calculating probability and odds for flush draws

[AnotherAddict]

Here are some questions about calculating probability of flush draws.

It is my understanding that in order to calculate the probability of the card you need to complete the hand (P), you divide the number of card unseen (X), by the number of outs you have (Y). Example: 4 card flush draw on the flop, there are 47 cards unseen, there are 9 cards that can help you complete the flush. Therefore, the formula is P=X/Y.
Is that correct?

I have read that when you have a 4 card flush draw on the turn card the probability and odds of completing the flush
Odds are: (47/9)= 5.22 to 1
Probability is: (9/47)= roughly 19.1%
Is that correct?

On the River, Odds are: (46/9)= 5.11 to 1
Probability is: (9/46)= roughly 19.56%

Now, if you have the 4 card flush on the flop, are the probability and odds of getting a flush by the river the roughly 19.56% and 5.11 to 1 ?
Is that correct?

Therefore you must have greater than roughly 5.11 to 1 pot odds in order for it to be a positive expectation and for you to continue profitably.
Is that correct?

I have read that you have a 3 card flush, in order for you to get a runner runner flush on the river, it is a 1 in 25 to complete the flush.

To calculate that it would be (10/47)*(9/46)= roughly 4.1% chance or 24 to 1 odds of completing the flush. Is that correct?

Finally, when is it proper to calculate the number of cards unseen (X) minus the number of cards that will complete your hand (Y), devided by the number of cards that will complete your hand (Y)?
The formula of ((X-Y)/Y). Example, you have a 4 card flush and there are 47 cards unseen, 38 of them will not complete your flush, 9 will. Therefore, it would be ((47-9)/9)= 4.22 to 1 odds.

[LetYouDown]

No to your 4 flush on the flop percentage. Here's a different way of looking at it. Calculate the odds that you DON'T make the flush, then subtract them from 1.

If you're not going to make it, there are 47 - 9 cards that can come in any order on the turn and the river...because neither of your flush cards can come. So you need the number of possible combinations of turn/river and the number of combinations for those 38 cards.

Total Combinations = 47 C 2
Total Non-Flush Combinations: 38 C 2

47 C 2 = 1081
38 C 2 = 703

Odds = 703/1081 that you won't make it.

therefore

Odds = 378/1081 that you will make it, or basically 35% or 1.86 to 1.

With only one card to come and a flush draw, you don't really need math, there's 9 "outs" in 46 cards so it's 9/46 or ~19.6% or 4.1 to 1.

[olavfo]

[ QUOTE ]
Here are some questions about calculating probability of flush draws.

It is my understanding that in order to calculate the probability of the card you need to complete the hand (P), you divide the number of card unseen (X), by the number of outs you have (Y). Example: 4 card flush draw on the flop, there are 47 cards unseen, there are 9 cards that can help you complete the flush. Therefore, the formula is P=X/Y.
Is that correct?


[/ QUOTE ]

It's the other way around. The number of outs (Y) divided by the total number of unseen cards (X) equals the probability to hit your out on the next card. P = Y/X.

If you have a suited hand and 2 of your suit on board on the flop, the probability of hitting the flush on the turn is P = 9/47 = 0.191 = 19.1%.

[ QUOTE ]

I have read that when you have a 4 card flush draw on the turn card the probability and odds of completing the flush
Odds are: (47/9)= 5.22 to 1
Probability is: (9/47)= roughly 19.1%
Is that correct?


[/ QUOTE ]

Not quite. The probability is P = 9/47 = 0.191 as you say, but in odds form it's (1/P - 1) to 1 = 4.22 to 1.

[ QUOTE ]

On the River, Odds are: (46/9)= 5.11 to 1
Probability is: (9/46)= roughly 19.56%

Now, if you have the 4 card flush on the flop, are the probability and odds of getting a flush by the river the roughly 19.56% and 5.11 to 1 ?
Is that correct?

Therefore you must have greater than roughly 5.11 to 1 pot odds in order for it to be a positive expectation and for you to continue profitably.
Is that correct?


[/ QUOTE ]
Same error as above. The probability is P = 9/46 = 0.196, but the odds are (1/P-1) to 1 = 4.11 to 1.

[ QUOTE ]

I have read that you have a 3 card flush, in order for you to get a runner runner flush on the river, it is a 1 in 25 to complete the flush.

To calculate that it would be (10/47)*(9/46)= roughly 4.1% chance or 24 to 1 odds of completing the flush. Is that correct?


[/ QUOTE ]

Same error again. The probability is P = (10/47)*(9/46) = 0.0416, and in odds form we get (1/P - 1) to 1 = 23.02 to 1 = 23 to 1

[ QUOTE ]

Finally, when is it proper to calculate the number of cards unseen (X) minus the number of cards that will complete your hand (Y), devided by the number of cards that will complete your hand (Y)?
The formula of ((X-Y)/Y). Example, you have a 4 card flush and there are 47 cards unseen, 38 of them will not complete your flush, 9 will. Therefore, it would be ((47-9)/9)= 4.22 to 1 odds.

[/ QUOTE ]

You can always do this to find the odds in "X to 1" form. This is what I've been doing in all the algebra above. (Note that (X-Y)/Y = X/Y-1, so it's the same formula as 1/P-1, which I used)

olavfo

[AnotherAddict]

That's it. I understand now. When calculating odds, you are dividing the cards that will not help you by the cards that will.

In probability, you are dividing all the unseen cards by the # of cards that will help you.

One more thing, is it true that you must have greater than roughly 4.11 to 1 pot odds in order for it to be a positive expectation and for you to continue profitably?

[LetYouDown]

Assuming you're ahead, and you the corrects odds are in fact 4.11 to 1, then yes. Minus any rake of course

[olavfo]

[ QUOTE ]
One more thing, is it true that you must have greater than roughly 4.11 to 1 pot odds in order for it to be a positive expectation and for you to continue profitably?

[/ QUOTE ]
That's the pot odds you need to continue from the turn to the river, since (9/46 -1) to 1 = 4.11 to 1.

On the flop things are a bit different, since there are two cards to come. The probability of making the flush with two cards to come is

P = 1 - (38/47)*(37/46) = 0.35 = 35%

and the corresponding pot odds are (1/P - 1) to 1 = 1.86 to 1.

This means that if you have two or more callers, you can bet the flush draw for value on the flop. You need 1.86 bets to enter the pot for each bet you put in, and with two callers you get 2 bets in for each of yours. So you can bet and raise the flush draw for value on the flop if more than one opponent comes along for the ride.

If you're in late position this might also give you a free card, so you can sometimes see the river without paying additional bets, if the flush doesn't come in on the turn.

On the turn, you use the probability with one card to come (9 outs and 46 unseen), which gives P = 9/46 = 0.196 and pot odds (1/P - 1) to 1 = 4.11 to 1. But note that you can not bet a flush draw for value on the turn, unless you have at least 5 callers (since you need 4.11 bets to enter the pot for each of yours in order to have a value bet).

olavfo

[Glenn K]

It is a bit hard to sit at the table and begin to do serious math. An easy way to count which is quite accurate up to ten outs is to multiply the number of outs by 2 on turn and another two on river. Open str8 draw after flop equals 8 outs. That means approx 32% to hit the str8 from flop to river and 16% from turn to river.