Ace2 Joined: Oct 2, 2017
• Posts: 1041
September 6th, 2018 at 7:26:19 PM permalink
I also calculated the Fire Bet with the Poisson process. This method is much less tedious than any other method I've seen.

The probability of winning with all 6 points is the integral from zero to infinity of:

((1 - e^(-x/a)) * (1 - e^(-x/b)) * (1 - e^(-x/c)))^2 * e^(-x/s) / s dx

Where:

a = 6,684 / 55 = average rolls to hit a point-8 winner
b = 1,671 / 22 = average rolls to hit a point-9 winner
c = 6,684 / 125 = average rolls to hit a point-10 winner
s = 1,671 / 196 = average rolls to seven out

The exact answer is 3700403899126040038831518494284887738125 / 22780863797678919004236184338193605974839452

=~ 0.000162.
It�s all about making that GTA
ChumpChange Joined: Jun 15, 2018
• Posts: 2356
December 11th, 2019 at 1:52:31 AM permalink
I've played at a table for 4 hours a couple times and the Fire Bet hit for 4 or 5 numbers. It doesn't mean the table wasn't cold; it's just one shooter broke through the ice.
DeMango Joined: Feb 2, 2010
• Posts: 2767
December 11th, 2019 at 9:50:18 AM permalink
The demise of the FireBet is that many gamblers realize how the odds are stacked against them. The demise of the ATS in Biloxi are the losses the casinos have sustained. First lower payouts. Then higher minimums. Some just say fugedaboutit. Three casinos now entirely gone.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Wizard Joined: Oct 14, 2009
• Posts: 23279
March 23rd, 2020 at 10:44:22 AM permalink
I just created a separate page for the which has much more information on how to analyze it.

As always, I welcome all comments.
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2 Joined: Oct 2, 2017
• Posts: 1041
March 23rd, 2020 at 2:47:51 PM permalink
My comments on the Calculus section of the Fire Bet page:

First off, I don't understand why you mention the light bulb problem. A large part of this section is an attempt to the explain the fire bet in terms of the light bulb problem, which is unnecessary and confusing IMO. Though they both use the Poisson process, they are not even the same basic problem/concept..one calculates average time for an event to happen and the other calculates the probability of an event ever happening in a single trial.

The formula:

(1-exp(-x/24))^2*(1-exp(-x/15))^2*(1-exp(-25x/264))^2*exp(-98x/165)/(165/98)

you posted about 2/3rd's down the section is correct, but it did not copy correctly into your large picture of the formula (the last term in the numerator shows (-e^(-x/24)) when it should be (*e^(-98x/165))

Your explanation of the formula is:
Quote: Wizard

Socket 4 light bulb has NOT burned out.
Socket 5 light bulb has NOT burned out.
Socket 6 light bulb has NOT burned out.
Socket 8 light bulb has NOT burned out.
Socket 9 light bulb has NOT burned out.
Socket 10 light bulb has NOT burned out.
Socket 7 light bulb HAS burned out.

Once again, I don't see why light bulbs are mentioned.

Anyway, you actually have this backwards. The formula says: what is the probability at any given time that all bulbs (points) 4,5,6,8,9,10 HAVE burned out at least once (HAVE been made at least once) while bulb 7 (Seven out) has NOT burned out (seven-out has NOT occurred).

For instance, the first part of the equation is (1-exp(-x/24)). What this says is: the average waiting time for a point 4 to be made is 24. Therefore, using Poisson, the probability of this happening zero times over time x is: exp(-x/24). The complement of this is (1-exp(-x/24)) which gives us the probability this event HAS happened more than zero times.

The last part of the formula exp(-98x/165) is the Poisson formula for the probability that zero seven-outs have occurred (or as you would say, socket 7 has NOT burned out).

Your reverse-logic might be due to thinking in terms of the light bulb problem, which is a different approach that does involve probabilities of desired events not happening.

Quote: Wizard

This entire integral should be divided by (165/98), the average number of Come Out Rolls (not counting those that resolve in one roll) the shooter will have.

I don't think that's the reason we multiply by 98/165 (probability of a seven-out). The shooter does not get multiple trials...the trial is over at the first seven out or when the fire bet is won, whichever comes first.

As mentioned, we are calculating the probability of being in a winning state (all 6 points HAVE been made at least once and a seven-out has NOT occurred) at any time. Let's say we did not include the 98/165 factor. Then, for instance, a winning state at t=30 (all points made >0 times, 7-out made 0 times) can be counted again at t=40 (starting with the t=30 string, some of the same points have been made again, while a 7-out still hasn't occurred).

We fix this issue by ending each winning string with a seven-out (98/165 probability). Therefore it ends there and can't be counted again at a future t value.

Quote: Wizard

The first step in this method is to calculate the probability of all 7 possible pertinent outcomes of a pass line bet after a point is established.

Point of 4 made and won = (3/24) × (3/9) = 1/24

Actually, these probabilities are BEFORE a point has been established. The first term above (3/24) is probability of going from "no point established" state to "point 4 established". Then 3/9 chance for making the point from that state.
Last edited by: Ace2 on Mar 23, 2020
It�s all about making that GTA
Wizard 