September 14th, 2016 at 9:14:33 AM
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Available at one table, $5 minimum.

When the dice are going to a new shooter, you have the opportunity to wager that he will make 4 consecutive rolls without a 7. Wager must be made prior to new shooter's initial comeout roll. Even money payout when it hits.

This seems like over time it should be a winner. Am I missing something?

When the dice are going to a new shooter, you have the opportunity to wager that he will make 4 consecutive rolls without a 7. Wager must be made prior to new shooter's initial comeout roll. Even money payout when it hits.

This seems like over time it should be a winner. Am I missing something?

September 14th, 2016 at 9:31:08 AM
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Yes. The probability of throwing four rolls without any 7s is (5/6)

In every 1296 bets, you are expected to win 625 and lose 671, for a HE of 46 / 1296 = 3.55%.

^{4}= 625 / 1296.In every 1296 bets, you are expected to win 625 and lose 671, for a HE of 46 / 1296 = 3.55%.

September 14th, 2016 at 9:36:13 AM
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It's one of the oldest "new" wagers around. It's a game called "Four the Money" by Naif Moore, Jr., from 1995:

https://www.google.com/patents/US5829748

From the patent, the house edge is 3.55%, which is correct:

p(no 7) = 5/6

p(no 7 four times) = (5/6)^4 = .48225

p(lose) = 1-.48225 = .51775

The edge is the difference or 3.55%.

https://www.google.com/patents/US5829748

From the patent, the house edge is 3.55%, which is correct:

p(no 7) = 5/6

p(no 7 four times) = (5/6)^4 = .48225

p(lose) = 1-.48225 = .51775

The edge is the difference or 3.55%.

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563

September 14th, 2016 at 9:43:25 AM
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Thanks, guys.

September 18th, 2016 at 12:40:42 AM
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Really is amazing that in all the years since introduction, no one else has it. All Tall Small certainly spread like wildfire and I believe Sam's town introduced it also about the same time.

When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.

September 18th, 2016 at 6:36:03 AM
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While it's easy to think of all the shooters that have at least 4 rolls, it's also very easy to forget that many of them roll a come-out seven. That's what you're missing.Quote:iamnomadThis seems like over time it should be a winner. Am I missing something?

When you thing about it, it's not really that amazing. The All/Tall/Small and Fire bets are similar to this in that you make them before the shooter's first roll. The difference is the potentially large payout.Quote:DeMangoReally is amazing that in all the years since introduction, no one else has it. All Tall Small certainly spread like wildfire and I believe Sam's town introduced it also about the same time.

The even money payout of the 4 in a row bet isn't all that compelling to a bettor. And therefore not compelling to a casino either.

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? 😁

September 18th, 2016 at 11:52:06 AM
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Quote:MathExtremistIt's one of the oldest "new" wagers around. It's a game called "Four the Money" by Naif Moore, Jr., from 1995:.

Is that what they're calling it at Sam's Town?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

September 18th, 2016 at 12:10:08 PM
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It seems like it'd be tough to make a bet that players would like based on # of no-7's.

1/((5/6)^20) = 38.3375999245

That's 20 rolls without a 7. 1 in 38.33 shooters will roll 20 no-7's.

Who's gonna bet money (even at 37-to-1, close to fair odds) that someone will get 20 rolls without a 7? It's damn near fair, but seems like a shooter going 20 rolls w/o a 7 is much harder than 1 in 38.33.

Maybe 30 rolls (1 in 237.37) at 220-to-1 payout could work (7.177% HE, I reckon), with a nice payout to induce players to make the bet. Of course, one problem with a high payout wager, where everyone would win at the same time, if there's $100 in action on the bet, that's $22k loss for the casino (not to mention, they're probably losing quite a bit for a 30-hand roll)....not something the casino may want to put up with (variance), not to mention, now you need someone (floor/boxman) to keep track of rolls (like they do at Fremont or one of those downtown stores).

1/((5/6)^20) = 38.3375999245

That's 20 rolls without a 7. 1 in 38.33 shooters will roll 20 no-7's.

Who's gonna bet money (even at 37-to-1, close to fair odds) that someone will get 20 rolls without a 7? It's damn near fair, but seems like a shooter going 20 rolls w/o a 7 is much harder than 1 in 38.33.

Maybe 30 rolls (1 in 237.37) at 220-to-1 payout could work (7.177% HE, I reckon), with a nice payout to induce players to make the bet. Of course, one problem with a high payout wager, where everyone would win at the same time, if there's $100 in action on the bet, that's $22k loss for the casino (not to mention, they're probably losing quite a bit for a 30-hand roll)....not something the casino may want to put up with (variance), not to mention, now you need someone (floor/boxman) to keep track of rolls (like they do at Fremont or one of those downtown stores).

September 18th, 2016 at 2:02:28 PM
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Quote:WizardIs that what they're calling it at Sam's Town?

Wiz, I don't think it has a fancy name, just something like "4 rolls no 7", box is right in the center of the table. Again, offered at only 1 table.

February 21st, 2020 at 11:39:13 PM
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Quote:ThatDonGuyYes. The probability of throwing four rolls without any 7s is (5/6)

^{4}= 625 / 1296.

In every 1296 bets, you are expected to win 625 and lose 671, for a HE of 46 / 1296 = 3.55%.

Would parlaying a win increase the house edge?

February 22nd, 2020 at 9:16:59 AM
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no,Quote:BigJWould parlaying a win increase the house edge?

the question shows a very possible non-understanding of what exactly a 'house edge' is.

and is equating house edge with expected value where they are different

(values for each in an example could be the same)

The OP 'thought' that this bet is beatable.

"This seems like over time it should be a winner."

the word 'seems' is a weasel word

this is exactly (a simple version) how the casino makes it's money on casino games offered.

they (casino) short pay you on a win based on the probabilities of winning

most have NO CLUE as to the probability of winning

take a fair coin flip

p=(1/2);\\probability of winning

q=1-p;

fair=q/p;

fairDec=1.*fair;

fair+1

fairDec+1

on a 1 unit wager, one SHOULD get back 2 units on a win (including the original wager)

for this to be a fair wager

that formula is simply q/p in the example

gp > p=(1/2);

gp > q=1-p;

gp > fair=q/p;

gp > fairDec=1.*fair;

gp > fair+1

%5 = 2

gp > fairDec+1

%6 = 2.0000000000000000000000000000000000000

IF, on a win, you get back 2 units

the difference 2-2=0 is the short pay (0)

he=difference/total fair return = 0/2 = 0%

this wager

p=(5/6)^4;

q=1-p;

fair=q/p;

fairDec=1.*fair;

fair+1

fairDec+1

should pay IF fair 1296/625 or 2.0736

the casino pays only 2 (including the original wager)

gp > p=(5/6)^4;

gp > q=1-p;

gp > fair=q/p;

gp > fairDec=1.*fair;

gp > fair+1

%11 = 1296/625

gp > fairDec+1

%12 = 2.0736000000000000000000000000000000000

the casino pays you 1250/625 or (2)

46/625 (0.0736) IS the short pay amount

IF, on a win, you get back ONLY 2 units (sure, the casino rounds down...seems FAIR)

he=difference/total fair return = 0.0736/2.0736

or (46/625)/(1296/625)= 46/625 * 625/1296 = 46/1296 = 23/648

some want the percentage: 3.5493827160493827160493827160493827161%

the expected value of a 1 unit wager = 23/648 *1

the expected value of a 3 unit wager = 23/648 *3

you should now be able to answer you own question.

the ev changes for multiple different wagers, but not the house edge)

winsome johnny (not Win some johnny)