# TEXAS HOLD'EM LESSONS #2

## Backdoor Draw

After the *Flop*, when it is required for the winning combination that both on the *Thorn*
and on the *River* (fourth and fifth community cards) 2 necessary cards to come,
it is called **Backdoor Draw** or **Runner-Runner** in Texas Hold'em. For example,
you have 2 *hearts* on your hand, and another one appeared on the *Flop*. Then for a *Flush*
it's necessary
that both *Thorn* and *River* cards to be Hearts. It's called **Backdoor Flush**.
Generally strong poker players "do not walk" for the two cards, indeed probability to obtain the cards is very small.

In the given
example it's necessary that one of 10 remained hearts to come on the *Thorn* from the rest (unknown)
47 cards. Probability of this event is 10/47. If this occurs, then on the *River* it's necessary that the extra
heart be dealt, the chance of appearance of such card is 9 of 46. The overall probability of the arrival of two
hearts is equal to (10/47)*(9/46), e.g. approximately 4.2%, or 1 to 23. On the basis of this math,
the most correct solution for *Backdoor Flush* is almost always **Pass**. However,
in some situations, especially when there are chances of improving your hand to the strongest
one by the purchase of other cards, it makes sense to gamble. This solution is also depended on the ratio
of the pot size to the necessary bet, your position in the game, the number of active
players, their force and your knowledge of their tactics, and other factors. The following examples illustrate the
overall idea. The examples are taken from real
*Limit Holdem* games and, although they are not so academic as the specially devised studies,
they're sufficiently significant.

### Example 1

(bet limit $15-$30,*blinds*$10-$15). We have A spade 7 spade on large

**Blind**. Serious player sitting after us

*Raises*. We know that he attacks with a

*pair of Tens*or higher, and also with A-K or A-Q. Player in the middle position makes the same bet, just as the

*Button*. Rest of the players

*Pass*. We

*Call*. 4 players remains in the game, and $130 in the pot. The

*Flop*brings 9 spade 8 spade 2 spade, giving us backdoor of possibility of higher

*Flush*and

*Straight.*We say

*Check*, the player who attacked befor the Flop

*Bets*, and other two players

*Pass*. What should we do?

**The answer:** *Call.* The pot is already at $145, and the
Call will cost $15,
e.g., approximately 10% of the pot (ratio ~ 10 to 1). In almost 13% of cases
*Thorn* or *River* will bring A,
giving us a pair. Thus, taking into account the strength of our opponent's
cards when he attacks before the *Flop*, it's
possible to assume that he (opponent) will win with two A (one possible combination), A-K (8 combinations)
and A-Q (8 combinations), i.e., in 17 cases. But in 24 cases there can be
a pair of K, Q, J or 10.
So, the probability of our win due to appearance of A is 13%*24/(24+17)=7.6%.
If 7 comes on the Thorn or the *River* we win with 24 possible combinations of
our opponent A-K or A-Q, and we lose to any of the assumed pairs
(27 combinations), i.e., this gives 13%*24/(24+27)=6.1% more. And even more than
4% is a chance of *Backdoor Flesh*, plus almost 4% for *Backdoor Straight*.
Sure, this is the game, and there
can be a lot of different scenarios, but with a probability to win of more than 20%
the bet of 1/10 of the pot is completely justified.

### Example 2

(bet limit $30-$60, blinds $20-$30). On a large blind we have 9 clubs 8 clubs. Player to the left begins with*Raise*, two others

*Call*, small blind -

*Pass*. The attacked player, according to our information, usually attacks in the early position with a pair of Jacks or higher, or A-K and A-Q. We Call, there are $260 in the bank, and on the

*Flop*10 clubs 5 spades 2 diamonds appear. We've got

**, correspondingly, and also chances to get**

*Backdoor Straight Flush**Flush*or simple

*Straight*. We say

*Check*, next player

*Bets*, other two say

*Pass*. What should we do?

**The answer:** *Call.*
Let us make a number of assumptions. Our opponent have either one of 24 possible pairs or A with K or Q
(32 variants). If **8** or **9** cards come, the best hand will be
probably our hand. Also,
there is a probability that on the *Thorn* one of our opponent's
non-suitable cards will come and he will not increase his bet before the *River*, and there is
a real chance to see the fifth card for free. In this case we repeat the possibility to
get **8** or **9**. If on the *Thorn* **Q** (and our opponent
do not hold A-Q and not have a pair of Q) or **6** are dealt, then we have
*Straight* with the hole, and there is a free chance to stick this hole on
the *River*. Also it is worthwhile to consider that after Calling on
Thorn it is possible to purchase arranging clubs (10 cards), J (not clubs, it's already calculated -
3 cards), 7 (analogously - 3 cards), and above mentioned 8 or 9 (6 additional
cards). Altogether 22 of 47,
without considering "hole" chances.

