What is the price 'break even' point in this game?
February 9th, 2013 at 5:18:58 PM
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I was playing with a new game of chance.. and its pretty much pure chance. I start out with a regular deck of cards and remove all the 10's. With the remaining cards the face cards are worth 0 and the ace's are worth 1. I take the remaining 48 cards- and through a process of elimination and the player getting to see an occiasional card to help their decision making we get down to the final 4 cards. The final 4 cards are revealed and put in order of lowest card to highest. And they are paid the amount that this comes out to.
So if they had a K-Q-8-7 they receive $78. If they have 2-2-4-A they get $1,224. All face cards get you $0. Of course 'perfect' would be all four 9's for $9,999.
I'm trying to found out at what price does the payback even out and in the long run you'd expect to have the same as when you started. 0% edge both ways.. and I can play with the numbers from there. Assuming its whole dollars and you could leave with anywhere from $0 to $9,999.
So if they had a K-Q-8-7 they receive $78. If they have 2-2-4-A they get $1,224. All face cards get you $0. Of course 'perfect' would be all four 9's for $9,999.
I'm trying to found out at what price does the payback even out and in the long run you'd expect to have the same as when you started. 0% edge both ways.. and I can play with the numbers from there. Assuming its whole dollars and you could leave with anywhere from $0 to $9,999.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 9th, 2013 at 5:57:44 PM
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I'm not sure there is an easy solution to this. I would try to run the 270,725 combinations of the last 4 cards and evaluate via "brute-force". This is assuming you didn't see any cards that will not be part of the final four.
February 9th, 2013 at 6:49:37 PM
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I don't have the math skills for a an exact answer but it will average close to the low end. 4 out of 9 times you will have a 1 which means the payout will be 1,xxx. The other 4 numbers are not near as important. 8 out of 9 times the last 4 cards will contain an A or 2 meaning the pay out maximum payout is 2,xxx 88% of the time.
Be careful when you follow the masses, the M is sometimes silent.
February 9th, 2013 at 7:26:45 PM
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if an A or a 2 is 8 out of 9 times- where does that leave from for the 12 face cards? - surely its not 9 out of 9 times youd have all of them...
if that's 12 out of the 48 cards in 0's alone- that's 1/4 of the deck. 4 cards drawn- is it really that close to 100% of the time you'd have at least one 0 value card immediately limiting your expectations to less then $1,000?
and 8 out of 9 times you have a 1 or 2 - meaning we are now limiting ourselves to less then $300 87% of the time.
Suddenly this is looking like a feasable gambling game. (small bet for small chance of large payoff, and a large chance of mediocre payoff)
if that's 12 out of the 48 cards in 0's alone- that's 1/4 of the deck. 4 cards drawn- is it really that close to 100% of the time you'd have at least one 0 value card immediately limiting your expectations to less then $1,000?
and 8 out of 9 times you have a 1 or 2 - meaning we are now limiting ourselves to less then $300 87% of the time.
Suddenly this is looking like a feasable gambling game. (small bet for small chance of large payoff, and a large chance of mediocre payoff)
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 9th, 2013 at 7:42:58 PM
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Quote: Malaruand through a process of elimination and the player getting to see an occiasional card to help their decision making we get down to the final 4 cards.
If you ignore this part we can calculate the exact EV of the game. It's equivalent to "deal the first 4 cards" and sort then pay, by your rules. This could be evaluated exactly, by brute force or by simulation.
If this decision making process is important though, it throws all the math out the window.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
February 9th, 2013 at 7:48:15 PM
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Quote: dwheatleyIf you ignore this part we can calculate the exact EV of the game. It's equivalent to "deal the first 4 cards" and sort then pay, by your rules. This could be evaluated exactly, by brute force or by simulation.
If this decision making process is important though, it throws all the math out the window.
