New poker game
February 22nd, 2012 at 8:54:39 AM
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Has anyone seen or heard of a 4 card poker game where each player and the dealer gets 4 cards. Each player Antes before the deal. All cards are dealt down. The players can look at their cards and bet to Play an additional minimum of their Ante or fold. The object is for the player to use any of his 4 cards to beat the dealer's 1 card that's turned over randomly by the dealer. So if the dealer turns over a 9 of Spades you have to have a 10 of Spades or better. The dealer’s Suit must be matched and higher to win which pays 1 to 1. There are also some side bets as pair or better, etc. which can be bet with the Ante and pay higher odds. It sounds interesting but I can’t figure the basic player vs dealer odds. Can anyone help?
February 22nd, 2012 at 9:24:53 AM
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Doesn't really sound like poker, but it does sound like an interesting game.
2's in your hand would be worthless. With 8 being the middle card in the deck, you can expect to have the advantage with an 8 or better, unless you have a card at a lower value then 8 (meaning the dealer could not have that card).
Just analyzing it quickly, it would seem to me the house edge would be largely in favor of the house.
This game would have to be analyzed much like video poker. The be hand dealt would be all 4 aces, as it would be 100% return. All 2's would be the worst hand dealt.
If you have all 4 different suits, then your probability of a card beating the dealer's card would work like this:
By multiplying that by 25%, and adding them up for each suit, you would get your probability of winning.
This obviously would change if you didn't have 1 of each suit.
2's in your hand would be worthless. With 8 being the middle card in the deck, you can expect to have the advantage with an 8 or better, unless you have a card at a lower value then 8 (meaning the dealer could not have that card).
Just analyzing it quickly, it would seem to me the house edge would be largely in favor of the house.
This game would have to be analyzed much like video poker. The be hand dealt would be all 4 aces, as it would be 100% return. All 2's would be the worst hand dealt.
If you have all 4 different suits, then your probability of a card beating the dealer's card would work like this:
| Card Value | Prob of winning |
|---|---|
| A | 1 |
| K | 0.916666667 |
| Q | 0.833333333 |
| J | 0.75 |
| 10 | 0.666666667 |
| 9 | 0.583333333 |
| 8 | 0.5 |
| 7 | 0.416666667 |
| 6 | 0.333333333 |
| 5 | 0.25 |
| 4 | 0.166666667 |
| 3 | 0.083333333 |
| 2 | 0 |
By multiplying that by 25%, and adding them up for each suit, you would get your probability of winning.
This obviously would change if you didn't have 1 of each suit.
February 22nd, 2012 at 9:46:17 AM
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Delete
In a bet, there is a fool and a thief.
- Proverb.
February 22nd, 2012 at 10:48:35 AM
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I am not sure why the dealer needs to take four cards. Sounds like a complicated version of "Casino War".
In any case, doesn't having multiples of the same suit in your hand actually reduce the chance that the dealer will turn the same suit? I don't know if that all comes out in the wash, since you proably can't increase or decrease the size of your "Play" bet after the deal.
In any case, doesn't having multiples of the same suit in your hand actually reduce the chance that the dealer will turn the same suit? I don't know if that all comes out in the wash, since you proably can't increase or decrease the size of your "Play" bet after the deal.
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 22nd, 2012 at 10:50:49 AM
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Probably because they are using a shuffling machine that spits out 4 card hands.Quote: AyecarumbaI am not sure why the dealer needs to take four cards.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 10:59:34 AM
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As Trip points out in his chart above, if you have one of each suit, you have exactly 50% chance of winning.
This is just an assumption, but, if you have 2 or more of a specific suit, and the dealer's card is the same suit, I think you still have a 50% chance.
Therefore, the odds are always 50/50 if you match the suit.
So matching the suit is important, and is where the house edge comes in. And it begs the questions:
What are the odds of the dealer having one of your suits, if you have: 3 suits; 2 each of 2 suits; 3 of one suit and one of another, or 4 of one suit? And what are the odds of getting each of those hand compositions?
This is just an assumption, but, if you have 2 or more of a specific suit, and the dealer's card is the same suit, I think you still have a 50% chance.
I'm pretty sure it does.Quote: Ayecarumbadoesn't having multiples of the same suit in your hand actually reduce the chance that the dealer will turn the same suit?
Therefore, the odds are always 50/50 if you match the suit.
So matching the suit is important, and is where the house edge comes in. And it begs the questions:
What are the odds of the dealer having one of your suits, if you have: 3 suits; 2 each of 2 suits; 3 of one suit and one of another, or 4 of one suit? And what are the odds of getting each of those hand compositions?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 11:08:49 AM
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Quote: DJTeddyBearAs Trip points out in his chart above, if you have one of each suit, you have exactly 50% chance of winning.
