Poker Digest Vol. 2, No. 26, December 17 - 30, 1999
Another realm in which elementary expectation can help us is in finding an optimum strategy. A game which illustrates this very well is Let It Ride. The game is straightforward and involves few decisions.
The player places three equal bets and is then dealt three cards. The dealer has two downcards, and before the dealer turns over any of her cards, the player has the option of either withdrawing his first bet or letting it ride. After each player has made the decision about the first bet, the dealer exposes one of her cards at which time the player now has the option of withdrawing his second bet or letting it ride. After each player makes this decision, the dealer exposes her other card. The player is then paid for each remaining bet if he has a winning hand, or loses all remaining bets if he does not have a winning hand.
Winning hands are based on the player's best poker hand has using his three cards together with the dealer's two cards. The payout is 1-to-1 for a pair of 10s or better, 2-to-1 for two pair, 3-to-1 for three-of-a-kind, 5-to-1 for a straight, 8-to-1 for a flush, 11-to-1 for a full house, 50-to-1 for four-of-a-kind, 200-to-1 for a straight flush, and 1,000-to-1 for a royal flush.
Recently I had a discussion with a Let It Ride player who had a misconception about determining an optimum strategy for the game and I want to mention it here so no readers have the same misconception. He believed that even though he might have negative expectation based on the three cards he starts with, the first card the dealer turns over might produce such a large positive expectation that it more than compensates the initial negative expectation. Therefore, he thought it might be better to let the first bet ride in certain negative expectation situations. He then wanted to perform some complicated analysis to determine whether or not such situations could arise.
He is looking at the game in the wrong way. The proper way to look at the game is as follows. The three bets are completely independent of each other. The player has no option with respect to the third bet in that this bet must stay until the end, so it has no effect on determining an optimum strategy. The first bet can be withdrawn only before the dealer has exposed any cards. If the player lets the first bet ride, it then stays until the end. Thus, the analysis of whether or not the first bet should be withdrawn depends only on the expectation of the three cards the player starts with.
The value of the first bet changes, sometimes dramatically, after the dealer exposes one card, and what the player is seeing is a branching conditional probability, but this already is factored into the expectation calculation performed before any cards were exposed. This means the decision on what to do with the first bet depends only on the expectation arising from the player's three cards.
The second bet is easier in that it is withdrawn if the expectation is negative after one card is exposed. Therefore, the optimum strategy is very simple: If the player has negative expectation at a particular stage of the hand, then withdraw the appropriate bet; if the player has positive expectation at that stage, let the bet ride (thereby contributing to the name of the game).
The preceding optimum strategy is easily stated but as stated is uninformative for most people because it is not particularly easy to calculate quickly the expectation for a given hand. The implication is that in order for a player to play an optimum strategy, he must know the situations in which the expectation is positive. Fortunately, there are few such situations so they are easy to remember.
As an example, let's look at a situation many players misplay. Suppose you look at your three cards and find the deuce, three and seven of clubs. I've seen many players in this kind of situation let the first bet ride looking to see if the dealer turns over a club for her first card. We shall see this is incorrect.
Altogether there are C(49,2) = 1,176 two-card hands for the dealer. If the player lets the first bet ride, in terms of calculating the expectation, it does not matter in which order the dealer's two cards are turned since this bet must stay until the end. Let's count how many ways the player wins.
The only possible winning hands in this case are a pair of 10s or better, two pair, three-of-a-kind or a flush. First, the only way a pair of 10s or better may occur is for the dealer to hold the pair. There are five possible ranks and six pairs of each rank giving us 30 possible ways of getting a pair of 10s or better. The only way the player can finish with two pair is for the dealer's two cards to form pairs with two of the player's cards. There are three choices for the ranks to pair and three choices for each of the pairing cards. This gives 27 ways to achieve two pair. Similarly, there are nine ways for the player to end up with three-of-a-kind.
C(10,2) = 45 ways for the dealer to have two clubs which
gives the player a flush. Altogether this gives us 111 dealer hands
which produce a win for the player so that the remaining 1,065 dealer
hands result in a loss for the player. Since each of the numbers just
determined divided by 1,176 is the probability of that particular
outcome occurring, in order to calculate the expectation of letting the
first bet ride, we multiply the numbers by the respective payoffs, add
them and divide by 1,176. This produces
Now suppose the player has three suited cards which allow two ways to catch a straight flush. For example, the deuce, three and five of clubs allow two ways of catching a straight flush. The numbers of pairs, two pair, three-of-a-kinds and flushes which may arise is the same as above, but two of the possible flushes are straight flushes and now the player also can catch straights.
Let's see if these additional possible wins are
enough to make the expectation of keeping the first bet positive. The
only new number we have to calculate is the number of ways of making
straights. There are two combinations of ranks yielding straights, and
for each combination there are 16 possible straights. This gives us
32 straights, but two of them are straight flushes which we count
separately. Similarly, of the 45 flushes, two are straight flushes.
The calculation for the expectation of letting the first bet ride now is
Again keeping the first bet for this player hand has negative expectation, but it isn't all that negative. It also shows us clearly that as soon as you add one more way to win with a beginning hand like the above, the expectation becomes positive. When is the latter the case? If the three suited cards allow two straight flushes and at least one of the cards has rank 10 or more, then let the first bet ride. If the three suited cards allow three straight flushes, that is, they have successive ranks and the smallest rank is in the range three through 10, then let the first bet ride. If all three cards have rank 10 or more, then let the bet ride. In the same way, it can be shown that three suited cards allowing one straight flush and having two cards of rank 10 or more have positive expectation for the first bet.
The best possible player hand which is not already a winner and neither
suited nor has a pair is 10-J-Q. (A little reflection on this hand will
convince you.) Thus, if this hand has negative expectation for the first
bet, then so does any other unsuited hand without a pair. The player can
win with this hand if it develops into either a big pair, two pair,
three-of-a-kind, or a straight. If you count in the same way we have
done above, you will find there are 372 dealer hands giving the player
a single big pair, there are 27 dealer hands giving the
player two pair; there are nine dealer hands giving the player
three-of-a-kind; and there are 48 dealer hands giving the
player a straight. This yields
A hand which contains a small pair (nines or smaller) leads to a winner
only if the player ends up with either two pair, three-of-a-kind, a
full house, or four-of-a-kind. Going through the same kind of counting
done above, the expectation calculation ends up looking like
The analysis for whether or not the player should withdraw his second bet is easier because there's only one card left to expose. If the player has any four suited cards, there are nine cards giving him at least a flush and only 39 which do not. Since a flush pays 8-to-1, this by itself already makes the expectation positive and other potential winning cards make the situation even better.
If the player has non-suited cards of four consecutive ranks and it is open at either end, then any of eight cards will give him a straight while 40 cards will not. Since a straight pays 5-to-1, the player's expectation is precisely zero. This means if he has any additional way of winning--such as at least one card of rank 10 or higher--he has positive expectation of winning and should let the second bet ride. If a player has an unsuited J-Q-K-A, then his expectation is again zero.
Another aspect of the game now becomes important. Let It Ride is a sociable game and some people like it for that reason. Part of the sociability is the fact that players sometimes show each other their cards. Thus, if a player has an open-ended straight draw and knows one of his neighbors has none of the cards he needs, he now has positive expectation and should let his bet ride. On the other hand, if his neighbor has one or more of the cards he needs, he should withdraw his second bet. Some cardrooms discourage this kind of information exchange, but a wise player will take advantage of it whenever possible.
Let us now summarize the optimum strategy for Let It Ride.
Let the first bet ride in the following situations and no others:
Let the second bet ride in the following situations and no others: