Street Dog

(SEE PAGE 2)

I will try to make this as short as possible :)

I go over friends house where he hosts a casino game and offers free beer...yum.

The casino game is a variation of Red Dog which we call Street Dog and it is a mix of Let It Ride, & Red Dog.

Rules:

2 Bets are posted in advance and the player has the option of pulling back 1 bet after the first 2 cards are dealt.

If the spread is zero (JT, KQ, TT, etc) both bets are immediatly moved to the Street (like a pass line in craps) where the bets are held captive until the player get a 3 of a kind which pushes the bets back to him.

Upon getting a 3 of a kind, a fourth card is dealt and, if the player gets 4 of a kind, he gets even money on all the money in the Street.

If the player does not play the next hand or leaves the table then all bets on the Street are forfeited.

1 Card Spread pays 9 to 1
2 or more Card Spread pays even money.
0 Card Spread money goes to the street.

Joker counts as 2 if it comes in the first two cards.

Joker counts as any card if it comes after the first 2 cards.

-The worst part of gambling with friends is that they like to leave as soon as they get up by a little bit.
-By capturing money on the street they are more apt to stay for the push, which they have to keep gambling in order to have a shot at.
-Getting paid even money on an ever growing stack size in the street (however bad the odds of getting 4 of a kind w/ 4 cards before getting bets pushed by a 3 of a kind) makes for an interesting jackpot.

We play it and our friends like it, but I would like to know if the edge is grossly in his favor so I can say that the Wizard thinks I should get comped 2 beers an hour ;)

I can attempt to analyze this, but I'm confused on how it is played. Maybe if you give a sample game or 2 (or more), I can understand it better. I will also need to know how many decks are used and if Aces are high or low or both.

Single Deck w/ 2 jokers, Ace is always high just as in Red Dog, deck is dealt 10 hands deep before reshuffle in our games.

$10 Game, 2 bets are posted for $20 total:

Hand 1: player is dealt a 5 and a 6, the spread is 0, dealer moves the $20 to the street.

Hand 2: Player is dealt an 8 and a 3, the spread is 4, player pulls back $10, player is dealt a Jack, Player loses $10.

Hand 3: Player is dealt QJ, the spread is 0, dealer moves $20 to the street, $40 total is in the street.

Hand 4: Player is dealt 77, the spread is 0, player is dealt an 8, dealer moves $20 to the street, $60 total in street.

Hand 5: Player is dealt a 2 and a jack, the spread is 8, player is dealt a 6, player wins $20.

Hand 6: player is dealt 97, the spread is 1, player is dealt a joker, player wins $200, player may want to leave at this point but sees his $60 in the street and continues to play.

Hand 7: Player is dealt T2, the spread is 7, player is dealt an ace, player loses $20.

Hand 8: Player is dealt T6, the spread is 3, player pulls back $10, Player is dealt a ten and loses $10.

Hand 9: Player is dealt JT, the spread is 0, dealer moves $20 to the street, $80 total is in the street.

Hand 10: Player is dealt 65, the spread is 0, dealer moves $20 to the street, $100 total is in the street.

Hand 11: player is dealt TT, the spread is 0, player is dealt a ten, dealer turns over a fourth card revealing an ace, $100 on the street + $20 in bets is pushed back to the player, $0 is in the street.

Had a joker or a ten come on the 4th card, player would win even money on everything including the street and total dollars in the street would be $0.

Hand 11 (another way): player is dealt TT, the spread is 0, player is dealt an Ace, dealer moves $20 to the street, $120 total is in the street.

Player tells dealer that he wishes to leave, dealer yells "money in the street, finders keepers!", player forfeits $120.

Player was up $180 and left $120 in the street, player walks away up $60.


I have played casino games where pushes go to the dealer and it is a total turn off to play those games.

At least when the money is stacked in front of me on the street I can get eventually get it back if I continued playing...and didn't run out of money.

Most players will buy in short and run out of money before saving those bets though; bring more money next time ;)

Ok I think I get it.

Quote:

2 Card Spread pays even money

should be

Quote:

2 or more Card Spread pays even money

What happends if a joker is one of the first two cards?

Sorry miplet, my English tends to go flying out the window when I am typing ;)

2 or more Card Spread does pay even money.

Joker is a 2 if it comes out in the first 2 cards.

thanks.

As we play it:

Joker, 2, 2 would make 3 of a kind.

Joker, 2, 2, 2 would make 4 of a kind.

Joker, 7, 7 is a loss.

Joker, 4, is a 1 card spread paying 10 to 1 when a 2 or joker lands.