This is just an assumption, but, if you have 2 or more of a specific suit, and the dealer's card is the same suit, I think you still have a 50% chance.
I'm pretty sure it does.
Therefore, the odds are always 50/50 if you match the suit.
So matching the suit is important, and is where the house edge comes in. And it begs the questions:
What are the odds of the dealer having one of your suits, if you have: 3 suits; 2 each of 2 suits; 3 of one suit and one of another, or 4 of one suit? And what are the odds of getting each of those hand compositions?
First off, your odds aren't always 50/50 if you have all four suits. If you had all 8's, you're chance of winning at that point (with the knowledge you have) would be 50/50.
Missing a suit would harm you, as it would decrease your chances of winning by 25% as well as decreasing the odds that the dealer would have your other suit.
Brings up my next point, I said having all 2's would be 0%. It should be noted that having 2,3,4,5 all in one suit would also have a 0% probability.
February 22nd, 2012 at 11:13:43 AM
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I think they are. The average of the numbers in your chart is 50%. So, having all 4 suits, without knowledge of the rank, means, on average, you have a 50% chance.Quote: TriplellFirst off, your odds aren't always 50/50 if you have all four suits.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 11:13:46 AM
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True. But similarly, having A, x, x, x of one suit is a 100% win - if the dealer shows that suit. I.E. Average of 50%Quote: TriplellBrings up my next point, I said having all 2's would be 0%. It should be noted that having 2,3,4,5 all in one suit would also have a 0% probability.
So we're back to the odds of the dealer having the same suit, depending on your hand composition.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 11:17:56 AM
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Actually, it's higher than 50% so I'll retract this statement.Quote: DJTeddyBearTrue. But similarly, having A, x, x, x of one suit is a 100% win - if the dealer shows that suit. I.E. Average of 50%
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 11:34:46 AM
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The whole point is that you have to make a bet blindly, and then you have a choice to double that bet, or surrender it.
By the time you know the suit of the dealer...it is too late, you have already had to make that decision.
Therefore, the strategy comes in to "with the cards in your hand, when is it worth it to double your bet, and when is it worth it to surrender"...which is what I'm trying to figure out.
Does it pay 1-1 on your total bet (ante plus raise). Finally, can the raise be all the way to the table maximum?
By the time you know the suit of the dealer...it is too late, you have already had to make that decision.
Therefore, the strategy comes in to "with the cards in your hand, when is it worth it to double your bet, and when is it worth it to surrender"...which is what I'm trying to figure out.
Does it pay 1-1 on your total bet (ante plus raise). Finally, can the raise be all the way to the table maximum?
February 22nd, 2012 at 11:38:31 AM
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OH!
I totally missed the part about the decision to play or fold.
I totally missed the part about the decision to play or fold.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 22nd, 2012 at 11:55:32 AM
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I've not put this into a spreadsheet but I think it's easiest to think of the cards left in the deck where you win and hence by subtraction those that you lose. The combos are 4 cards in one suit 4000, 3100, 2200, 2110, 1111. The chances of winning are the number of cards, in suits you have, that are lower than the highest one you have divided by 48.
A quick thought suggests that 1111 hands are 50-50, as on average you'll have 6 cards in each suit where the dealer gets a matching suit with lower total and loses (i.e. you win). Whereas when you have more cards in one suit (say spades no hearts), yes you have more chances in spades (best scenario is 11) but there are 13 hearts where you definitely lose.
(I am assuming that if the dealer has a heart, and you have none, then you lose Ante and Raise.)
Out of interest where did you see the game?
A quick thought suggests that 1111 hands are 50-50, as on average you'll have 6 cards in each suit where the dealer gets a matching suit with lower total and loses (i.e. you win). Whereas when you have more cards in one suit (say spades no hearts), yes you have more chances in spades (best scenario is 11) but there are 13 hearts where you definitely lose.
(I am assuming that if the dealer has a heart, and you have none, then you lose Ante and Raise.)
Out of interest where did you see the game?
February 22nd, 2012 at 12:06:59 PM
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The dealer takes 4 cards down and selects 1 to turn over. I guess that is to add some suspense to the game. There is a round of betting after the cards are dealt before the dealer turns his card over. To bet the player must bet at least 1x his ante and can bet up to 3x the ante.
What I can't figure out is what are the house odds if that since the dealer card contorls the suit and eliminates all cards of the other 3 suits. If there are 6 palyers that means that 28 cards (including 4 for the dealer) are dealt from a 52 deck. That leaves 24 un-played cards with the dealer's turn card controling the Suit which has to be a large advantage for the house.