For the purpose of this question it is not important- act as if it dont even occur. I just want to know the break even point in a pure by random/luck point of view.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 9th, 2013 at 8:22:05 PM
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Just written a quick sim for this :
// Define the card list
cardList = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9]
cSize = cardList.size()
value = 0
int runTimes = 1000000
(1..runTimes).each{
// Shuffle Cardlist in place, use Fisher-Yates
Random randomGen = new Random()
int i = 0
for(i = (cSize-2); i>0; i--){
j = randomGen.nextInt(i)
temp = cardList
cardList = cardList
cardList= temp
}
// Sort the first four cards into a seperate list
valueList = cardList[0..3].sort()
// Work out value of the hand
value += valueList[0]*1000+valueList[1]*100+valueList[2]*10+valueList[3]
}
println value/runTimes
I may have screwed it up. Don't claim it's efficient code either.
1 million runs gave 936.26 as the average hand value. Second time gave 935.07. So somewhere around 932-938, I'd guess.
Can run it for longer if you need a more precise answer."Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
February 9th, 2013 at 8:36:56 PM
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It just sounds high. Did it take into account the reordering of the cards to place the lowest denomination first and highest last?
AKA- 0-4-5-1 = 0145 and 3454 = 3445
AKA- 0-4-5-1 = 0145 and 3454 = 3445
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 9th, 2013 at 8:42:27 PM
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I did a sort on the first 4 cards.
Just checked, looks about right to me. 12 0's in the deck, chance of getting a 0 is about 70%. So 70% of hands will be less than $1000, but when you don't have a leading 0, the rest will be worth a lot more.
I could well be wrong though.
Just checked, looks about right to me. 12 0's in the deck, chance of getting a 0 is about 70%. So 70% of hands will be less than $1000, but when you don't have a leading 0, the rest will be worth a lot more.
I could well be wrong though.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
February 9th, 2013 at 8:54:58 PM
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very iiinnntttteeerrreeesssttttiiinnnnggggg....... so with a 930 avg- if the house asked for a bet of 1 unit to win 10 units with a payback of .93 units; that would make the house edge 7-8%? Or am i way way off on that one lol.
If I consider it as 7-8%- and then consider the small adjustments that can be made both with the player input (getting to decide on a card a couple times) plus a small adjustment to payouts here or there... I could slim that house edge down to a reasonable game. Target of 2-3%.
It could be offered in several different price amounts- simple math ones would be feasible such as $1, $2, $5, $10, $20, $50, $100, $200, $500, & $1,000
as it is betting 1 to win 10 or 10 to win 100 don't sound too exciting.. but there are possibilities such as bonus amounts for 3 of a kind in final 4 or suited final four or whatnot. Ill have to play with the idea some.
Input, thoughts, ect welcomed- this is just a game I have kicked around a for a bit in a simple form.
If I consider it as 7-8%- and then consider the small adjustments that can be made both with the player input (getting to decide on a card a couple times) plus a small adjustment to payouts here or there... I could slim that house edge down to a reasonable game. Target of 2-3%.
It could be offered in several different price amounts- simple math ones would be feasible such as $1, $2, $5, $10, $20, $50, $100, $200, $500, & $1,000
as it is betting 1 to win 10 or 10 to win 100 don't sound too exciting.. but there are possibilities such as bonus amounts for 3 of a kind in final 4 or suited final four or whatnot. Ill have to play with the idea some.
Input, thoughts, ect welcomed- this is just a game I have kicked around a for a bit in a simple form.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 10th, 2013 at 4:46:15 AM
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A $100 bet can get you $999.90
A $10 bet can get you $99.99
A $1 bet can get you $9.99 9/10
Always round up to the nearest dollar. Witha 7 or 8 % HE it won't bite into it that much, as the end numbers are by definition the highest.
Problem.... every hand will be paid out, and in unusual amounts... making it very slow to deal....
To spice it up.... any payment over $500 (on the $100 bet, as an example) is paid double. Any hand with a picture is a loser (paid 0).
Good luck.
A $10 bet can get you $99.99
A $1 bet can get you $9.99 9/10
Always round up to the nearest dollar. Witha 7 or 8 % HE it won't bite into it that much, as the end numbers are by definition the highest.
Problem.... every hand will be paid out, and in unusual amounts... making it very slow to deal....
To spice it up.... any payment over $500 (on the $100 bet, as an example) is paid double. Any hand with a picture is a loser (paid 0).
Good luck.
February 10th, 2013 at 7:32:52 AM
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Pay only the first digit.