Quote: MightyOne

Sorry miplet, my English tends to go flying out the window when I am typing ;)

2 or more Card Spread does pay even money.

Joker is a 2 if it comes out in the first 2 cards.

thanks.

Ok I'll attempt to analyze this in my free time. Might take a week.

Quote: MightyOne

As we play it:

Joker, 2, 2 would make 3 of a kind.

Joker, 2, 2, 2 would make 4 of a kind.

Joker, 7, 7 is a loss.

Joker, ten, Joker is a loss.

Joker, 3, is a 1 card spread paying 10 to 1 when a 2 lands.

Joker, 2 Joker =3 of a kind?
7 7 Joker = 3 of a kind?
A joker is a 2 as a 3rd (and 4th) card ,if it doesnt make a 3 (Or 4) of a kind?

If a joker comes in the first 2 cards it counts as a 2 (as if there were six 2's in the deck).

Hand 1: First two cards are a joker and a 2, the hand is 22. The third card is a joker making 3 of a kind, dealer turns over a 4th card and checks for 4 of a kind (two jokers and one 2 was used leaving three 2's in the deck to complete 4 of a kind)

Hand 2: First two cards are 77, the dealer turns over a joker, the hand is 777, dealer turns over a 4th card and checks for 4 of a kind (two sevens and one joker were used leaving two 7's and one Joker to complete 4 of a kind).

Hand 3: First two cards are a joker and a seven, the hand is 72, dealer turns over a 7, player loses.

Thank you for your time miplet.

Quote:

Joker, ten, Joker is a loss.

You have this as a loss, but isn't it a wn? Joker (2), 10 ,Joker(any card in the middle)

typo sorry,

Jokers confuse my brain.

Joker, ten, joker is a win.

Joker & ten is T2 & a 3rd card joker is any card

I wish I would proof read before posting...

Here is the recap:

Single Deck w/ 2 jokers, Ace is High as in Red Dog.

A joker is a 2 if dealt in the first 2 cards but is any card if dealt as a 3rd (or 4th) card.

A 1 card spread pays 9 to 1.
A 2+ card spread pays even money.
A 0 card spread immediately moves both player bets to the street.

Failure to play next hand results in forfeiting the chips in the street.

Player posts 2 bets before cards are dealt.

If there is a spread, player has the option of pulling back 1 bet after the first 2 cards are dealt.

Upon being dealt trips, bets are pushed back to the player including the chips on the street.

If a player is dealt quads then even money is payed on the money in the street and player bets.

Joker, 4 is the lowest 1 card spread and AQ is the highest.

The Deal:

http://i47.tinypic.com/2uoszd3.jpg

1. Players post $20 (2 bets)

2. Two cards are dealt face up and the jackpot card is dealt face down.

3. Players pull a bet back or let it ride

4. If there is a spread then a card is dealt to the gap.

5. If there is a pair then the jackpot card is turned over.

6. Bets are settled.

I've done a partial analysis. Basic strategy is to pull back spreads of 2 to 5. Spreads 6 have no player or dealer advantage if a joker is not one of the first 2 cards, but a dealer advantge if a joker is used as the first 2 cards.
Here is a table of player ev for the first 2 cards. I'll fill in the blanks when I figure them out. Note: These evs are based on how much is bet after 2 cards are delt. I should redo them based on before 2 cards are delt. Ok I may have an error in here. I'll double check later.

	Joker	2	3	4	5	6	7	8	9	10	j	q	k	a
Joker				0.05769	-0.65385	-0.5	-0.34615	-0.19231	-0.03846	0.11538	0.26923	0.42308	0.57692	0.73077
2				0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0	0.15385	0.30769	0.46154	0.61538	0.76923
3					0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0	0.15385	0.30769	0.46154	0.61538
4	0.05769	0.26923				0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0	0.15385	0.30769	0.46154
5	-0.65385	-0.61538	0.26923				0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0	0.15385	0.30769
6	-0.5	-0.46154	-0.61538	0.26923				0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0	0.15385
7	-0.34615	-0.30769	-0.46154	-0.61538	0.26923				0.26923	-0.61538	-0.46154	-0.30769	-0.15385	0
8	-0.19231	-0.15385	-0.30769	-0.46154	-0.61538	0.26923				0.26923	-0.61538	-0.46154	-0.30769	-0.15385
9	-0.03846	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923				0.26923	-0.61538	-0.46154	-0.30769
10	0.11538	0.15385	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923				0.26923	-0.61538	-0.46154
j	0.26923	0.30769	0.15385	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923				0.26923	-0.61538
q	0.42308	0.46154	0.30769	0.15385	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923				0.26923
k	0.57692	0.61538	0.46154	0.30769	0.15385	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923
a	0.73077	0.76923	0.61538	0.46154	0.30769	0.15385	0	-0.15385	-0.30769	-0.46154	-0.61538	0.26923

In a $10 game, the bet will always be $20 before cards are dealt.