What I can't figure out is what are the house odds if that since the dealer card contorls the suit and eliminates all cards of the other 3 suits. If there are 6 palyers that means that 28 cards (including 4 for the dealer) are dealt from a 52 deck. That leaves 24 un-played cards with the dealer's turn card controling the Suit which has to be a large advantage for the house.
February 22nd, 2012 at 12:13:00 PM
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I assume when you say selects do you mean pick at random or is the largest valued card chosen? Also does the dealer always qualify? As to other players having cards - that is irrelevant (unless you know what they are or if you know whether they raised or folded).Quote: Carl1946The dealer takes 4 cards down and selects 1 to turn over...
February 22nd, 2012 at 12:15:30 PM
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Ok, so I get the probability of each card winning as follows:
4 suits:
chance of winning = (((1/4)*d)+((1/4)*s)+((1/4)*h)+((1/4)*c))
where d,s,h,c are the values in the table above.
3 suits(2,1,1,0):
chance of winning = (((1/4)*s1)+((1/4)*s2)+((11/48)*(os-(1/12))))
s1 and s2 are the values of the cards that don't match, and os is the top valued card of the matching suits
2 suits(2,2,0,0):
chance of winning = ( ((11/48)*(os1-(1/12)))+((11/48)*(0s2-(1/12))))
again, these are the top values of those suits
2 suits (3,1,0,0)
chance of winning = ( ((5/24)*(os1-(1/6)))+((1/4)*(s2)))
os1 corresponds to the suit with 3 cards, s2 corresponds to the single card in suit
1 suit(4,0,0,0)
chance of winning = ( ((3/16)*(s1-(1/4)))
s1 is the top valued card in the suit.
I'm not 100% on this, but i'm pretty sure it's right...
| Card Value | Prob of winning |
|---|---|
| A | 1 |
| K | 0.916666667 |
| Q | 0.833333333 |
| J | 0.75 |
| 10 | 0.666666667 |
| 9 | 0.583333333 |
| 8 | 0.5 |
| 7 | 0.416666667 |
| 6 | 0.333333333 |
| 5 | 0.25 |
| 4 | 0.166666667 |
| 3 | 0.083333333 |
| 2 | 0 |
4 suits:
chance of winning = (((1/4)*d)+((1/4)*s)+((1/4)*h)+((1/4)*c))
where d,s,h,c are the values in the table above.
3 suits(2,1,1,0):
chance of winning = (((1/4)*s1)+((1/4)*s2)+((11/48)*(os-(1/12))))
s1 and s2 are the values of the cards that don't match, and os is the top valued card of the matching suits
2 suits(2,2,0,0):
chance of winning = ( ((11/48)*(os1-(1/12)))+((11/48)*(0s2-(1/12))))
again, these are the top values of those suits
2 suits (3,1,0,0)
chance of winning = ( ((5/24)*(os1-(1/6)))+((1/4)*(s2)))
os1 corresponds to the suit with 3 cards, s2 corresponds to the single card in suit
1 suit(4,0,0,0)
chance of winning = ( ((3/16)*(s1-(1/4)))
s1 is the top valued card in the suit.
I'm not 100% on this, but i'm pretty sure it's right...
February 22nd, 2012 at 12:20:38 PM
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Yes. The dealer's suit is the only suit that can compete and you must beat the dealer's high card in that suit or you lose.
I first saw the game a while back at a casino off the strip but can't remember where and then in an Indian casino in Oklahoma but I was able to play it a little at a card room. I lost...
I first saw the game a while back at a casino off the strip but can't remember where and then in an Indian casino in Oklahoma but I was able to play it a little at a card room. I lost...
February 22nd, 2012 at 12:22:09 PM
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Do you get paid 1/1 on the ante and the raise? or does the ante push on a win, and 1/1 on a raise?
February 22nd, 2012 at 12:23:07 PM
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Does the dealer pick a high card out of the 4? Or does he randomly flip over 1 of his 4 cards? If he picks his highest card, then that changes everything...
February 22nd, 2012 at 12:28:53 PM
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The dealer selects 1 card at random from his 4 cards while they are face down. There is no dealer qualify like in other similar games.
I think having some information from the other players raise or fold could be helpful but probably not that much. You are playing against the dealer not the other players. However many players beat the dealer those players get paid.
I think having some information from the other players raise or fold could be helpful but probably not that much. You are playing against the dealer not the other players. However many players beat the dealer those players get paid.