If before the *River* we would have something potentially better than one Pair, we will attack
(Raise) themselves,
with the pair - simply Check and Call at the bet site of our opponent. We assume that the opponent approximately
in half of the cases will answer to our Bet because he have originally higher Pair
(after purchasing it or
suspecting our bluff). Being very pessimistic, let us assume also, that we will lose, i.e., let us
Fold, if nothing appears on the *Thorn*, and the opponent Raises. Nothing
means one of
25 remained unnecessary to us cards, and in this case our $30 for the answer will be
lost anyway.
Statistically this is $30*(25/47), i.e., approximately $16. But there is a probability of large troubles
- we buy necessary card on the Thorn, answer the Bet of our opponent and buy nothing on
the *River*. For *Flush* this is 10/47
(clubs on Thorn) * 31/46 (without the clubs and the paired to our cards), i.e., of about 14%.
For *Straight *-
6/47 (J or 7, non-clubs) * 32/46 (without *Straight and Pairs*) = 9%. Altogether in 23% more cases we
will lose $90 ($30 after the *Flop* and $60 after the *Thorn*) - statistically approximately $20.

Our chances to get *Flush, Straight, two Pairs* or *Three-of-a-Kind*
are a total of about 8%. If this occurs, we'll win $290, that are already in the bank, plus $60,
bet on *Thorn*, and in half of the cases, as we assumed,
$60 more on the *River*, which we will now consider as $30. Altogether $380*8% =
approx. $30. Besides
these combinations, with probability 22% we'll have one pair, which in 43% of cases will lose to
opponent,
and in 57% it'll win. (there is another chance, that we will purchase this Pair on Thorn, and
our opponent -
on the *River* and he win, but this can happen less than in 1% of cases, and it is not taken
into the account in this conditional calculation.) Respecting the opponent, we won't expect that
he will
answer our Bet after the *River* with combination A-K or A-Q, correspondingly, in 22%*57% of cases our
winnings
will compose $350, i.e., statistically $44. For us, having one Pair we'll have
to make extra bet after the *River*,
and we will lose $30 on the *Flop* + $60 on the *Thorn* and + $60 more on the *River*, i.e., $150, in 43% of cases. Result
of 22%*43%*$150=$14, i.e., with one pair on hands our average win will compose $44-$14=$30.

Summary:
-$16-$20+$30+$30=$24, i.e., with *Calling* we have positive math. expectation.

### Example 3

(limit $30-$60, blinds $20-$30). K spade 10 spade in the late position. Nobody enters the game, and attacking to blinds we make Raise. Player on the batton Raises, both blinds Pass, we Call. There are $230 in the pot and two players. Flop - 7 clubs 6 diamond 5 spade, we have Backdoor Flush and Highest Straight, and also chances to get higher Pair. We Check, the opponent Bets. What should we do next?**The answer:** *Pass.* The
relationship of the necessary Bet to the Pot is approximately 1 to 9, and 6
cards would allow us to get the higher Pair. But the problem is in the fact that it's absolutely unknown,
whether this pair, if it comes,
is actually higher, or aggressively attacked opponent has A-K with extra K or A -10
with 10 dealt on the Thorn. Or probably he already has highest Pair. 4% of chances to purchase two
spades
to Flush + 1.5% for 8 and 9 to Straight - that's not enough to gamble additional money. In this
case, the Bet on the Flop won't our last bet - if we get one of the suitable
cards on the Thorn, we will have to answer to our opponent's bet again, moreover from the earlier, with respect
to him, position.

### Example 4

(limit $10-$20, blinds $5-$10). In the early position on hands A diamond Q diamond, and we begin with*Raise*. The next player Raises, 2 additional players in the average position and large blind

*Call.*There are five players and $105 in the pot. The Flop with K spade 8 diamond 3 clubs let up hope for higher Flush or Pair. Large blind bets. What do we do?

**The answer:** *Pass.* The player that made a bet on the
first hand in the situation, when there were a mob of 4 players, who, moreover,
Raised before the Flop,
sure has a Pair, or even set K, especially with this many-colored and non-Straight
Flop. Only A
(and that's not for sure) can save us, or our two diamonds. Thus far the ratio of
the Bet to the Pot almost
1 to 12, and the probability of getting A is 1 to 15, but the main problem is that
we have three additional players after us,
who quietly Raised before the Flop, and it is completely possible that someone will increase
the bet now too.
And we'll have to pay again. Unjustifiably.

### Example 5

(limit $10-$20, blinds $5-$10). We are sitting on the large blind and playing free Call with 10 clubs 6 clubs. 5 players entered into the game, including the small blind, and formed a pot of $50. The Flop brought 9 spade 7 diamond 3 clubs, proposing Backdoor Flush and Straight with one hole. We Check, Bet-Call-Call-Pass. What should we do?**The answer:** *Pass.* There are $80 in the pot and we should put $10
to make a bet, i.e., 1 to 8. We can purchase 8 on the
Thorn with probability of 1 to 11, and our Backdoor Flush, even if we'll see our two
necessary clubs,
has a small chance to be the highest. Moreover, somebody of our three opponents can hold J -10
and with this 8 (on the Thorn) he'll obtain higher Straight, and this will be very
expensive for us.

It is evident from the given examples that in the situations, when it is necessary to purchase extra two cards, one should play very deliberately and prudently. However, this can be said about all aspects of the game of poker.

*Oleg Granovsky*

First published: Gambling Magazine "Casino Player", Russia, translated by SmartPlayers.net, © July 2005.

Texas Holdem Poker Lessons #3 ►

**Recommended Articles:**

1. What Do You Do When You Flop Second Pair? (20/06/2005)

2. Loosing With Pocket Kings: Why Does It Only Happen To You? (04/07/2005)

3. What Do You Do When There Are Too Many Pre-Flop Callers in Every Pot? (17/05/2005)