Blackjack promotion
My local casino is offering a blackjack promotion next Wednesday, $50 maximum bet. They are placing 4 jokers in a 6 deck shoe. If you get a joker as one of your first 2 dealt cards, you get your bet paid back extra as a bonus, then you play the hand as normal with your original bet. If you get 2 jokers, each as one of your first 2 cards, then you get 2X your original bet back. Split is considered 2 normal hands, with joker bet bonus available on first 2 cards of each. Bonus available on double down THIRD card but you only get your initial bet paid back (not the double). All other times when joker appears e.g. as your third card, to dealer it is burnt. Can I get an estimate of house / player edge, over and above basic strategy, on this game please. I believe that this game gives the player an advantage off the top, but don't have an exact number
Quote: sorcererofoddsHi,
My local casino is offering a blackjack promotion next Wednesday, $50 maximum bet. They are placing 4 jokers in a 6 deck shoe. If you get a joker as one of your first 2 dealt cards, you get your bet paid back extra as a bonus, then you play the hand as normal with your original bet. If you get 2 jokers, each as one of your first 2 cards, then you get 2X your original bet back. Split is considered 2 normal hands, with joker bet bonus available on first 2 cards of each. Bonus available on double down THIRD card but you only get your initial bet paid back (not the double). All other times when joker appears e.g. as your third card, to dealer it is burnt. Can I get an estimate of house / player edge, over and above basic strategy, on this game please. I believe that this game gives the player an advantage off the top, but don't have an exact number
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sorcererofodds,
I see this is your first post, so Welcome to the Wizard of Vegas board!
Your post left some questions (If you get a joker and receive the bonus, is the joker replaced with another card so you can play your hand? What is the normal house edge for the game?), so I can give just a rough estimate of the edge.
Adding 4 jokers to a 6D shoe brings the total to 316 cards, so the probability of any random card being a joker is 4/316. That means the probability of getting a joker in your first two cards is (a bit less than but really close to) 2*4/326, or 8/316, or about 1 time per 40 rounds. This alone will give you a 2.5% edge over the usual house edge for the game. If we assume that house edge is about 0.5% (so NOT a 6:5 table!), then you will have a healthy 2% edge over the house.
I know you have additional chances for the bonus (3rd card on DD; second card on split hands; double bonus for joker-joker) so your edge will be slightly more than 2% (again, for a NON-6:5 game), so I recommend that as long as you have the bankroll, just flat bet $50 on as many spots as you can.
And if you notice all 4 jokers have been used, maybe take a bathroom break until the new shoe ;-)
Hope this helps!
Dog Hand
We can just assume in a half of the 2-card joker-containing hands, player does not have an edge because dealer uses up the other half. This will drop the player edge to 1%. This is a little tricky. I'm not sure if this is correct, but one thing is for sure, player needs to use up as many jokers as possible.
Quote: acesideWhen dealer gets a joker or player uses up a joker as a hit card, the joker is burnt. This just means the player's edge depends strongly on the number of available jokers in the remaining cards.
We can just assume in a half of the 2-card joker-containing hands, player does not have an edge because dealer uses up the other half. This will drop the player edge to 1%. This is a little tricky. I'm not sure if this is correct, but one thing is for sure, player needs to use up as many jokers as possible.
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aceside,
Considering only the player and dealer, I agree with your implied assertion that, for 2-card, joker-containing hands, half of them will go to the dealer and half to the player. This is similar to saying that, on average, the dealer and the player each receive one-half of the BJs.
However, I do not understand why you claim, "This will drop the player edge to 1%." The calculation I showed is based on the player receiving a joker in his first two cards approximately once in forty rounds. Sure, we can also say the same for the dealer, but that does not invalidate the calculation: in the long run, the player will receive an addition 2.5% from this promo.
Dog Hand
There are some other pair splits that may be optimum as the remaining jokers start to become a higher and higher fraction of the shoe. I haven't really surveyed all the possible splits. 66vs7 may be one. Maybe 99vs7 and 99 vs A? 44vs4?
Let me do a rough calculation first. Player getting a joker probability is 12x4/312=15%. This means the players edge gain from this promotion is 15%. This is crazy! Is this correct?
I assume the promo bonus is only paid on any one hand that has the joker. You would not get paid for all six hands, so no change to EV. Maybe I don't understand the full rules.Quote: acesideIf the player can play 6 hands simultaneously, can you guys help calculate the player’s edge from this promotion?
Let me do a rough calculation first. Player getting a joker probability is 12x4/312=15%. This means the players edge gain from this promotion is 15%. This is crazy! Is this correct?
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I think your overall EV would be higher playing one hand. The later hands might turn negative EV based on jokers being consumed on the earlier hands. It is too late to reduce your bet. You have more flexabilty if you play one hand at a time, especially close to the cut card.
My instinct is players must catch these jokers before they leak out, so that means that a player needs to play simultaneous multiple hands, but I just cannot get the math right. Please help.
I am only coming at it from getting the highest percentage RTP. If you are flat max betting (unless the joker count goes -EV), every one of the 6 hands has the same chance of getting a joker as if you had only played one hand from the top of the shoe. You can adjust the splitting strategy, but you lost the ability to decrease your bet, if needed.Quote: acesideThe basic rule is that when dealer gets a joker, that joker is burned; when dealer or player uses a joker to hit, that joker is burnt too. Burnt means removed. The main edge of this game is in catching jokers, but they can easily leak out.
My instinct is players must catch these jokers before they leak out, so that means that a player needs to play simultaneous multiple hands, but I just cannot get the math right. Please help.
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On the other hand, if the deck is rich in jokers, you are probably right that you want to spread more hands to catch as many of the jokers in your hands versus the dealer hands.
From the point of max hourly profit, you want to get more hands in per hour on a +EV game. Still, I would rather be betting 6 units on one hand heads up versus one unit on six hands for max profit.
Quote: MentalI am only coming at it from getting the highest percentage RTP. If you are flat max betting (unless the joker count goes -EV), every one of the 6 hands has the same chance of getting a joker as if you had only played one hand from the top of the shoe.Quote: acesideThe basic rule is that when dealer gets a joker, that joker is burned; when dealer or player uses a joker to hit, that joker is burnt too. Burnt means removed. The main edge of this game is in catching jokers, but they can easily leak out.
My instinct is players must catch these jokers before they leak out, so that means that a player needs to play simultaneous multiple hands, but I just cannot get the math right. Please help.
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The issue here is that this is not an infinite-deck game. It’s a 6-deck game with only 4 deers hiding in here for us to hunt in this trap. This calls for a different strategy. I guess we need to consider the RTP for one whole card shoe.
Since it is a extreme low volatility game that has significantly positive EV during the promo, nobody cares about RTP. It was just an academic question that I was interested in.Quote: acesideQuote: MentalI am only coming at it from getting the highest percentage RTP. If you are flat max betting (unless the joker count goes -EV), every one of the 6 hands has the same chance of getting a joker as if you had only played one hand from the top of the shoe.Quote: acesideThe basic rule is that when dealer gets a joker, that joker is burned; when dealer or player uses a joker to hit, that joker is burnt too. Burnt means removed. The main edge of this game is in catching jokers, but they can easily leak out.
My instinct is players must catch these jokers before they leak out, so that means that a player needs to play simultaneous multiple hands, but I just cannot get the math right. Please help.
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The issue here is that this is not an infinite-deck game. It’s a 6-deck game with only 4 deers hiding in here for us to hunt in this trap. This calls for a different strategy. I guess we need to consider the RTP for one whole card shoe.
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Max EV per hour is probably to max bet as many spots as possible. then min bet whenever the count goes negative so as to get to the top of the next shoe ASAP. After the count indicates negative EV, the EV can go positive again if no jokers fall and penetration is deep enough. This strategy will make it super obvious that you are counting. Perhaps it is better to just flat bet.
As has been said you may split or double more often as drawing a joker has the bonus. So yes things like 22,33,77 vs 8, 99 v 7 etc will come into play. Also you'll be doubling things like 9 vs 2, and possibly even 9 vs 7 and more soft doubles. If you do the caclulation it may even be worth splitting 10s, and if it was right you'd always do so and it wouldn't look obvious. I would guess it's similar logic to when the count gets favourable.
Ignoring the additional benefit of the ideas above, it does seem that there is an added advantage of receiving the bonus 8/316 times, which is about 2.5%. Even if there's only one joker left to come it's over 2/316 (since there are fewer than 316 cards to go), which at 0.6% is about the same as the basic BJ House Edge (assuming reasonable rules). So most the time, unless all four come out early, you would want to play on in any case.
I looked at how often the last joker gets used up in the first decile of the cards in the shoe, then 2nd decile, etc. So, 12.96% of the time that you are 60% through the shoe, all the jokers will be gone. This rises to 23.90% at 70% penetration.Quote: charliepatrickIgnoring the additional benefit of the ideas above, it does seem that there is an added advantage of receiving the bonus 8/316 times, which is about 2.5%. Even if there's only one joker left to come it's over 2/316 (since there are fewer than 316 cards to go), which at 0.6% is about the same as the basic BJ House Edge (assuming reasonable rules). So most the time, unless all four come out early, you would want to play on in any case.
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| Decile | Jokers Gone | Cum % |
| 1 | 0.01% | 0.01% |
| 2 | 0.15% | 0.16% |
| 3 | 0.61% | 0.77% |
| 4 | 1.76% | 2.53% |
| 5 | 3.81% | 6.34% |
| 6 | 6.62% | 12.96% |
| 7 | 10.94% | 23.90% |
| 8 | 16.93% | 40.83% |
| 9 | 24.59% | 65.41% |
| 10 | 34.59% | 100.00% |
Quote: charliepatrickIf you assume the cards are laid out in a long line of boxes, then your hand next round will receive cards from two of those boxes. At the start of the shoe the chances of any box having a joker, i.e. the each one corresponding to your hand, is the same (4/316). Thus it doesn't really matter if you play multiple hands or many single ones, except playing multiple hands flat betting means you get more hands/hour. As you go through the shoe and jokers appear, or not, the chances go down or up accordingly. Obviously it's fairly easy to see when all four jokers have gone and stop playing/take a break.
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I question this. Player getting the joker is a large positive event; dealer getting the joker depletes the shoe of a joker without any benefit to the player. Playing multiple hands (say 6 hands) minimizes the number of jokers/shoe that will be dealt to the dealer
Let's try some numbers:
6 deck shoe with 75% penetration; expected number of jokers available to player is 3 jokers per shoe.
If playing dealer heads up, player will get 1.5 jokers per shoe while playing approx. 86 rounds (deals). If playing 6 hands vs the dealer, player will get 2.571 jokers per shoe while playing 33 rounds at 6 hands per round (200 player hands per shoe)
From an EV standpoint, you may be better off playing heads up, but to maximize winnings from a session of finite length I suspect the player is better off playing 6 hands than playing heads up vs the dealer.
I think you are missing Charlie's point if you disagree with him. If you flat bet, your EV per hand is unchanged whether you bet one or six hands. Your EV per hand only depends on the probability of getting one or two jokers. How does the dealer using up jokers affect this at all?Quote: gordonm888Quote: charliepatrickIf you assume the cards are laid out in a long line of boxes, then your hand next round will receive cards from two of those boxes. At the start of the shoe the chances of any box having a joker, i.e. the each one corresponding to your hand, is the same (4/316). Thus it doesn't really matter if you play multiple hands or many single ones, except playing multiple hands flat betting means you get more hands/hour. As you go through the shoe and jokers appear, or not, the chances go down or up accordingly. Obviously it's fairly easy to see when all four jokers have gone and stop playing/take a break.
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I question this. Player getting the joker is a large positive event; dealer getting the joker depletes the shoe of a joker without any benefit to the player. Playing multiple hands (say 6 hands) minimizes the number of jokers/shoe that will be dealt to the dealer
Let's try some numbers:
6 deck shoe with 75% penetration; expected number of jokers available to player is 3 jokers per shoe.
If playing dealer heads up, player will get 1.5 jokers per shoe while playing approx. 86 rounds (deals). If playing 6 hands vs the dealer, player will get 2.571 jokers per shoe while playing 33 rounds at 6 hands per round (200 player hands per shoe)
From an EV standpoint, you may be better off playing heads up, but to maximize winnings from a session of finite length I suspect the player is better off playing 6 hands than playing heads up vs the dealer.
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Your total EV/hr obviously increases at six hands, but six times as much. If you are changing your bet or even sitting out, then the EV per hand can be improved by playing one hand per round. You have more information at the start of each hand.
Online casinos can count your hands with no real effort. Many online promotions are limited to a specific number of games/hands. This is why I found it interesting to think about optimizing EV per hand versus just max betting as many spots as possible.
Head to Head
Player: Avg #cards used per hand = 2.723, Avg #hands/shoe = 45.751
Dealer: Avg #cards used per hand = 2.554, Avg #hands/shoe = 44.592
Average number of cards used per shoe = 238.459 (note you play out the hand after the cut card, so this is above the 234 figure).
Ignoring doubling the Player sees 45.751*2 "first two" cards in a shoe. A deck averages one joker per 78 cards. So on average the player sees 91.502/78 jokers per shoe = 1.173. This is 1.173 jokers per 44.592 rounds = 1 in 38.015.
Five hands per round
This assumes the averages are the same. The round uses 5*2.723+2.554 cards per round = 16.169 cards per round, so 14.748 rounds per shoe. The Player sees 5*2*14.748*1.026 "first two" cards = 151.31. So on average the player sees 151.31/78 jokers per shoe = 1.940. This is 1.940 jokers per 73.74 (14.748*5) initial hands = 1 in 1 in 38.010.
The only slight advantage I can see is playing head-to-head you see 238 cards but playing multiple hands, because you draw more cards per round, you have more chances to see cards beyond the cut cards, hence see more cards per shoe.
btw I've also ignored the times Dealer gets a Blackjack and you get a Joker.
six decks 75% pen
Overall Result: Exp: -0.004943411491803911 Hands: 44592181 Win: 20146686 Lose: 23392241 Tie: 3927649 CHY: 0 BJk: 2016745
Player Hand Size: 2 20906615
Player Hand Size: 3 18024250
Player Hand Size: 4 5571190
Player Hand Size: 5 1097919
Player Hand Size: 6 138109
Player Hand Size: 7 12106
Player Hand Size: 8 759
Player Hand Size: 9 33
Average Player Hand Size: (124580100/45750981) 2.723004
Dealer Hand Size: 1 9161333
Dealer Hand Size: 2 11322302
Dealer Hand Size: 3 16022899
Dealer Hand Size: 4 6601301
Dealer Hand Size: 5 1318923
Dealer Hand Size: 6 153557
Dealer Hand Size: 7 11318
Dealer Hand Size: 8 533
Dealer Hand Size: 9 15
Average Dealer Hand Size: (113879420/44592181) 2.553798
Parms: ndx:6 sh:1000000 ch 88 cards pay 1 pen:75% dstops:99cards maxspl:99 strat:0 Time:21:8:7:813
Quote: charliepatrick
Five hands per round
This assumes the averages are the same. The round uses 5*2.723+2.554 cards per round = 16.169 cards per round, so 14.748 rounds per shoe. The Player sees 5*2*14.748*1.026 "first two" cards = 151.31.
What is this factor "1.026" in the above equation?
Also, if these five hands are played by five different players, your result says that each player will have the same amount of edge as that in head-to-head playing with the dealer. Is this Correct? In the real world, players always spread horizontally into multiple hands to catch good cards when the situation is advantageous, though.
How long does the promo run for?Quote: sorcererofoddsHi,
My local casino is offering a blackjack promotion next Wednesday, $50 maximum bet. They are placing 4 jokers in a 6 deck shoe. If you get a joker as one of your first 2 dealt cards, you get your bet paid back extra as a bonus, then you play the hand as normal with your original bet. If you get 2 jokers, each as one of your first 2 cards, then you get 2X your original bet back. Split is considered 2 normal hands, with joker bet bonus available on first 2 cards of each. Bonus available on double down THIRD card but you only get your initial bet paid back (not the double). All other times when joker appears e.g. as your third card, to dealer it is burnt. Can I get an estimate of house / player edge, over and above basic strategy, on this game please. I believe that this game gives the player an advantage off the top, but don't have an exact number
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Whatever the case.
Just Do It.
Every player hand has the same EV if played with the same strategy. A lot of people have a problem sorting this out, so let me try to explain it.Quote: acesideAlso, if these five hands are played by five different players, your result says that each player will have the same amount of edge as that in head-to-head playing with the dealer. Is this Correct? In the real world, players always spread horizontally into multiple hands to catch good cards when the situation is advantageous, though.
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Imagine a different promo game where splitting is not allowed and the dealer deals from two shoes. However, one shoe is used for the first two cards of every hand including the dealer hand. The other shoe is used for all hit card draws. The cut cards are inserted after exactly 280 cards, not including the burn card. Four jokers are randomly inserted into each shoe. You are not allowed to see the cut card or the joker insertion. This means that the dealer can deal exactly 20 rounds to a six spots from the first shoe (7*2*20=280). Likewise, he can deal 70 rounds heads up to one spot.
This setup means that prior to the joker insertion you know exactly which cards in the first shoe that you will get for every single hand in the round. This is true whether one spot or six spots are being dealt
If you believe that there is a difference in the EV per hand when you play one hand heads up, or one to six hands at a full table, then you must believe that the jokers were inserted in front of one or both of the cards in the hands that you believe have higher EV. But, the jokers were inserted randomly. There is no pair of cards in the shoe that is any more likely to catch the jokers. Every hand has the same 'a priori' EV.
After each round, you can surmise that the first shoe is richer or poorer in jokers then when you started, but you still don't have any reason to think that any remaining jokers will hit any particular spot or the dealer's hand. All hands have the same EV.
Players tend to drop their bet amount on each of these multiple-hands when they spread from 1-hand into 6-hand playing. However, if we lock the bet amount to be one unit max on every spot you play, the profit you will gain from the 6-hand playing is just 6 times of the 1-hand playing. Gamblers are smarter than mathematicians. Is this correct?
Live dealer is a different story.
