Fire Bet Math

If you were to review all fire bets won in Vegas, what is the expected number of total points won?

Obviously it’s at least 6 since all 6 points need to be won at least once, but in most cases I’m sure some points would be won more than once.

I know this doesn’t answer your question, but neither does your question reflect the title of the thread.

It is my understanding that the math for the fire bet is so horribly complex, that It is usually solved by simulation rather than calculation.

Ditto for the ice / dark side bet.

Quote: DJTeddyBear

I know this doesn’t answer your question, but neither does your question reflect the title of the thread.

It is my understanding that the math for the fire bet is so horribly complex, that It is usually solved by simulation rather than calculation.

Ditto for the ice / dark side bet.

Feels like a Markov chain would work for this question. But it’s beyond my skill level.

Quote: unJon

Feels like a Markov chain would work for this question. But it’s beyond my skill level.

It is, but it becomes 64 simultaneous equations in 64 unknowns.

Example for the Fire bet:
Let P(x4,x5,x6,x8,x9,x10) be the probability of winning where xN = 1 if point N has been made and 0 if it has not.
If 4, 5, 6, and 8 have been made, but 9 and 10 have not:
P(1,1,1,1,0,0) = 3/24 x 1/2 x P(1,1,1,1,0,0) + 4/24 x 2/5 x P(1,1,1,1,0,0) + 5/24 x 5/11 x P(1,1,1,1,0,0) + 5/24 x 5/11 x P(1,1,1,1,0,0) + 4/24 x 2/5 x P(1,1,1,1,1,0) + 3/24 x 1/2 x P(1,1,1,1,0,1)
Note this reduces to:
P(1,1,1,1,0,0) x (1 - 3/24 x 1/2 - 4/24 x 2/5 - 5/24 x 5/11 - 5/24 x 5/11) = 4/24 x 2/5 x P(1,1,1,1,1,0) + 3/24 x 1/2 x P(1,1,1,1,0,1)
P(1,1,1,1,0,0) = 2640/1799 x (1/15 x P(1,1,1,1,1,0) + 1/16 x P(1,1,1,1,0,1))
The solution is P(0,0,0,0,0,0), which is the starting point

After some serious number crunching, I get the exact probability for the Fire bet is:
3,700,403,899,126,040,038,831,518,494,284,887,738,125 / 22,780,863,797,678,919,004,236,184,338,193,605,974,839,452
which is about 1 / 6156.3.

If the probability of winning the fire and ice bet are 1 in 6172 and 1 in 238 respectively, and the probability of a seven out vs a point made is 98/165, then the answer might be

Ln(6172/238) / Ln(98/67) = 8.56 points

8.56 sounds reasonable, but not sure it the logic is valid

Slightly revised answer

Ln(6156.3/238.91) / Ln(98/67) = 8.54 points

Does anyone know how to simulate this ?

Quote: Ace2

Slightly revised answer

Ln(6156.3/238.91) / Ln(98/67) = 8.54 points

Does anyone know how to simulate this ?

My total guess estimate is:

(Chance of making any six points before seven out) / (chance of fire bet) * 6

Quote: unJon

My total guess estimate is:

(Chance of making any six points before seven out) / (chance of fire bet) * 6

The chance of any 6 before a sevenout is (67/165)^6 = 0.00448. Divide by chance of fire bet win of 0.000162 and you get about 28 total points hit. I highly doubt that

Quote: Ace2

The chance of any 6 before a sevenout is (67/165)^6 = 0.00448. Divide by chance of fire bet win of 0.000162 and you get about 28 total points hit. I highly doubt that

Agree. Way too high. Should have run numbers! Your 8.5 estimate feels too low to me. I would have guessed something more like 10 but have no reasoned basis for thinking that.

I ran a simulation, and either my RNG is acting up again, or the average number of points made in a won Fire Bet is somewhere around 7.413.
In fact, 30% of the time that a Fire Bet is won, it is won on six points, 30% of them needed seven, and 20% needed eight.
It does seem counterintuitive, but remember, we are limiting this to won Fire Bets.

The question may be, what is the probability of winning a Fire Bet in 6, 7, 8, 9, ... points, and then compare those numbers.

Quote: ThatDonGuy

I ran a simulation, and either my RNG is acting up again, or the average number of points made in a won Fire Bet is somewhere around 7.413.
In fact, 30% of the time that a Fire Bet is won, it is won on six points, 30% of them needed seven, and 20% needed eight.
It does seem counterintuitive, but remember, we are limiting this to won Fire Bets.

The question may be, what is the probability of winning a Fire Bet in 6, 7, 8, 9, ... points, and then compare those numbers.

Wow.

This math is over my head, but...

Quote: Original post - Ace2

... Obviously it’s at least 6 since all 6 points need to be won at least once. ...

I *think* all the math and discussion has been about the elusive 6 point fire bet.

Um....

You only need to win 4 unique points to get a return.

And for some paytables, you only need to win 3 unique points.

Quote: DJTeddyBear

This math is above my head, but...

I *think* all the math and discussion has been about the elusive 6 point fire bet.

