implied odds carry a heavy weight.

[Cobra]

Boogster,

When determining if you have the right Pott odds to call a situation you do not add your bet into the pot. Your bet is not part of the pot untill you actually put it in. In your example of $30 in the pot and it cost you $5 to see it your pott odds are 6:1. And with $30 in the pott and $10 to call your pott odds are 3:1.

Cobra

[boogster]

You're right.

[ QUOTE ]
Boogster,

When determining if you have the right Pott odds to call a situation you do not add your bet into the pot. Your bet is not part of the pot untill you actually put it in. In your example of $30 in the pot and it cost you $5 to see it your pott odds are 6:1. And with $30 in the pott and $10 to call your pott odds are 3:1.

Cobra

[/ QUOTE ]

[YoureToast]

[ QUOTE ]

I estimate my outs on the flop to be any ace or ten -- 7 outs. Formula: 7x4=28. 28% chance of seeing one of my cards on the turn or river. Translated into pot odds, this is roughly 4 to 1 odds of hitting one of my outs. If the pot + bets is giving me better than 4 to 1 odds, I call on the flop (or raise, if the situation dictates).

[/ QUOTE ]

This is very bad advice. Yes, you have a 28% of hitting your hand by the river with 7 outs, but you are not using pot odds properly because you are failing to include the cost of calling on the turn and river. To use the pot size to determine whether to call on the flop, you should consider only whether the pot size justifies the call for the next card only and then make a small adjustment for the implied odds you may get if you hit. In the current example, you are justified in calling with 7 outs if the pot is laying about 6 to 1 (not 4 to 1) (if the draw is for the nuts with very little chance of a redraw, then you can require slightly less odds due to implied odds).

[SpeakEasy]

[ QUOTE ]
[ QUOTE ]

I estimate my outs on the flop to be any ace or ten -- 7 outs. Formula: 7x4=28. 28% chance of seeing one of my cards on the turn or river. Translated into pot odds, this is roughly 4 to 1 odds of hitting one of my outs. If the pot + bets is giving me better than 4 to 1 odds, I call on the flop (or raise, if the situation dictates).

[/ QUOTE ]

This is very bad advice. Yes, you have a 28% of hitting your hand by the river with 7 outs, but you are not using pot odds properly because you are failing to include the cost of calling on the turn and river.

[/ QUOTE ]

You are correct that I am only calculating the "Express Odds" in this situation. The wisdom of the remainder of my advice is open for vigorous debate based on a variety of possible assumptions. However, let's return to the initial question of the first post:

[ QUOTE ]
If i had two overcards and a flush draw on the flop, how would i go about calculating my odds of hitting one of my outs by the river?

[/ QUOTE ]

My advice answered this question precisely. If you want Lesson 2 on "Implied Odds," that will cost you extra...

If I call, there are several possible scenarios:
1. One of my outs hits on the turn. Will he pay me off when I bet? Unknown. If yes, my implied odds back on the flop call increases.
2. One of my cards does not hit on the turn. He bets. Should I call? My decision here will be based on a new calculation of the pot odds, plus a slight adjustment for the implied odds on the river.
3. One of my cards did not hit on the turn but hits on the river. If I bet, will he pay me off? If yes, this becomes part of the implied odds calculation back on both the flop and the turn.

Because there are so many variables after the flop, I will calculate my pot odds on the flop (like in my post above) and factor in an adjustment for my implied odds on the turn and river based on quick assumptions about what I think my opponent will do if one of my cards hits (which is, of course, usually wild-ass speculation, especially when you are trying to quickly assess whether to make a call on the flop).

My advice above accurately calculates the chances of hitting one of my outs by the river. In my example, I will call with 4 to 1 odds, recalculate on the turn if I don't hit, and make an adjustment based on whether I think he will pay me off if I hit.

Anything more than this will sprain my brain for a simple flop decision.

[binions]

You have it backwards.

It's 1 - the chance you won't hit an out.

So, 1-((32/47)*(31/46)) = 54.1% to hit at least one of your outs.

[top2pear]

Does anyone else see how we have failed to resolve the difference between Goodnews's calculation and his friend's calculation?

We all just accepted Goodnews's faulty odds count: two overcards and a flush draw (presumably 4-to-the-flush) is only 13 outs twice because two of the helpful overcards are also part of the flush draw.

