Fire Bet in Craps

If I read his scenario correctly, he was looking for the odds of hitting the remaining four after hitting 4 and 10, as well as the odds of the remaining four after hitting 6 and 8.

i witnessed a firebet going for the 6th point at redrock casino last night.

three players had $5 on the firebet.

the 5 points already made were 4,5,6,8,9.

so all that was needed was a 10.

they hit the 10 to become the point but ended up sevening out.

so starting from the comeout roll on this hand they were 23:1 to get paid 1000:1 on their firebet.

3/24 to establish 10 as a point x 3/9 of making the 10 as the point.

1/8 x 1/3 = 1/24 or 23:1

Just to be nit picky, I suppose.... I think their chances were better than 1 in 24 at that point. In addition to the opportunity you describe, they also had the (perhaps slim) possibility of setting and making one or more other points for a second (third, fourth, etc.) time and finally setting and making the 10. Your 3/24 fraction rules out that possibility.

It also rules out the far more likely possibility of establishing one of the other points, and seven out on that point.

So arent the odds LOWER than 1 in 24?


Hmmm... Actually, I think it is correct to rule out both possibilities. There's a 1 in 24 chance of "X", and 23 in 24 of something else.

The chance of hitting the 10 then 7 out, or hitting something else or 7 out or hitting something else and making that point are ALL covered in the 23 in 24 'other possibilities'.

No, I think rudeboyoi was correct that there is a 1 in 24 chance of setting and making the 10 (sixth fire bet point) as the very next point that is set. I think I was correct that there are additional opportunities of setting and making the 10 after having made additional points. Combined, this gives something greater than 1 in 24, but it is far from easy to calculate.

I agree that things mentioned in your last paragraph are covered in the 23 of 24 "other possibilities." However, some of those other possibilities include making the sixth fire bet point. That is why the chance of the big win is greater than 1 in 24, given that all of the points except the 10 have already been made.

On the come out roll there are three possible outcomes at this point.

1. Sevening out.
2. Repeating a point already made (4 to 9).
3. Rolling a 10 on the come out roll, and then making it.

We need to quantify the second and third probabilities only. The shooter will eventually make a point, and then eventually make it or seven out. The probability that the point established and then made is 4 to 9 is:

(3/24)*(3/9) + (4/24)*(4/10) + (5/24)*(5/11) + (5/24)*(5/11) + (4/24)*(4/10) = 0.364394.

The probability of establishing a 10 point and then making it is (3/24)*(1/3) = 0.041667.

Let p be the probability of making a 10 point before sevening out. If the player makes any other point he is right back to where he started from. So...

p = 0.364394 × p + 0.041667
p × (1-0.364394) = 0.041667
p = 0.041667/(1-0.364394)
p = 0.065554

Quote: Doc

... but it is far from easy to calculate.

Well, it was far from easy for ME to calculate. Then the Wizard went and made it look almost trivial. (I hate it when that happens.)

Quote: Wizard

p = 0.065554

So the Wizard states that the last point, the 10 and its conversion is 15.25 to 1. (1 divided by .065554)

Interesting -- does anyone check his work??

Do you suppose that a Ph D candidate could do the research? How long would it take to validate that conclusion? and
Would a Ph D be awarded for such research?

Just wondering.

tuttigym

Could someone show the calculations again to determine how much to lay against the 5th and 6th points? I was playing last weekend and the shooter established point number 5, which happened to be the 5. I had $1 on the fire bet, so I laid $120, with $4 vig to win $80.

Quote: tuttigym

So the Wizard states that the last point, the 10 and its conversion is 15.25 to 1. (1 divided by .065554)

Interesting -- does anyone check his work??.....

Just wondering.

tuttigym

Yep, I did. It is exacly 55/839.