So 2359 pays 2 units. 0099 pays zero. Add a bonus for 9999, or possibly any triple, and you might have something?
I can re run a simulation with tweaked rules, and/or send list of results for distribution analysis.
So 2359 pays 2 units. 0099 pays zero. Add a bonus for 9999, or possibly any triple, and you might have something?
I can re run a simulation with tweaked rules, and/or send list of results for distribution analysis.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
February 10th, 2013 at 10:43:37 AM
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For those people that intuitively thought the value was high, remember any face card will average your hand in the lower hundreds, two face cards give you a hand in the teens. I didn't verify this, but the number sounds about right to me at least.
Hands with at least 1 face card = 1 - C(40,4)/C(52,4) = 0.6624
And more importantly...hands with at least 1 10-value card:
1 - C(36,4)/C(52,4) = 0.7824
Hands with at least 1 face card = 1 - C(40,4)/C(52,4) = 0.6624
And more importantly...hands with at least 1 10-value card:
1 - C(36,4)/C(52,4) = 0.7824
February 10th, 2013 at 11:51:18 AM
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Assuming Aces are low, pictures come first, and others cards in ascending order...
No Ten decks - (1) 981.91 (2) 1011.50 (4) 1026.12 (999999) 1040.61
[$0-$999: 69.727%, $1111-1999: 11.792% $2222+: 18.481% - so there's a fair chance of a nice prize]
Normal decks - (1) 745.82 (2) 770.51 (4) 782.75 (999999) 794.91
Initially I misread your rules and assumed it was normal cards put in order from Ace downwards.
Assuming Ace is high, then any Ace guarantees a payout of over $1000 even if the hand has a face card.
Next if the hand doesn't have an Ace AND has at least one Picture then the payout is less than £1000
Next, the remainder hands must exceed $2222.
So few hands are less than $1000.
With normal 52-card decks
(1 deck) 1751.83 (2 decks) 1784.20 (4 decks) 1800.00 (6 decks) 1805.21 (999999 decks) 1815.56
With no tens, which increases the payouts
2251.56, 2288.72, 2306.75, 2312.69, 2324.45
No Ten decks - (1) 981.91 (2) 1011.50 (4) 1026.12 (999999) 1040.61
[$0-$999: 69.727%, $1111-1999: 11.792% $2222+: 18.481% - so there's a fair chance of a nice prize]
Normal decks - (1) 745.82 (2) 770.51 (4) 782.75 (999999) 794.91
Initially I misread your rules and assumed it was normal cards put in order from Ace downwards.
Assuming Ace is high, then any Ace guarantees a payout of over $1000 even if the hand has a face card.
Next if the hand doesn't have an Ace AND has at least one Picture then the payout is less than £1000
Next, the remainder hands must exceed $2222.
So few hands are less than $1000.
With normal 52-card decks
(1 deck) 1751.83 (2 decks) 1784.20 (4 decks) 1800.00 (6 decks) 1805.21 (999999 decks) 1815.56
With no tens, which increases the payouts
2251.56, 2288.72, 2306.75, 2312.69, 2324.45
February 10th, 2013 at 3:18:51 PM
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I got around 930 with a single deck, so I suspect I made a mistake... how did you get to 981?
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
February 10th, 2013 at 5:23:17 PM
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Quote: charliepatrickAssuming Aces are low, pictures come first, and others cards in ascending order...
No Ten decks - (1) 981.91 (2) 1011.50 (4) 1026.12 (999999) 1040.61
[$0-$999: 69.727%, $1111-1999: 11.792% $2222+: 18.481% - so there's a fair chance of a nice prize]
Normal decks - (1) 745.82 (2) 770.51 (4) 782.75 (999999) 794.91
981.91 = 1.9% house edge. thats perfect.. unfortunalty as has been pointed out- the game is imperfect.. and needs some streamlining.. and I like the ideas provided.
In a bit Ill try to give a streamlined idea of the game.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 10th, 2013 at 6:59:39 PM
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Quote: charliepatrick
Normal decks - (1) 745.82 (2) 770.51 (4) 782.75 (999999) 794.91
I get 745.82 for one deck as well.
Specifically, 201913083 / 270725.