{This is the first and only time that i will ever be the first one to spot a mathematical mistake.}

[Cobra]

I was assuming he had something like AK /images/graemlins/heart.gif and the flop came 97 /images/graemlins/heart.gif, 3 /images/graemlins/club.gif. In this situation he would have 9 outs to the flush and an additional 6 outs to top pair and kicker.

Cobra

[KCFire105]

huh?

[KCFire105]

Hopefully there is a simple reply to my question.

Would the following example be correct?

Flop = 9/images/graemlins/heart.gif 5 /images/graemlins/heart.gif 2 /images/graemlins/spade.gif

I hold K /images/graemlins/heart.gif J /images/graemlins/heart.gif

I would have 13 outs equaling a 52% chance of catching.

No matter the pot size I would be able to call any size bet given that I would automatically receive better that 1-1 on my call?

Slightly different situation with no flush draw on the Flop of- 9 /images/graemlins/spade.gif 5 /images/graemlins/diamond.gif 2 /images/graemlins/heart.gif

I have K /images/graemlins/club.gif J /images/graemlins/diamond.gif

So I have 6 outs (minus the backdoor straight draw, or should that be figured in somehow?)

I now have a 24% chance of hitting, so I had best be getting 3 to 1 on my money (pot size + plus their bets?)

(On a side note, in early position, would you come out betting or check with the 13 outer flush draw and two overs?)

Thanks!

[damaniac]

[ QUOTE ]
Hopefully there is a simple reply to my question.

Would the following example be correct?

Flop = 9/images/graemlins/heart.gif 5 /images/graemlins/heart.gif 2 /images/graemlins/spade.gif

I hold K /images/graemlins/heart.gif J /images/graemlins/heart.gif

I would have 13 outs equaling a 52% chance of catching.

No matter the pot size I would be able to call any size bet given that I would automatically receive better that 1-1 on my call?

[/ QUOTE ]

First off, you would have 9 flush outs plus 6 unseen K/J (the KJ/images/graemlins/heart.gif (although one normally discounts overcard outs), so that's 15.

Second, you may be 52% (or whatever it would be, I'm not calculating it now) to hit...by the river. You really should just calculate your odds to hit on the next card and compare that to the pot size, because you will usually have to call a turn bet which changes your effective odds. Then you reevaluate on the turn if you miss. Knowing your chances of hitting by the river are really only good for equity purposes and if your opponent is all-in.

[ QUOTE ]
Slightly different situation with no flush draw on the Flop of- 9 /images/graemlins/spade.gif 5 /images/graemlins/diamond.gif 2 /images/graemlins/heart.gif

I have K /images/graemlins/club.gif J /images/graemlins/diamond.gif

So I have 6 outs (minus the backdoor straight draw, or should that be figured in somehow?)

I now have a 24% chance of hitting, so I had best be getting 3 to 1 on my money (pot size + plus their bets?)

(On a side note, in early position, would you come out betting or check with the 13 outer flush draw and two overs?)

Thanks!

[/ QUOTE ]

Again, your odds are really less than half this. The overcards aren't always good, but even ignoring that, you will have to pay a big bet on the turn to see the river, so you really only have about a 12% chance of hitting on the turn, so factor that in.

As far as flush draws go, if you are multiway on the flop and have a flush draw, you should usually bet and raise. You will make your flush 35% by the river, and usually will win when you do. If you have two or more opponents, you are putting in less than 35% of the money, so each bet (as long as 2 or more opponents call) earns you money.

[top2pear]

You're right, Cobra. My bad. (And i thought i was being all smart and stuff. /images/graemlins/wink.gif)

[Goodnews]

[ QUOTE ]
You will make your flush 35% by the river, and usually will win when you do. If you have two or more opponents, you are putting in less than 35% of the money, so each bet (as long as 2 or more opponents call) earns you money.

[/ QUOTE ]

Really? 35% to hit backdoor flushes? I don't think thats right, isnt it more like 3%?

[Hellmouth]

[ QUOTE ]
[ QUOTE ]
You will make your flush 35% by the river, and usually will win when you do. If you have two or more opponents, you are putting in less than 35% of the money, so each bet (as long as 2 or more opponents call) earns you money.

