…………PAYS…………………………………………
BET …(FOR 1) …PROBABILITY …HOUSE EDGE
..2 ………40 ……..0.020408 ……..0.183673
..3 ………50 ……..0.015625 ……..0.218750
..4 ………65 ……..0.012346 ……..0.197531
..5 ………80 ……..0.010240 ……..0.180800
..6 ………90 ……..0.008820 ……..0.206209
..8 ………90 ……..0.008820 ……..0.206209
..9 ………80 ……..0.010240 ……..0.180800
.10 ……..65 ……..0.012346 ……..0.197531
.11 ……..50 ……..0.015625 ……..0.218750
.12 ……..40 ……..0.020408 ……..0.183673

As with any "specialty" bet that a casino offers to go along with the basic game, the house edge is always much much higher. Rose and Doc simply avoid all of these bets whether they be fire bets, ATS, PaiGow fortune prop bets, million dollar bets on Three card poker, prop bets at BJ, etc. We always stick with the low edge bets. Hope that others will as well.

Doc,
Agreed. The money that we spend on entertainment bets is better served being placed on our best edge box number bet or odds on the Come bet. That is what will grow our bankrolls. Skip the entertainment/gambling bets and use the low edge smart bets—that is advantage play and the GTC way!

actually the 3 and 11 only pay 50 to1

That makes them an even worse bet with a house edge of 0.21875. I changed the table above accordingly.

Skinny I am missing some thing here . With the chart the edge is less then one and on the pass line bet the casino has a greater edge you are saying ? It just seems to me that this bet should have a greater edge for the casino .Or am I looking at it wrong .
I am not a math guy but why would the casino change from a bad bet to a better bet . I thought the fire gave them a great edge along with the STA bet .

The chart does not express the house edge in percentages but the actual number.

To change it to a percentage you must multiply it by 100.

For example, 0.183673 = 18.37%

Now I see that . Boy I thought that I was missing some thing .I had no idea that I needed to do the Xs by 100 .Did think that it was a good bet for the casinos tho .
Good Rolling. 😀

What is the probability of throwing the hard 6 or 8, 2 times before a seven or a 6 or 8 and the same for the 4 and 10, please skinny.

There are 6 ways to make a seven and 4 ways to make a soft six or eight with 1 way to make it hard. With 1 way to win and 10 ways to lose, there is 1 way out of 11 To win a bet on the hard six or eight. The probability of winning it once is 1/11=.090909

The probability of winning it twice is (1/11) X (1/11) = 1/121 = 0.008264.

Since there are 2 ways to make a soft four or ten do you think you can figure out the probability of winning a hard four or ten once and twice? Follow what I did for the hard six or eight.

So, if you toss a come out 7 you lose your Repeater bet? Can you wait until the Come Out sequence is over before placing the bet? Can you place the Repeater bet at any time during the hand or just at the beginning of a hand? Just a few questions that came to me regarding this feature bet. Inquiring minds and all that!

"Skinny" wrote: There are 6 ways to make a seven and 4 ways to make a soft six or eight with 1 way to make it hard. With 1 way to win and 10 ways to lose, there is 1 way out of 11 To win a bet on the hard six or eight. The probability of winning it once is 1/11=.090909
The probability of winning it twice is (1/11) X (1/11) = 1/121 = 0.008264.
Since there are 2 ways to make a soft four or ten do you think you can figure out the probability of winning a hard four or ten once and twice? Follow what I did for the hard six or eight.

So, following tit for tat what you did above for the hard six or eight,
There are 6 ways to make a seven and 2 ways to make a soft four or ten with 1 way to make it hard.
With 1 way to win and 8 ways to lose, there is 1 way out of 9 to win a bet on the hard four or ten.
The probability of winning it once is 1/9 = 0.111111
The probability of winning it twice is (1/9) X(1/9) = 1/81 = 0.0123456

Very well done.

A+

You can place the bet at any time up until it has hit once at that time the bet becomes closed to new action assuming that someone was on it if no one is on it then the count for the number stays at 0 and only starts after someone makes the bet, and the number has hit once up until that point it is still open to new action.

Now lets assume that you open with a 6 that would mean that you would have to throw the 6 seven times to get paid I am sure that jumps the probability greatly.

Consider this at least on the come out seven you win on your line bet, so if you are to play this bet at lest make sure your line bet is enough to cover the repeater bet.

Bman,

In your example, you imply that the come out six doesn’t count. I’m fairly certain that you are wrong on that. Are you saying that two come 12s would not count toward a bet on the 12?
But a comet seven loses the bet. Logic says that doesn’t make any sense.

I may be wrong but after seeing it at Gold Coast last trip I had a roll where someone made $2,250 off of my roll with a $10 bet on each of the numbers.

Noah

Noah,

I don’t think he means it that way. My reading of his explanation is that if no one is betting on a number it does not matter how many times you throw that number towards the repeater bet. It also does not matter how many come out sevens you throw regarding the repeater bet on a number that does not have any bets on it.

But once a repeater bet is placed on a number, that number is live and from that point forward you need to throw the number the specified number of times before a seven to win your repeater bet.

Bman,

Regarding your question about coming out with a 6 if you do not have a repeater bet on it. But after the come out decide to place a repeater bet on the 6. The odds on that bet do not change from the original table.

Let’s look at how I determined the probabilities for the 6 for example. By the way, from that you should be able to see how to calculate the probabilities for this bet with the other numbers.

There are five ways to make a 6 and six ways to make a 7. Therefore the probability of throwing a 6 before a 7 is 5/11. Five ways to win the bet out of the eleven throws
(five 6’s and six 7’s) (5 + 6 = 11) that are relevant to the bet.

In order to win the bet one must throw six 6’s before throwing a 7. Therefore the probability of doing this six times is the product of doing it once times itself six times, ie.
(5/11) X (5/11) X (5/11) X (5/11) X (5/11) X (5/11) = 15625/1771561 = 0.008820.

It does not matter how many 6’s you throw before making the bet. The odds of your throwing a 6 before a 7 are always 5/11 on each and every throw. The dice do not have any memory. They do not know that you threw a 6 before making a repeater bet on the 6. The only thing that matters is every time you prepare to throw the dice, there is a 5/11 chance of throwing a 6 instead of a 7. Once you make the bet, that number becomes live and that is the point at which you determine the probability of doing it six times in a row.

In fact, the probability of establishing 6 as your point number is 5/(3+4+5+5+4+3) = 5/24. Five ways to throw a 6 out of the 24 ways to establish any point number. So if no one has a repeater bet on the 6 what does the fact that you established it as your point have to do with anything? Furthermore, the probability of throwing a 6 on any given roll is 5/36. Five ways to throw a 6 out of the 36 ways to throw any possible number from 2 thru 12. If you establish the 6 as your point number, the chance of throwing a 6 on the very next roll is still 5/36. The probability of throwing any number on any given roll does not change based on what was thrown before that roll.

Kind of off topic but…Damn Skinny! I love reading your posts. I learn something valuable every time I do. Keep posting. Hope to meet you some day at a GTC event.

Yeah Skinny, rereading Bmans post, I think you are right in what he is saying.

Noah

Thank you Skinny that makes perfect sense to me.

Good looking out acpa.