Let It Ride Alternative Paytable

Hello, everyone!

I have been working hard on my new strategy website and I have been trying to find pay tables not covered by the Wizard of Odds website. I found a youtube video (lost it lol) from really recently that displays an interesting pay table for Let It Ride.

Royal Flush 500 to 1
Straight Flush 100 to 1
Four of a Kind 40 to 1
Full House 10 to 1
Flush 6 to 1
Straight 4 to 1
Three of a Kind 3 to 1
Two Pair 2 to 1
Tens or Better 1 to 1

According to the ordinary basic strategy for this game, the house edge should be 6.47%... but what if we need a slightly different basic strategy?

My question is: has anyone has tried to find basic strategy for alternate Let It Ride pay tables?

Thanks!

Quote: harris

Hello, everyone!

I have been working hard on my new strategy website and I have been trying to find pay tables not covered by the Wizard of Odds website. I found a youtube video (lost it lol) from really recently that displays an interesting pay table for Let It Ride.

Royal Flush 500 to 1
Straight Flush 100 to 1
Four of a Kind 40 to 1
Full House 10 to 1
Flush 6 to 1
Straight 4 to 1
Three of a Kind 3 to 1
Two Pair 2 to 1
Tens or Better 1 to 1

According to the ordinary basic strategy for this game, the house edge should be 6.47%... but what if we need a slightly different basic strategy?

My question is: has anyone has tried to find basic strategy for alternate Let It Ride pay tables?

Thanks!
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Your 6.47% is right, and a slightly different strategy does help, but not by much. I ran it by enumerating all 2,598,960 hands. The same code gives the usual 3.51% on the standard table, so I’m happy with the method. I don’t know of a published strategy for this particular pay table.

Per unit of one starting bet, standard strategy on this table gives a house edge of 6.4702%. Optimal strategy gives 6.3332%. So you save 0.137 percentage points, or about 1.4 cents a hand at $10 a spot. Element of risk goes from 5.25% to 5.19%.

Optimal strategy on 3 cards: ride any pair of 10s or better, any three of a kind, and three suited cards where the number of high cards is greater than the number of gaps. A high card is a 10 or better. Gaps are the missing ranks between the cards, so J-10-8 suited has one gap and two high cards, which is a ride. Aces only count high here, so A-2-3 suited is not a ride.

On 4 cards: ride any made paying hand, any four to a flush, and four consecutive cards from J-10-9-8 up to K-Q-J-10.

Compared with standard strategy, on 3 cards you stop riding these suited hands: 3-4-5 through 7-8-9, then 7-8-10, 7-9-10 and 8-9-J, then 7-10-J, 8-10-Q, 8-J-Q, 9-10-K, 9-J-K and 9-Q-K.

On 4 cards, assuming no flush draw, you stop riding 7-8-9-10, 2-3-4-5 through 6-7-8-9, and the four-high-card hands that need one specific rank for a straight (A-K-Q-J, A-K-Q-10, A-K-J-10 and A-Q-J-10).

Most of the saving comes from those last two 4-card groups. They’re exact ties on the standard table, with an EV of 0, so riding them costs nothing there and strategy cards tend to include them. On this table they lose 1/6 and 1/12 of a unit respectively. Your 6.47% assumes you ride them, and just folding those ties gets you to 6.36%.

The lower flush and straight payouts drive all of this. For example, 7-8-9-10 goes from +1/8 to -1/24 of a unit: (8×4 + 3×1 - 37×1) / 48.

The same method works for any pay table. For each possible 3-card and 4-card holding, count the final hands in each category and ride if the EV is positive. The two decisions are independent, so you can solve them separately. There are only about 22,100 and 270,725 holdings, and it takes about a minute even in plain Python.