One thing about the freerolls on Full Tilt is that I’m starting to become immune to the bad beats. Just for that alone, I loves me some freerolls. I was in a SnG short stacked and pushed with 44. I had one caller who was very loose and aggressive, and I hoped he would call. We turned over our cards and we had ourselves a race, 4♠4♥ vs A♦T♦. The flop was music to my ears, 2♠6♦4♣. The turn brought X♦ and the river gave the LAG a flush, bouncing me in 7th.

So that got me started on the odds of making various hands before or on the river. Yes these topics are covered many places, but I am doing this for two reasons. 1) It’s not covered on MattahFahtu and 2) I understand things better after I write them down. So this is for me, damnit!

The first installment is the Flush. Fitting, no? :)

Q. You hold any two cards of the same suit. What are the odds that you flop a flush?

A. Out of the 52 cards in the deck, there are 13 cards in each suit and we know about 2 of them, so there are 11 left. There are 50 unknown cards left that have 117,600 (50 * 49 * 48 ) different combinations to generate the flop. To make our flush there are 990 ( 11 * 10 * 9 ) possibilities, giving us a 0.84% ( 990 / 117,600 ) chance that we will flop a flush.

Q. You hold any two cards of the same suit, and one of your suit hits on the flop. What are the odds that you will complete the flush by the river?

A. There are 13 cards in each suit we know about 3 of them, so there are 10 left. One must hit on the turn and one on the river. So there are 90 different combinations of the same suit falling (10 * 9), now the inverse of that is there there are a total of 5 known cards (you hold two of then, and the board shows three of them), so there are a total of 47 unknown cards left. The possible number of combinations of the turn and river are 2,162 (47 * 46) and of those 2,162 only 90 of them will help you. So the odds are 90 / 2,192, or 4.16%, that the perfect two cards will fall. Now if we are in an all-in situation and your opponent(s) do not hold any cards of your suit, the odds increase (duh).

Q. You hold any two cards of the same suit, and two of your suit hits on the flop. What are the odds that you will complete the flush by the river?

A. There are 13 cards in each suit we know about 4 of them, so there are 9 left. One must hit on either the turn or the river. We have a 9 outs twice. Phil Gordon has a nice rule about this one, the rule of 4 and 2. Basically on the flop you multiply your total outs by 4 for the percentage of hitting your out by the river. On the turn, you multiply your outs by 2 for the percentage of hitting your out on the river. So we have 9 * 4, or roughly 36% that one of your suit will fall on the turn or the river and 9 * 2, or roughly 18% that the card will fall on the river. You will see why Phil’s rule of 4 and 2 is really good to use.

Since we only need one card to fall on either the turn or the river, we have 9 cards to help at each round out of a total of 47, so the odds that the card will fall on the turn is 9 / 47, or 19.15%. As for the river, there is one less card to hurt us, to the the fraction increases to 9 / 46, or 19.57%. The 19.57% is close to Phil’s 18%, one reason why Phil’s Quickie Math works.

Those are the odds that one of our suit will fall at either the turn or at the river, so if we wanted to find the odds that the card will fall after the flop and by the river after the flop falls as we have 9 outs twice. From the paragraph above, we have 9 / 47 for the turn and 9 / 46 for the river. That is the total number possibilities that the cards can fall that will help us. (9 / 47) + ( 9 / 46) or 38.71%, which also includes hitting our cards on both streets. Again, really close with Phil’s quickie math for 36%.

We can also do the inverse with the following equation, which is generic enough to figure out odds for any hand:
Flop to River % = 1 - ( ((47 - Outs) / 47) * ((46 - Outs) / 46) )

( ((47 - Outs) / 47) * ((46 - Outs) / 46) ) = the odds that we do not hit our card on the turn or river

Take 1 away from that and it’s the odds that we will hit our card on either street, not both.

So the math is now:
1 - ( ((47 - 9) / 47) * ((46 - 9) / 46) )
1 - ( (38 / 47) * ( 37 / 46) )
1 - ( .809 * .804 )
1 - .65
.35
35%

35% percent is even closer to Phil’s Rule at 36%. Now add in the 4.12 % that our suit hits both streets and you have 39.12%, which is close to 38.71% that we had above. If we round the numbers to the nearest integer, we get 39% in both cases.

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