February 22nd, 2012 at 12:37:12 PM
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You get paid 1/1 on both ante and raise. The higher payouts come from the side bets. The side bets have nothing to do with you winning or losing the basic hand against the dealer. The side bet is what you wager your 4 card hand will compose ie: pair, flush, 3 of kind, etc. Side bet must be made at the same time as the ante.
February 22nd, 2012 at 12:42:18 PM
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Quote: Carl1946You get paid 1/1 on both ante and raise. The higher payouts come from the side bets. The side bets have nothing to do with you winning or losing the basic hand against the dealer. The side bet is what you wager your 4 card hand will compose ie: pair, flush, 3 of kind, etc. Side bet must be made at the same time as the ante.
If you get paid 1/1 on everything, then the strategy seems simple to me.
Raise the max when your % of winning is >= 50% and fold everything else.
February 22nd, 2012 at 12:42:51 PM
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Quote: TriplellThe whole point is that you have to make a bet blindly, and then you have a choice to double that bet, or surrender it.
By the time you know the suit of the dealer...it is too late, you have already had to make that decision.
Therefore, the strategy comes in to "with the cards in your hand, when is it worth it to double your bet, and when is it worth it to surrender"...which is what I'm trying to figure out.
Does it pay 1-1 on your total bet (ante plus raise). Finally, can the raise be all the way to the table maximum?
Most raise allowed I have seen is 3x the ante but I think that would be up to the house.
February 22nd, 2012 at 12:57:44 PM
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No I think you fold if less than 25% of winning. Between 25% and 50% raise 1. The logic is that sometimes you'll win 2 (admittedly not very often) but it's better than always losing 1: you are 3 better off more than 25% of the time, worth spending the extra 1 on.Quote: TriplellRaise the max when your % of winning is >= 50% and fold everything else.
(Note in qualifying games the percentages are different as sometimes you only win 1, i.e. only 2 better off.)
February 22nd, 2012 at 1:02:54 PM
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Quote: TriplellOk, so I get the probability of each card winning as follows:
Card Value Prob of winning A 1 K 0.916666667 Q 0.833333333 J 0.75 10 0.666666667 9 0.583333333 8 0.5 7 0.416666667 6 0.333333333 5 0.25 4 0.166666667 3 0.083333333 2 0
4 suits:
chance of winning = (((1/4)*d)+((1/4)*s)+((1/4)*h)+((1/4)*c))
where d,s,h,c are the values in the table above.
3 suits(2,1,1,0):
chance of winning = (((1/4)*s1)+((1/4)*s2)+((11/48)*(os-(1/12))))
s1 and s2 are the values of the cards that don't match, and os is the top valued card of the matching suits
2 suits(2,2,0,0):
chance of winning = ( ((11/48)*(os1-(1/12)))+((11/48)*(0s2-(1/12))))
again, these are the top values of those suits
2 suits (3,1,0,0)
chance of winning = ( ((5/24)*(os1-(1/6)))+((1/4)*(s2)))
os1 corresponds to the suit with 3 cards, s2 corresponds to the single card in suit
1 suit(4,0,0,0)
chance of winning = ( ((3/16)*(s1-(1/4)))
s1 is the top valued card in the suit.
I'm not 100% on this, but i'm pretty sure it's right...
Just realized that all of these cannot be right, as it compounds some of the events. For instance, if you only have 1 suit, I subtracted 1/12 for each suit that would be under the top valued card, and then multiplied it by the probability that you and the dealer share a suit. The problem with the formula, is if you have an ace, it should be simply the probability you and the dealer have the same suit. Looks like there would need to be a table made up for each instance (I was trying to avoid that)....
February 22nd, 2012 at 1:21:30 PM
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Quote: charliepatrickNo I think you fold if less than 25% of winning. Between 25% and 50% raise 1. The logic is that sometimes you'll win 2 (admittedly not very often) but it's better than always losing 1: you are 3 better off more than 25% of the time, worth spending the extra 1 on.
(Note in qualifying games the percentages are different as sometimes you only win 1, i.e. only 2 better off.)
If the ante is 1 unit, and you can raise 1 unit, 2 units, or 3 units, it would correspond to betting 1 to win 2, betting 2 to win 3, and betting 3 to win 4.
So if you are below 33%, you should fold. 33% to 40%, you should raise 1 unit. 40% to 43%, you should raise 2 units. and everything above, you should raise 3 units.
Is that right?
February 22nd, 2012 at 1:38:09 PM
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I tweaked the formulas. I think they are accurate now.