February 10th, 2013 at 7:29:20 PM
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table seats 7
Players make a wager and optional bonus wager (bonus wager must be less then or equel to play wager) and then every player is delt 4 cards.
After the players look at their 4 cards they are allowed to make a second (optional) 'discard' wager of equal value to the first, but must discard a single card by placing it next to this wager.
The dealer will pick up all discards and deal the players out one additional card to replace the discard. At this point the dealer will go through the players hands one by one- revealing the hand and will payout accordingly
the hand ranks are from highest to lowest 9-8-7-6-5-4-3-2-A-J-Q-K (J's, Q's and K's make the hand a loser, and an Ace is low)
The player will be paid according to the cards they have plus any bonuses applicable
The payouts will be: (initial bet - discard bet if made)
Any hand with a face card = loss - loss
Lowest card in the hand is a:
A or 2 = 1x - push
3 or 4 = 3x - 1x
5 or 6 = 5x - 1x
7 or 8 = 10x - 2x
All 9's = 50x - 50x
Bonuses: (These pay out regardless of if there are face cards)
3 of a kind 1x
Straight 2x
Flush 3x
4 of a kind 50x
Straight Flush 100x
Royal Flush 1000x
If the bonus is not paid it is a loss. However the initial bet does not need to win for the bonus to win.
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The following are random examples:
Player bets $100 on initial bet and $10 on the bonus bet, Their cards are: 8c, Kc, 4d, As- they elect to make the second bet to discard the Kc-
they get the Qs - so their final hand is a loss. no bonus is payable- they loose $210.
Player bets $20 on initial bet and $20 on bonus- cards come out- 5s, 8d, kd, 7h, they elect to make the discard wager and place another $20 to
get rid of the Kd, they get the 6c. As for main and discard bets- they have a 5 low- so they get paid $100 on initial and $20 on discard- the bonus
bet gets paid $40. Player turns $60 in bets into $220
Player bets $10 on initial bet, and elects not to play bonus- the initial cards are K-J-J-9; elects not to discard a card because there is no way to
make a winning hand.
Player places $50 on initial bet, and $20 on bonus bet- initial cards are Q-Q-Q-Q - player elects ot stay with these cards and looses the initial
bet- but wins $1,000 on bonus bet.
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I did 5 rounds by hand for 7 players each round with these rules and the 7 players betting $10 a hand on initial and $5 a hand on bonus- electing
to take a discard if they were 1 card away from a paying hand (and not for bonus potential) and my 7 players ended up:
One player won a bonus (3 of a kind) - though the 4's three landed a hand away from giving one of them quad-3s. End result for each position:
p1- $25
p2- ($15)
p3- ($25)
p4- $85
p5- $25
p6- $25
p7- $35
HOUSE- ($155)
I know that 5 rounds means less then a grain of sand on the beach- but I felt like it was paying off too easily. What if I make the bonus bet required and must match the initial bet? That should rub off a lot of that player advantage. In fact if I recalculate the results with this rule change it would look like this:
p1- $10
p2- ($40)
p3- ($50)
p4- $60
p5- $0
p6- $0
p7- $10
HOUSE- $10
I dont know if thats a fair balance or if its too harsh. Basicly I had the players draw another card if they were 1 card away from having a winning hand. The lone exception was when player 1 had A-3-3-3. I had him discard the A ideally to pickup the case 3 but figured if he gets anything from 3 to 9 would be good; and figured it was a good strategy play. He got a 6 I think. Also I could reduce the payout for 3-4 from 3x-1x to 2x-1x and that would help.
Also thinking of a name... '4 card 9' or 'Hi-9 poker' comes to mind.. but Im not good with names.
Players make a wager and optional bonus wager (bonus wager must be less then or equel to play wager) and then every player is delt 4 cards.
After the players look at their 4 cards they are allowed to make a second (optional) 'discard' wager of equal value to the first, but must discard a single card by placing it next to this wager.
