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Casino Games - Red Dog

Author: Yui-Bin Chen, Ph.D.

Introduction

First the dealer shows two cards. Aces are high and deuces are low. You win if the third card falls between the first two cards.

• If the first two cards are consecutive, this is a tie.
• If the first two cards are not consecutive and are not a pair, then players can raise their bets up to the original bets and the dealer later shows the third card. If its value falls between the values of the first two cards, players win and the payout is determined by the following chart.
• If the first two cards are a pair, no raise is allowed and a third card is drawn immediately. If they make three-of-a-kind, players receive 11 times of their bets. Otherwise, it is a tie.
• Rules

The dealer flips 2 cards: X, Y (assuming X <= Y). Then, if necessary, he/she flips a third card (Z) after players' raise.

• If (X = Y), then
(a) no raise, and
(b) if (Z = X), then players win, else Tie.
• If (X+1 = Y), then Tie.
• If (X+1 < Y), then
if (X < Z < Y), then players win, else players lose.
• Payout Schedule

Raise Allowed Raise Not Allowed
Spread 1 2 3 4 - 11 Consecutive Pair 3-of-a-kind
Payout 5:1 4:1 2:1 Even Tie Tie 11:1

Strategy

In general, you should raise when the spread is greater than or equal to 7. A perfect strategy indicates that you should raise when

Probability(win) × Return(win) > 1.

Expected Gain/Loss

Assume this game is dealt from n decks (n >= 1). We consider four different cases and let Pr() and V() be the probability and winning of each case.

```(a) X = Y = Z :
Pr(a) = C(13,1)*C(4n,3)/C(52n,3)
V(a) = 11
(b) X = Y <> Z :
Pr(b) = C(13,1)*C(4n,2)*C(48n,1)
/ ( C(52n,3)*(3!/2!1!) )
V(b) = 0
(c) X + 1 = Y :
Pr(c) = C(12,1)*C(4n,1)*C(4n,1)/C(52n,2)
V(c) = 0
(d) X + 1 < Y : (let k = Y - X - 1)
Pr(d,k) = C(12-k,1)*C(4n,1)*C(4n,1)/C(52n,2)
Pr(d,k,win) = Pr(d,k)*C(4kn,1)/C(52n-2,1)
Pr(d,k,lose) = Pr(d,k)*C(52n-4kn-2,1)/C(52n-2,1)
V(d,1,win) = 5
V(d,2,win) = 4
V(d,3,win) = 2
V(d,4,win) = V(d,5,win) = V(d,6,win) = 1
V(d,7,win) = V(d,8,win) = V(d,9,win) = V(d,10,win) = V(d,11,win) = 2
(because of the raise bet)
V(d,i,lose) = 1, when i = 1,2,3,4,5,6
V(d,i,lose) = 2, when i = 7,8,9,10,11
```
The expected gain/loss per hand is
```  E(n) = Pr(a)*V(a)
+ sum(Pr(d,k,win)*V(d,k,win)
+ Pr(d,k,lose)*V(d,k,lose))
= (-456n^2-210n+143) / (13*(52n-1)*(26n-1)).
```
However, the expected bet size is higher than 1, because of the raise. The expected bet size should be
```  B(n) = 1 + Pr(d,7) + Pr(d,8) + Pr(d,9)
+ Pr(d,10) + Pr(d,11)
= (796n-13) / (13*(52n-1)).
```
Then the expected gain/loss per dollor becomes
```  E(n)/B(n) = (-456n^2-210n+143)/((26n-1)*(796n-13)).
```
Using these formulae, we know the house has a (-E(n)/B(n)) edge, where n is the number of decks used. We list the house edges for different situations below:
Number of Decks House Edge
1 2.6718 %
2 2.6090 %
4 2.4472 %
6 2.3749 %
8 2.3353 %
Infinite 2.2033 %

Counting Method

When a Duce or an Ace is removed from the shoe, players have a better expected return. When an extra 6, 7, 8, 9, or 10 is added to the shoe, players also have a better expected return. Therefore, after carefully examining the change of expected return, we propose the following counting method (called it "C" Counting System -- when you plot the Count-Card in 2-dimensional space, it looks like a "C"):
 Card Count 2 3 4 5 6 7 8 9 10 J Q K A +5 +1 0 -1 -2 -2 -2 -2 -2 -1 0 +1 +5

The true count is defined as

```                     Running count
True Count = ----------------------
Number of unused set
```
where a set is a quarter of deck (13 cards).

When the running count is positive, the game is more favorable to players. However, in order to get a positive edge on Red Dog game, the true count must be higher than +5. To support the theory, we show a computer simulation of 200 million hands.

Simulation Results

Total number of rounds = 200,000,000 (some data are ignored because of the lack of sampling points). It is simulated from an 8-deck game with 75% penetration.
True Count Number of Hands Total Bet Size Gain/Loss Percentage
-8 21411 26400 -1104 -4.182%
-7 72182 89084 -5764 -6.470%
-6 243010 298022 -12041 -4.040%
-5 713879 868765 -37757 -4.346%
-4 1994749 2412983 -93382 -3.870%
-3 5295081 6361955 -231228 -3.635%
-2 13886986 16573472 -523450 -3.158%
-1 37704603 44689349 -1247534 -2.792%
0 80092644 94344958 -2204379 -2.337%
1 37373857 43751974 -849609 -1.942%
2 14201195 16509125 -234536 -1.421%
3 5431668 6271514 -55569 -0.886%
4 2000302 2292841 -6190 -0.270%
5 679187 773411 +2885 +0.373%
6 205999 232899 2397 +1.029%
7 59378 66676 925 +1.387%
8 13309 14802 68 +0.459%

In the following table, we show the simulation results from 4-, 6-, and 8-deck games (after 200 million hands):

True Count Gain/Loss Percentage
4-Deck 6-Deck 8-Deck
-8 -5.278% -5.631% -4.182%
-7 -5.392% -4.707% -6.470%
-6 -4.569% -4.946% -4.040%
-5 -4.627% -4.396% -4.346%
-4 -4.050% -4.105% -3.870%
-3 -3.704% -3.657% -3.635%
-2 -3.255% -3.272% -3.158%
-1 -2.848% -2.827% -2.792%
0 -2.415% -2.409% -2.337%
1 -1.959% -1.912% -1.942%
2 -1.414% -1.407% -1.421%
3 -1.023% -0.995% -0.886%
4 -0.464% -0.477% -0.270%
5 +0.352% +0.316% +0.373%
6 +0.918% +0.712% +1.029%
7 +1.492% +1.600% +1.387%
8 +2.053% +1.469% +0.459%
Expected Return -2.4472% -2.3749% -2.3353%
Expected
Return
-2.4472% -2.3749% -2.3353%

It seems that this counting method can be used at all kinds of games, because when the true count reflects players advantage. However, in order to make a profit while consistantly playing. You have to do a 1-900 spreading. Or you can do the back-count and jump into the game when the true count is +5 or higher (1 out of 200 chance, 0.48%). Since each round uses only 2 or 3 cards, when the count reaches +5, it may stay high for a while. This gives an advantage to back counters. But, on the other hand, the chance of getting such a positive stream is far below 0.48%.

Shuffle tracking is also useful to determine whether the incoming rounds are favorable.