If there is a spread then there is the option of reducing this bet to $10.

If there is no spread then both bets ($20) are moved to the street (I got hit with 3 of these in a row today..gosh!)

I was trying to play this strategy:

Pull $10 if <7 spread but leave both bets up on a spread of 1 for the possible 10 to 1 payoff.

Based on what you show here I will start playing a non-joker spread 6.

I couldn't catch a 3 of a kind today to save my life but the $200 spread 1 payoff leveled things a bit.




Thank you for your help miplet :o)

PS: I've added all the positive numbers & subtracted the negative numbers for an EV of -0.06518 + any positive EV that I can get out of an even money jackpot on the street.

Hopefully I did this right :)

I calculated the odds of making a:

3 of a Kind: 221 to 1

52/52 * 3/51 * 4/52

4 of a Kind: 3756 to 1

52/52 * 3/51 * 4/52 * 3/51

Would the odds of getting a 4 of a Kind before a 3 of a Kind be calculated:

3756/221 * 3756 = 63,835 to 1?

Of course I didn't factor in these combinations:

3 of a Kind (138 to 1)

Joker, Joker, 2

Joker, 2, 2

4 of a Kind (2342 to 1)

Joker, Joker, 2, 2

Joker, 2, 2, 2

Stake Size:

I didn't realize that I needed a few thousand dollars just to beat the street, I was just donking it out :o)

Hope I don't have any brainos or types. Here is my final analysis. I get a player ev of -0.35982705720035 assuming no money in the street. For example you make 2 bets of $10 to start($20 total), you would expect to lose about $3.60
Money in the street increase the player ev by 0.004951762998378 per bet. You will have a positive ev when there are 73 bets in the street. ($730 if you are betting 2 bets of $10.)

Hand	combos	% combos	pays	ev
street	1486752	0.19588	-2	-0.39176
lose 1	2249672	0.2964	-1	-0.2964
lose 2	1258312	0.16578	-2	-0.33157
win 1	932688	0.12288	1	0.12288
win 2	1015920	0.13385	2	0.2677
0 ev win	254592	0.03354	2	0.06709
0 ev lose	254592	0.03354	-2	-0.06709
win 20	102000	0.01344	20	0.26877
trips	33408	0.0044	0	0
quads	2088	0.00028	2	0.00055
Total	7590024	1	0	-0.35983

This analysis looks good for a single bet... but what if a player was to sit down with a sufficiently high bankroll and never leave with money on the street. That is, they would play until they freed the street money. Can we sill consider the street line as -2 EV? I am leaning towards no, but haven't thought it all the way through yet.

What I'm wondering about is the expectation of the long-term game. Simulation may be necessary, considering the sheer number of possible states to be in ($ in street). However, a Markov analysis might work...

Here's what I'm thinking: you have calculated the expectation of being in state 0 (where there are no bets on the street). You also have a simple formula (-.3598 + n * 0.00495) for the expectation of being in state n (n bets on the street). Now we need the long term steady state probabilities of being in any given state n. Multiply probs by EVs to get the long term expectation of the game.

If a player stays for the push then he can expect an EV of +0.03193?

If the push happens in 221 hands then the player would have invested $860 in the street (220 * 0.19588 * $20).

Some things to point out are:

Unlike Black Jack, you cannot leave during a bad streak as that would result in forfeiture of your street bets and a -36% edge.

The positive EV of $0.32 per $10 bet is easily offset by the negative EV of $3.60 from one person leaving money on the street.

If you buy in short then you are dead.

Unlike Black Jack you cannot exploit an edge as increasing your bets 10 fold will only result in adding another zero to the amount of money you need to survive.

Despite your heavy investment, it is vastly more likely that you will get a push than a jackpot.

I think that this might be a theoretically positive but realistically negative game :o)

I am frequently hit with a series of zero spreads which sucks my money into the street like a giant vortex and I am left reaching for my wallet.

I still think that it is a fun game.


Again, thank you for your generosity miplet.