Um....

You only need to win 4 unique points to get a return.

And for some paytables, you only need to win 3 unique points.

correct. This is all about the 6 point win

I did 1,000,742,154 runs (a "run" ends when either a point is missed or all six different points are made)
594,385,245 lost on the first point
The columns in the table are the number of different point numbers made before missing a point (or getting the sixth point without a miss, for column 6)
The rows are the total number of points made in that run
Note that there is no column zero, as "obviously" if the first point is made, then the total number of points made in that run is at least 1.

Points Made	1	2	3	4	5	6
1	241,346,611	0	0	0	0	0
2	18,009,835	79,996,784	0	0	0	0
3	1,447,256	17,597,433	20,750,661	0	0	0
4	123,393	3,130,771	8,962,636	3,946,860	0	0
5	10,861	525,976	2,749,912	2,783,365	489,286	0
6	962	87,232	743,840	1,296,997	507,271	50,169
7	83	14,535	189,887	505,208	328,618	50,370
8	6	2,463	46,463	179,759	173,463	32,410
9	0	455	11,495	60,334	81,651	16,644
10	0	67	2,784	19,767	35,239	7,742
11	0	16	691	6,289	14,557	3,274
12	0	0	159	1,912	5,777	1,349
13	0	2	40	560	2,260	497
14	0	0	11	208	861	185
15	0	0	0	63	326	62
16	0	0	0	17	114	32
17	0	0	0	3	47	11
18	0	0	0	3	18	3
19	0	0	0	0	5	1
20	0	0	0	0	1	0
21	0	0	0	0	1	0

Thanks for running the simulation

There must be a way to calculate this

Quote: Ace2

Thanks for running the simulation

There must be a way to calculate this

My math skills are not up to it. It almost seems like a Feynman path integral would work well as a path.

Or a very complicated Markov chain.

Quote: Ace2

Thanks for running the simulation

There must be a way to calculate this

There may be, but as I said before, it may have to be done "the long way." I tried using a Markov chain, but it didn't take into account, for example, the probability that, if you have made every point but a 10, the probability of making a 10 before missing any point.

Let's look at the probability of the first six points all being made, and all different numbers.
Assume the points are 4, 5, 6, 8, 9, and 10, in order.
The probability is (3/24 x 1/3) x (4/24 x 2/5) x (5/24 x 5/11) x (5/24 x 5/11) x (4/24 x 2/5) x (3/24 x 1/3).
There are 720 permutations of the six points, so the probability of making six different points in a row is 720 x (3/24 x 1/3) x (4/24 x 2/5) x (5/24 x 5/11) x (5/24 x 5/11) x (4/24 x 2/5) x (3/24 x 1/3), or about 1 / 20,000.
For all six points in the first 7 points made, one of them is made twice, but the seventh one cannot be the duplicate (as otherwise the first six would be different, but we are counting exactly 7 points made, not "at most 7"). For each of the six numbers, there are 15 ways they can be made twice in the first two six points, and for each one, there are 120 permutations of the other five point numbers. If the duplicate number is 4, the probability is 15 x 120 x (3/24 x 1/3)³ x (4/24 x 2/5)² x (5/24 x 5/11)². Repeat this for 5, 6, 8, 9, and 10, changing the cubed number accordingly, then add them up.
Repeat for 8, then 9, then 10, and so on.

Let’s say there was a multiplier that increased your fire bet payout by a factor of X if you won it with exactly 6 points. What would be a fair value for that multiplier ?

Depends how much you wanna pay in taxes. I'm gonna hope my regular betting wins me $5000 on 6 points without the Fire Bet bet. There might be a $500 table minimum though.

After some serious number crunching, here are the probabilities that a maximum-win Fire Bet needs 6, 7, 8, ..., 15 points:

Points	Success Rate
6	30.6704%
7	31.1351%
8	19.868%
9	10.2685%
10	4.7139%
11	2.0101%
12	0.8163%
13	0.3206%
14	0.1229%
15	0.0463%

The expected number of points needed to win a maximum-win Fire Bet (calculated through 30 points needed) is 7.3941586

For those of you interested in exact numbers, the fraction of maximum-win Fire Bets that needed exactly 6 points is
19,175,811,277,507,507,579,323,387,490,061,957,891,279 / 62,522,024,279,633,572,496,097,336,479,437,463,223,360

Quote: ThatDonGuy

After some serious number crunching, here are the probabilities that a maximum-win Fire Bet needs 6, 7, 8, ..., 15 points:

Points	Success Rate
6	30.6704%
7	31.1351%
8	19.868%
9	10.2685%
10	4.7139%
11	2.0101%
12	0.8163%
13	0.3206%
14	0.1229%
15	0.0463%

The expected number of points needed to win a maximum-win Fire Bet (calculated through 30 points needed) is 7.3941586

For those of you interested in exact numbers, the fraction of maximum-win Fire Bets that needed exactly 6 points is
19,175,811,277,507,507,579,323,387,490,061,957,891,279 / 62,522,024,279,633,572,496,097,336,479,437,463,223,360

Thanks TDG