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Ace2
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June 9th, 2020 at 3:24:37 PM permalink
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.
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DJTeddyBear
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June 10th, 2020 at 6:32:20 AM permalink
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.
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? 😁
unJon
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June 10th, 2020 at 7:36:24 AM permalink
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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
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June 10th, 2020 at 8:09:31 AM permalink
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.
Last edited by: ThatDonGuy on Jun 10, 2020
Ace2
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June 10th, 2020 at 8:14:58 AM permalink
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
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Ace2
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June 10th, 2020 at 6:24:33 PM permalink
Slightly revised answer

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

Does anyone know how to simulate this ?
It’s all about making that GTA
unJon
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June 11th, 2020 at 5:24:55 AM permalink
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
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ace2
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June 11th, 2020 at 8:35:33 AM permalink
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
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unJon
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June 11th, 2020 at 9:01:09 AM permalink
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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
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June 11th, 2020 at 10:52:28 AM permalink
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.
unJon
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June 11th, 2020 at 11:05:22 AM permalink
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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
DJTeddyBear
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June 11th, 2020 at 11:48:28 AM permalink
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.
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? 😁
Ace2
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June 11th, 2020 at 11:52:52 AM permalink
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
It’s all about making that GTA
ThatDonGuy
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ChumpChange
June 11th, 2020 at 12:35:31 PM permalink
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 Made123456
1241,346,61100000
218,009,83579,996,7840000
31,447,25617,597,43320,750,661000
4123,3933,130,7718,962,6363,946,86000
510,861525,9762,749,9122,783,365489,2860
696287,232743,8401,296,997507,27150,169
78314,535189,887505,208328,61850,370
862,46346,463179,759173,46332,410
9045511,49560,33481,65116,644
100672,78419,76735,2397,742
110166916,28914,5573,274
12001591,9125,7771,349
1302405602,260497
140011208861185
150006332662
160001711432
1700034711
180003183
19000051
20000010
21000010
Ace2
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June 11th, 2020 at 12:50:20 PM permalink
Thanks for running the simulation

There must be a way to calculate this
It’s all about making that GTA
unJon
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June 11th, 2020 at 1:04:32 PM permalink
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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
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June 11th, 2020 at 1:18:31 PM permalink
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.
Ace2
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June 12th, 2020 at 8:28:31 PM permalink
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 ?
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ChumpChange
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June 12th, 2020 at 9:09:59 PM permalink
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.
ThatDonGuy
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June 16th, 2020 at 9:17:12 AM permalink
After some serious number crunching, here are the probabilities that a maximum-win Fire Bet needs 6, 7, 8, ..., 15 points:
PointsSuccess Rate
630.6704%
731.1351%
819.868%
910.2685%
104.7139%
112.0101%
120.8163%
130.3206%
140.1229%
150.0463%

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
Ace2
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June 16th, 2020 at 6:55:52 PM permalink
Quote: ThatDonGuy

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

PointsSuccess Rate
630.6704%
731.1351%
819.868%
910.2685%
104.7139%
112.0101%
120.8163%
130.3206%
140.1229%
150.0463%

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
It’s all about making that GTA
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