Quote: miplet

Hope I don't have any brainos or types. Here is my final analysis. I get a player ev of -0.35982705720035 assuming no money in the street. For example you make 2 bets of $10 to start($20 total), you would expect to lose about $3.60
Money in the street increase the player ev by 0.004951762998378 per bet. You will have a positive ev when there are 73 bets in the street. ($730 if you are betting 2 bets of $10.)

Hand	combos	% combos	pays	ev
street	1486752	0.19588	-2	-0.39176
lose 1	2249672	0.2964	-1	-0.2964
lose 2	1258312	0.16578	-2	-0.33157
win 1	932688	0.12288	1	0.12288
win 2	1015920	0.13385	2	0.2677
0 ev win	254592	0.03354	2	0.06709
0 ev lose	254592	0.03354	-2	-0.06709
win 20	102000	0.01344	20	0.26877
trips	33408	0.0044	0	0
quads	2088	0.00028	2	0.00055
Total	7590024	1	0	-0.35983

Hand	combos	% combos	pays	ev
street	1486752	0.19588	-2	-0.39176
lose 1	2288024	0.301452	-1	-0.301452
lose 2	1219960	0.160728	-2	-0.321456
win 1	932688	0.12288	1	0.12288
win 2	1015920	0.13385	2	0.2677
0 ev win	254592	0.03354	2	0.06709
0 ev lose	254592	0.03354	-2	-0.06709
win 18	97920	0.012902	18	0.232236
win 9	4080	0.000537	9	0.004833
trips	33408	0.0044	0	0
quads	2088	0.00028	2	0.00055
Total	7590024	1	0	-0.38647
Stay/Push				+0.00529

After showing miplet's data to my friend we are going to take turns dealing a slightly different game.

The change: Pay 9 to 1 instead of 10 to 1 on spread 1 bets and change from Red Dog style "one and done" dealing to dealing a card for each player (discussed in my next post).

New method of dealing accomplishes:

1. There are player wins and player losses instead of just one or the other.
2. Not all of the players will get a push or jackpot at the same time.
3. Opens seats and keeps table full thus attracting new players (I always think from a casino POV).


Player gains 0.99% every time $20 is moved to the street.
Positive EV after 40 bets ($800) is moved to the street (could ramp up this number to $1,200 by dealing a $15 game).

WIN 18 is still a positive expectation play (8.9999 to 1 & pays 9 to 1) and encourages double bets on spread one

Staying for the push results in an EV of +0.00529

Leaving money in the street results in an EV of -0.38647.

Positive EV is more easily offset by the majority who do not have proper bankrolls.

A positive EV game is the only way you are going to get a recreational player to sit down at a game with a 38.6% house edge.

The game is pretty much a coin toss now with a giant vortex attacking insufficient bankrolls.

If you are paying 9 to 1 on 1 spreads, then the ev of Joker 4 is -0.038461538461538 so you would take back one of your 2 bets. Changes 4080 of win 18 to 4080 win 9, 38352 lose 2 to lose 1.

Quote: miplet

If you are paying 9 to 1 on 1 spreads, then the ev of Joker 4 is -0.038461538461538 so you would take back one of your 2 bets. Changes 4080 of win 18 to 4080 win 9, 38352 lose 2 to lose 1.

Thanks miplet, I edited the above post to reflect this information.

It seems that this rule change increases the players EV by 0.000211 giving them a "stay until push" EV of +0.005291.

Basic strategy is to play 2 bets on XX or worse:

T3, J4, Q5, K6, A7, non-joker spread 1's, & A-T + Joker

Suggested hand dealing:

http://i47.tinypic.com/2uoszd3.jpg

1. Players post $20 (2 bets)

2. Two cards are dealt face up and the jackpot card is dealt face down.

3. Players pull a bet back or let it ride

4. If there is a spread then a card is dealt to the gap.

5. If there is a pair then the jackpot card is turned over.

6. Bets are settled.

With a hand shuffled deck there is more card clumping. To my dismay I dealt three spread 1 wins in under 10 hands because I didn't shuffle well enough (they were all 4,6,5). That is when I got the idea of dealing the jackpot card face down so that if the cards were clumped then the 5 would be the jackpot card and some other card (hopefully losing) will be in the middle.

I think it would increase the EV for the players if I dealt a card in the gap for each player.

Here is an idea to make the game more interesting.

The deck is cut into 3 parts and players pick which part of the deck to get a hit from.

Cards are dealt in this order:


http://i50.tinypic.com/2qb7mtc.jpg


This further breaks up any card clumping in hand shuffled games.

or you could do this:

http://i45.tinypic.com/2s16o1z.jpg