[/ QUOTE ]

Really? 35% to hit backdoor flushes? I don't think thats right, isnt it more like 3%?

[/ QUOTE ]

Runner Runner flushes are like 1.5 outs on the flop. This means that if you have 3 hearts out and need two more it is like 3%. However if 4 hearts are out by the flop then the percentage of one more coming on the turn or river is about 35%.

Greg

[qwegybo]

that is great calculations. i really like the kiss formula myself. i cant calculate that fast at the table.
i wish i could. awesome stuff.

[damaniac]

Sorry I was thinking of a regular flush draw. My mistake.

[buriedbeds]

This is from the Full Tilt Poker newsletter a while back (yes, I'm enough of a poker nerd to keep them for future reference... /images/graemlins/smile.gif ), discussing the same basic concept:

"It is very difficult to calculate the exact odds of hitting a drawing hand when you're sitting at the poker table. Unless you're a genius with a gift for mathematics like Chris Ferguson, you will not be able to do it. That leaves two options for the rest of us: The first option is to sit at home with a calculator, figure out the odds for every possible combination of draws, and then memorize them. That way, no matter what situation comes up, you always know the odds. But for those of us without a perfect memory, there's an easier way. Here is a simple trick for estimating those odds.

The first thing you need to do is to figure out how many "outs" you have. An "out" is any card that gives you a made hand. To do this, simply count the number of cards available that give the hand you are drawing to. For example: suppose you hold Ac 8c and the flop comes Qh 9c 4c. You have a flush draw. There are thirteen clubs in the deck and you are looking at four of them -- the two in your hand, and the two on the board. That leaves nine clubs left in the deck, and two chances to hit one.

The trick to figuring out the approximate percentage chance of hitting the flush is to multiply your outs times the number of chances to hit it. In this case that would be nine outs multiplied by two chances, or eighteen. Then take that number, multiply times two, and add a percentage sign. The approximate percentage of the time you will make the flush is 36%. (The exact percentage is 34.97%.) Now let's say that on that same flop you hold the Jd Th. In this case you would have an open ended straight draw with eight outs to hit the straight (four kings and four eights). Eight outs with two cards to come gives you sixteen outs. Multiply times two and you will hit the straight approximately 32% (31.46% exactly) of the time.

One important thing to keep in mind is that the percentage stated is merely the percentage of the time that you will hit the hand you are drawing to, NOT the percentage of time that you will win the pot. You may hit your hand and still lose. In the first example, the Qc will pair the board and may give somebody a full house. In the second example both the Kc and the 8c will put a possible flush on the board, giving you the straight, but not necessarily the winning hand. Still, knowing the approximate likelihood of making your hand is a good beginning step on the road to better poker."

Hope that's helpful...

-jake.

[KCFire105]

Actually no one has answered my original question-
[ QUOTE ]
Would the following example be correct?

Flop = 9 5 2

I hold K J

I would have 13 outs equaling a 52% chance of catching.

No matter the pot size I would be able to call any size bet given that I would automatically receive better than 1-1 on my call?


[/ QUOTE ]

And by the way, I believe I only have 13 outs, not 15 as was said earlier. Your double counting the overs along with the same cards for the flush draw.

[top2pear]

Your example is correct but you made the same mistake that i made. You have 15 outs to improve (no fear of double-counting hearts because you can't catch a heart that pairs your hole cards: both your hole cards are hearts): all nine /images/graemlins/heart.gifs, K/images/graemlins/spade.gif, K/images/graemlins/club.gif, K/images/graemlins/diamond.gif, J/images/graemlins/spade.gif, J/images/graemlins/club.gif, and J/images/graemlins/diamond.gif.

So your example is what Goodnews was asking about and i believe that you can call any bet with such odds.

[Balla2hot4u]

i want to thank everyone for all their imput.. this has helped me tremendously.. not only did i learn outs on certain hands..and how to convert those outs to percentages(was getting very frustrated playing with my friends with the knowledge that i knew what percentage i had to hit my hand but what was bothering me was i was sitting there saying to myself.. great i know what percentage i have to hit my hand but what good does that do me.. i had no clue how to combine the pot odds and percentage to make my hand.. to know to call or fold or raise someone).. now with everyones imput i beilieve i have a good grasp on it and will actually apply it to my game.. thanks everyone
-Adam