Four to a flush with suited hand and two community cards, opponent raises, what to do?

Question

I had two suited hole cards in a nine-handed cash game, 50 chips entrance.

The flop opened up with two more cards of the same suit, meaning I had four cards to a flush. My opponent raised six on the flop and I called.

The turn card came down without my suit. My opponent raised another 10, so I folded.

Did I do the right thing? How do you calculate odds for a "four to a flush" situation?

Answer 1

In general to calculate your percentage of hitting you can do the following:

• Count your outs. In your case: 13 cards of your suit minus the 4 you already see make 9 cards in the deck which will make your flush.
• Calculate the amount of cards left. Since we can not know the cards of our opponents, we include them in our calculation. Hence there are 52 -2 (your cards) - 3(flop) = 47 cards left.
• The probability you will make your flush on the turn is therefore 9/47. The probability you will make the flush on the river is 9/46.
• Since the events are dependent, combining the probability goes as follows: For the second event, we should calculate the probability that event 2 happens AND event 1 did not. Therefore: 9/46 * (1-9/47) = 0.16. Combining this with the probability of event 1 gives: 9/47 + 0.16 = 0.349 and hence about 35%
• On the turn, your hitting chance (if the flush did not come on the turn) would be 9/46 = 0.19, or 19%

Your strategy on the flop should therefore be: call when the pot-odds are lower than 35% (you have to pay less than 35% of the pot to call).

However, this calculation is quit cumbersome in practice. Therefore, here is a quick rule of thumb to approximate your hitting chance:

• count your outs. In your case 9
• Multiply with 4 on the flop to get 36%
• On the turn, you multiply with 2 and add 2 to get 20%

Note that both values are one percentage point larger than the actual odds. Call this optimistic evaluation.

Answer 2

At the turn your opponent bet 10 into a pot of 62 so you were getting proper odds to call.

What is scary here is that your opponent was giving you odds to stay in the pot. They may have been on a bigger flush draw. If you have the bigger flush draw you have a chance of stacking them. If you were like 7, 8 and your opponent makes a big bet on the river you have to think about getting away. If you make a flush and the board pairs you may be beat. Still you should have called - in my opinion.

Another way to looks at combined odds is to formulate the null hypothesis. Not hit on turn and river you can multiply.
1 - ( (47-9)/(47) * (46-9)/46) ) = 34.9676%

OR in statistics gets kind of messy so one approach is to flop is to the null hyposis so it becomes an AND. With an AND you can multiply.

It comes out the same as the accepted answer
I actually thought accepted answer wrong until I tested it out

Using pot odds ratio versus % is different.

• Ratio
  Odds once you opponent bet 10 into the pot the pot was up to 72
  Your pot odds are 72:10
• %
  With % you need to add in you bet so your % pot odds is 10(62+10+10) = 12.2%
  You want pot % to be less than your card % so you were getting very good odds to call

Ratio to me is easier
Let's look at the ratio to make the flush on the turn alone
9 cards make the flush (13 - 2 (in hand) - 2 (on board))
38 cards don't make the flush (52 - 5 (exposed) - 9 (that make the flush))
Your hand odds are 38:9
Your pot odds are 72:10
72:10 > 48:9
You were getting proper odds just to call turn (let alone the river)
You should have called
Based on the turn alone it would have taken a bet of 20 to price you out

If you consider your odds with two cards to come it would take roughly a pot size bet to price you out. If opponent had gone all in with a pot size bet and you had an ace to the flush you should have called.

Since your bet becomes part of the pot it take more to price someone out than you might guess
Once your hand odds are 1:1 (or better) you cannot be priced out
(pot + bet) / bet is always > 1