They can be found in this spreadsheet.
https://docs.google.com/spreadsheet/ccc?key=0AneP5iBQCCXIdDdmM2QwM1RONE90Ymx6d19PeDN5R3c
They can be found in this spreadsheet.
https://docs.google.com/spreadsheet/ccc?key=0AneP5iBQCCXIdDdmM2QwM1RONE90Ymx6d19PeDN5R3c
February 22nd, 2012 at 1:48:21 PM
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(i) Raise X1 if between 25% and 50%.
There are 48 cards left in the deck for the dealer.
Say you had As2h2d2c then there are 12 chances to win (any spade) and 36 chances to lose (the other suits).
Option1 fold - every 48 possibilities lose - total loss = -48
Option 2 play - 12 Spades win 2 units, 12 Hearts lose 2, 12 Diamonds lose 2, 12 Clubs lose 2 - total payback = +24 -24 -24 -24 = -48.
Consider As3h2d2c playing means winning 13 times and losing 35 - payback = +26-70 = -44 (better than -48).
A very quick calculation shows with option to fold or raise X3 the house edge is 5.69%. This is because the house wins most hands where they have four cards in a suit, whereas the player should always fold them. This doesn't compensate for when the player has one or two suits and can only raise three times.
There are 48 cards left in the deck for the dealer.
Say you had As2h2d2c then there are 12 chances to win (any spade) and 36 chances to lose (the other suits).
Option1 fold - every 48 possibilities lose - total loss = -48
Option 2 play - 12 Spades win 2 units, 12 Hearts lose 2, 12 Diamonds lose 2, 12 Clubs lose 2 - total payback = +24 -24 -24 -24 = -48.
Consider As3h2d2c playing means winning 13 times and losing 35 - payback = +26-70 = -44 (better than -48).
A very quick calculation shows with option to fold or raise X3 the house edge is 5.69%. This is because the house wins most hands where they have four cards in a suit, whereas the player should always fold them. This doesn't compensate for when the player has one or two suits and can only raise three times.
February 22nd, 2012 at 1:55:38 PM
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I disagree. Raise x1 means you are getting paid at 2 to 1, which means that anything above 33% would be a good bet. Raise x2 means you are getting paid 3 to 2, which means that anything above 40% would be a good bet. And raise x3 means you are getting paid at 4 to 3, which means that anything above 43% would be a good bet. You're payout would increasingly approach the 50% odds, however never reach it (due to the fact that you are getting paid 1/1, however you must blindly submit 1 unit)....
My spreadsheet calculates the chances of winning. Now the question is what is the odds of being dealt each combination, and calculate your return.
I know for a fact that you should always fold if you are dealt a flush...
My spreadsheet calculates the chances of winning. Now the question is what is the odds of being dealt each combination, and calculate your return.
I know for a fact that you should always fold if you are dealt a flush...
February 22nd, 2012 at 2:13:27 PM
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I am assuming you either (i) fold and lose the Ante (ii) call and win the Ante and Raise (iii) call and lose the Ante and Raise. The way I worked it out was working out each combination of cards (for four suits 1111, three suits 211 etc). For each combination
(a) count how many cards the dealer can have where I get paid the Ante and Raise : W
(b) count how many cards the dealer can have where I lose the Ante and Raise : L
(For instance with AAAA W=12+12+12+12 L=0+0+0+0 AAAK W=12+12+12+11 L=0+0+0+1 etc. Where you have >1 cards in the suit, I assume the chances are how many cards are in the suit less than your highest card, so Ax has 11, Kx 10 etc.)
Compare the total of (W-L)*(A+R) against -48 and do the best (at some stage you're just better to fold).
If you like similarly compare the values where the raise is 3x, but you can see that if W<L you want to minimize (A+R) and if W>L you want to maximize (A+R).
Yes I agree that you always fold a flush.
(a) count how many cards the dealer can have where I get paid the Ante and Raise : W
(b) count how many cards the dealer can have where I lose the Ante and Raise : L
(For instance with AAAA W=12+12+12+12 L=0+0+0+0 AAAK W=12+12+12+11 L=0+0+0+1 etc. Where you have >1 cards in the suit, I assume the chances are how many cards are in the suit less than your highest card, so Ax has 11, Kx 10 etc.)
Compare the total of (W-L)*(A+R) against -48 and do the best (at some stage you're just better to fold).
If you like similarly compare the values where the raise is 3x, but you can see that if W<L you want to minimize (A+R) and if W>L you want to maximize (A+R).
Yes I agree that you always fold a flush.
February 22nd, 2012 at 2:21:12 PM
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Exactly, however you are simply just accounting for when you should minimize and maximize. If you calculate the actual probability of you winning, and you know the payout (which is based on your raise), you can define it down to when each denomination of raise fits best.
If it was allowed, I would agree, anything over 50%, and you would want to go all in on the ante...however using the denominations, it can clearly be seen that anything over 43% would be a good idea to bet 3x the ante. similiarly if 4x was allowed, you would bet it at a probability greater then 44.4% and bet 5x at 45.45%...
If it was allowed, I would agree, anything over 50%, and you would want to go all in on the ante...however using the denominations, it can clearly be seen that anything over 43% would be a good idea to bet 3x the ante. similiarly if 4x was allowed, you would bet it at a probability greater then 44.4% and bet 5x at 45.45%...
February 22nd, 2012 at 4:22:40 PM
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After 4 pages of discussion...
@Carl1946... where did you see this game... what Casino?
@Carl1946... where did you see this game... what Casino?
Some people need to reimagine their thinking.
February 22nd, 2012 at 4:26:46 PM
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If my math is correct, the house edge if playing 1x play blind is 24.5%.
For each of the 13 numbers (ranks) the dealer can have, other than an ace, there is a:
1 - (14 - rank)/51 *
1 - (14 - rank)/50 *
1 - (14 - rank)/49 *
1 - (14 - rank)/48
chance to draw 4 cards from the deck and not have one that is the same suit and higher in rank. For example, when the dealer has a 2 there is a 32.91% chance you won't have a card of the same suit (any rank would beat a 2). When the dealer has a jack there is a 77.86% chance that you won't have a card of the same suit that outranks a jack. The ace is a 100% probability you won't win.
The average across all cards is 62.27%. This means that playing blind:
(0.6227 * -1) + ((1 - 0.6227) * 1) = -0.2454, or a house edge of 24.5%. There's a rounding error there, it's actually closer to 24.55% but I think the chance of me just being completely wrong outweighs the significance of the rounding error.
For each of the 13 numbers (ranks) the dealer can have, other than an ace, there is a:
1 - (14 - rank)/51 *
1 - (14 - rank)/50 *
1 - (14 - rank)/49 *
1 - (14 - rank)/48
chance to draw 4 cards from the deck and not have one that is the same suit and higher in rank. For example, when the dealer has a 2 there is a 32.91% chance you won't have a card of the same suit (any rank would beat a 2). When the dealer has a jack there is a 77.86% chance that you won't have a card of the same suit that outranks a jack. The ace is a 100% probability you won't win.
The average across all cards is 62.27%. This means that playing blind:
(0.6227 * -1) + ((1 - 0.6227) * 1) = -0.2454, or a house edge of 24.5%. There's a rounding error there, it's actually closer to 24.55% but I think the chance of me just being completely wrong outweighs the significance of the rounding error.
February 22nd, 2012 at 6:25:27 PM
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Here is a Similar Game: Block
The Block bet is based on the player's first two cards and the dealer's up card. It can be found in Africa, South Africa, Egypt, Latvia, Estonia, Ireland, Morocco and Eastdend-EU. The bet wins if the dealer's up card matches the suit of one of the player's cards, and the player's card is higher. There are higher pays if the player's cards are a pair, suited, or both. Here is how the various winning hands are defined.Read More
The Block bet is based on the player's first two cards and the dealer's up card. It can be found in Africa, South Africa, Egypt, Latvia, Estonia, Ireland, Morocco and Eastdend-EU. The bet wins if the dealer's up card matches the suit of one of the player's cards, and the player's card is higher. There are higher pays if the player's cards are a pair, suited, or both. Here is how the various winning hands are defined.Read More
Stephen Au-Yeung (Legend of New Table Games®) NewTableGames.com
February 22nd, 2012 at 7:18:07 PM
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I get a player EV of -0.007957183 . That seems kind of low for such a simple game. I wonder if I have a typo or braino somewhere? I better double check my work.
“Man Babes” #AxelFabulous
February 23rd, 2012 at 6:22:19 AM
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Quote: mipletI get a player EV of -0.007957183 . That seems kind of low for such a simple game. I wonder if I have a typo or braino somewhere? I better double check my work.
I get a player EV of -0.381944778 . This seems much too bad for any casino game...
I understand that after the the player sees his cards he may either fold and lose the ante or put up a raise bet of 1x or 3x the ante, and that he will either lose the ante and the raise or win 1:1 on both the ante and the raise. I agree that the player should fold with less than a 25% probability of winning and do a 3x raise if he has more than a 50% chance of winning. Between 25% and 50%, he should just put up a 1x raise.
Edit: I agree with charliepatrick about the folding point. I changed the above to read "fold with less than 25% probability of winning." Also, raise 1x between 25% and 50% probability of winning.