The dealer will pick up all discards and deal the players out one additional card to replace the discard. At this point the dealer will go through the players hands one by one- revealing the hand and will payout accordingly
the hand ranks are from highest to lowest 9-8-7-6-5-4-3-2-A-J-Q-K (J's, Q's and K's make the hand a loser, and an Ace is low)
The player will be paid according to the cards they have plus any bonuses applicable
The payouts will be: (initial bet - discard bet if made)
Any hand with a face card = loss - loss
Lowest card in the hand is a:
A or 2 = 1x - push
3 or 4 = 3x - 1x
5 or 6 = 5x - 1x
7 or 8 = 10x - 2x
All 9's = 50x - 50x
Bonuses: (These pay out regardless of if there are face cards)
3 of a kind 1x
Straight 2x
Flush 3x
4 of a kind 50x
Straight Flush 100x
Royal Flush 1000x
If the bonus is not paid it is a loss. However the initial bet does not need to win for the bonus to win.
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The following are random examples:
Player bets $100 on initial bet and $10 on the bonus bet, Their cards are: 8c, Kc, 4d, As- they elect to make the second bet to discard the Kc-
they get the Qs - so their final hand is a loss. no bonus is payable- they loose $210.
Player bets $20 on initial bet and $20 on bonus- cards come out- 5s, 8d, kd, 7h, they elect to make the discard wager and place another $20 to
get rid of the Kd, they get the 6c. As for main and discard bets- they have a 5 low- so they get paid $100 on initial and $20 on discard- the bonus
bet gets paid $40. Player turns $60 in bets into $220
Player bets $10 on initial bet, and elects not to play bonus- the initial cards are K-J-J-9; elects not to discard a card because there is no way to
make a winning hand.
Player places $50 on initial bet, and $20 on bonus bet- initial cards are Q-Q-Q-Q - player elects ot stay with these cards and looses the initial
bet- but wins $1,000 on bonus bet.
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I did 5 rounds by hand for 7 players each round with these rules and the 7 players betting $10 a hand on initial and $5 a hand on bonus- electing
to take a discard if they were 1 card away from a paying hand (and not for bonus potential) and my 7 players ended up:
One player won a bonus (3 of a kind) - though the 4's three landed a hand away from giving one of them quad-3s. End result for each position:
p1- $25
p2- ($15)
p3- ($25)
p4- $85
p5- $25
p6- $25
p7- $35
HOUSE- ($155)
I know that 5 rounds means less then a grain of sand on the beach- but I felt like it was paying off too easily. What if I make the bonus bet required and must match the initial bet? That should rub off a lot of that player advantage. In fact if I recalculate the results with this rule change it would look like this:
p1- $10
p2- ($40)
p3- ($50)
p4- $60
p5- $0
p6- $0
p7- $10
HOUSE- $10
I dont know if thats a fair balance or if its too harsh. Basicly I had the players draw another card if they were 1 card away from having a winning hand. The lone exception was when player 1 had A-3-3-3. I had him discard the A ideally to pickup the case 3 but figured if he gets anything from 3 to 9 would be good; and figured it was a good strategy play. He got a 6 I think. Also I could reduce the payout for 3-4 from 3x-1x to 2x-1x and that would help.
Also thinking of a name... '4 card 9' or 'Hi-9 poker' comes to mind.. but Im not good with names.
"Although men flatter themselves with their great actions, they are not so often the result of a great design as of chance." - Francois De La Rochefoucauld
February 11th, 2013 at 12:19:55 AM
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Quote: thecesspitI got around 930 with a single deck, so I suspect I made a mistake... how did you get to 981?
My Fisher-Yates shuffle had a mistake in it, which stopped the last 9 being shuffled in. I now get around the 981 mark.
Sorry for my incorrect answer. I should add the F-Y shuffle to my libraries.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
February 11th, 2013 at 6:23:39 PM
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Quote: MalaruI was playing with a new game of chance.. and its pretty much pure chance. I start out with a regular deck of cards and remove all the 10's. With the remaining cards the face cards are worth 0 and the ace's are worth 1. I take the remaining 48 cards- and through a process of elimination and the player getting to see an occiasional card to help their decision making we get down to the final 4 cards. The final 4 cards are revealed and put in order of lowest card to highest. And they are paid the amount that this comes out to.
This has game show written all over it. (Let's Make A Deal, perhaps?)
What I want to know is: Why show the player cards that don't make up their final payment? Does the player get an offer along the way to bail prior to seeing their final win?
'Brian