February 23rd, 2012 at 6:33:23 AM
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Quote: TriplellI disagree. Raise x1 means you are getting paid at 2 to 1, which means that anything above 33% would be a good bet. Raise x2 means you are getting paid 3 to 2, which means that anything above 40% would be a good bet. And raise x3 means you are getting paid at 4 to 3, which means that anything above 43% would be a good bet. You're payout would increasingly approach the 50% odds, however never reach it (due to the fact that you are getting paid 1/1, however you must blindly submit 1 unit)....
My spreadsheet calculates the chances of winning. Now the question is what is the odds of being dealt each combination, and calculate your return.
I know for a fact that you should always fold if you are dealt a flush...
Well you would not fold with a flush if you had a side bet. Side bet will pay a flush. So you could loose your ante and bet but 100% win your side bet.
February 23rd, 2012 at 6:47:58 AM
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I would assume that side bets are resolved even if you fold the main bet.Quote: Carl1946Well you would not fold with a flush if you had a side bet. Side bet will pay a flush. So you could loose your ante and bet but 100% win your side bet.
It's a terminology thing.
"Fold" may imply to surrender the cards, into the discard rack.
The correct thing is to simply not make the Raise bet, but hold on to the cards.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 23rd, 2012 at 7:27:10 AM
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Well I get it as -6.46% if the player decides to fold when there's less than 33% chance of winning and -5.69% if using 25%.
One of things to remember is the dealer always qualifies and the payout is always Ante+Raise.
Raise 1 only, Raise 3 optimum, Raise if >33%.
1111 28561 364.000 000 3 7553.000 000 7202.000 000
211 158184 -4688.000 000 504.000 000 -64.000 000
22 36504 -5544.500 000 -5544.500 000 -5876.500 000
31 44616 -15051.000 000 -15051.000 000 -15891.000 000
4 2860 -2860.000 000 -2860.000 000 -2860.000 000
270725 -27779.500 000 -15398.500 000 -17489.500 000
-10.261 151% -5.687 875% -6.460 246%
One of things to remember is the dealer always qualifies and the payout is always Ante+Raise.
Raise 1 only, Raise 3 optimum, Raise if >33%.
1111 28561 364.000 000 3 7553.000 000 7202.000 000
211 158184 -4688.000 000 504.000 000 -64.000 000
22 36504 -5544.500 000 -5544.500 000 -5876.500 000
31 44616 -15051.000 000 -15051.000 000 -15891.000 000
4 2860 -2860.000 000 -2860.000 000 -2860.000 000
270725 -27779.500 000 -15398.500 000 -17489.500 000
-10.261 151% -5.687 875% -6.460 246%
February 23rd, 2012 at 7:53:23 AM
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Quote: ChesterDogI get a player EV of -0.381944778 . This seems much for too bad for any casino game...
I understand that after the the player sees his cards he may either fold and lose the ante or put up a raise bet of 1x or 3x the ante, and that he will either lose the ante and the raise or win 1:1 on both the ante and the raise. I agree that the player should fold with less than a 1/3 probability of winning and do a 3x raise if he has more than a 50% chance of winning. Between 1/3 and 50%, he should just put up a 1x raise.
Found my braino. Had an extra divide by 48. I now get the same EV. What a horrible bet! There must be more to it.
“Man Babes” #AxelFabulous
February 23rd, 2012 at 8:28:15 AM
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]Quote: charliepatrickWell I get it as -6.46% if the player decides to fold when there's less than 33% chance of winning and -5.69% if using 25%.
It doesn't make sense to put more money into the pot if you have less then a 33% chance of winning...ever.
If I bet $10, I would need to make a minimum $10 raise to play. My total payout would be $20 on a win.
The original $10 bet is dead. There is nothing you can do about it. The raise, worth $10, receives $20 on a win, which is 2 to 1, which corresponds to true odds of 33%.
The calculation of house edge, using optimal strategy, would be to calculate how often you would get a <33% hand.
Think of it like the Don't pass line in craps. The initial wager has -EV for the player. After the point is established, the bet has +EV.
This is the same way. Assuming that the probability of receiving a hand with <=33% chance of winning is >50%, then the casino will have +EV.
Drawing a flush only accounts for 3,681 hands of a possible 270,725, which is roughly 1.4%. The other 98.6% is a little bit more difficult to calculate.
February 23rd, 2012 at 8:46:31 AM
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My next finding is that if you receive all different suits, your probability of winning will be 33% or greater when your card values total is >=24. This is using Numbers as their values and J-A as 11-14 respectively.
>=50% seems to correspond with card values of 32. This all leads to the reasoning the casino limits your raise. The following table shows the raise and corresponding payouts:
>=50% seems to correspond with card values of 32. This all leads to the reasoning the casino limits your raise. The following table shows the raise and corresponding payouts:
| Raise | Payout | Odds |
|---|---|---|
| 1 | 2 | 33.33% |
| 2 | 3 | 40.00% |
| 3 | 4 | 42.86% |
| 4 | 5 | 44.44% |
| 5 | 6 | 45.45% |
| 6 | 7 | 46.15% |
| 7 | 8 | 46.67% |
| 8 | 9 | 47.06% |
| 9 | 10 | 47.37% |
| 10 | 11 | 47.62% |
| 20 | 21 | 48.78% |
| 50 | 51 | 49.50% |
| 100 | 101 | 49.75% |
| 1000 | 1001 | 49.98% |
February 23rd, 2012 at 9:07:05 AM
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Of the 270725 possible hands:
46150 are positive EV (raise x3)
9269 have 0 EV (atleast call)
159302 are negative EV but you lose less by calling
11765 have an EV of -.5 so you can either call or fold
44239 have an EV of less than -.5 so you should fold
46150 are positive EV (raise x3)
9269 have 0 EV (atleast call)
159302 are negative EV but you lose less by calling
11765 have an EV of -.5 so you can either call or fold
44239 have an EV of less than -.5 so you should fold
“Man Babes” #AxelFabulous
February 23rd, 2012 at 9:08:54 AM
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Quote: MrCasinoGamesHere is a Similar Game: Block
The Block bet is based on the player's first two cards and the dealer's up card. It can be found in Africa, South Africa, Egypt, Latvia, Estonia, Ireland, Morocco and Eastdend-EU. The bet wins if the dealer's up card matches the suit of one of the player's cards, and the player's card is higher. There are higher pays if the player's cards are a pair, suited, or both. Here is how the various winning hands are defined.Read More
Does the player bet an ante to play and a side bet on the odds of a pair, suited, etc. ?
February 23rd, 2012 at 9:24:05 AM
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Quote: mipletOf the 270728 possible hands:
Not sure where you got the extra 3 possible hands (52c4 is 270,725)
February 23rd, 2012 at 9:27:41 AM
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Copy and paste error :) fixxed previous post.
“Man Babes” #AxelFabulous
February 23rd, 2012 at 9:32:47 AM
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Where did your numbers come from? It doesn't make sense to me that you could have -EV but benefit from calling, unless you are saying you have less then 50% to win, but making a x1 bet makes you lose less, in which case I have already showed that anything above or equal to 33% would have positive EV.
February 23rd, 2012 at 9:35:01 AM
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Quote: Carl1946Quote: MrCasinoGamesHere is a Similar Game: Block
The Block bet is based on the player's first two cards and the dealer's up card. It can be found in Africa, South Africa, Egypt, Latvia, Estonia, Ireland, Morocco and Eastdend-EU. The bet wins if the dealer's up card matches the suit of one of the player's cards, and the player's card is higher. There are higher pays if the player's cards are a pair, suited, or both. Here is how the various winning hands are defined.Read More
Does the player bet an ante to play and a side bet on the odds of a pair, suited, etc. ?
Block is a blackjack sidebet.
February 23rd, 2012 at 9:40:16 AM
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As I wrote before I first saw the game around 8yrs ago at a casino down town off the strip. I tried to understand it then but was just looking around and moved on. I then saw it in an AZ Indian casino on my way back from Mexico a year later but not sure if it was played the same although the side bets started with any pair paid 1.5:1. I recently played it for several hours at a private card room with the rules as I remember when I first described them in the begining of this discussion but the side bets started with a pair of Js that paid 2:1 .
What I am trying to figure out is what is the house advantage of the basic ante/play bet. When i played just the basic bet for a period of time I lost more $ than won $ but when I played the basic ante/play and a side bet each hand for a period of time I won more $ overall.
What I am trying to figure out is what is the house advantage of the basic ante/play bet. When i played just the basic bet for a period of time I lost more $ than won $ but when I played the basic ante/play and a side bet each hand for a period of time I won more $ overall.
February 23rd, 2012 at 10:34:38 AM
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I thought I found it, but the game seems a little different. It's called "High Survivor"...
Here is the patent: http://www.patentgenius.com/patent/7568701.html
However, in this game, it looks as if you play against the other players at the table, and the dealer's only purpose is to take a "rake" and determine the suit...
Here is the patent: http://www.patentgenius.com/patent/7568701.html
However, in this game, it looks as if you play against the other players at the table, and the dealer's only purpose is to take a "rake" and determine the suit...
