Fire bet: Calculating the odds.
November 15th, 2011 at 2:29:01 PM
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IRS REGULATIONS.
Reportable Gambling Winnings
Report gambling winnings on Form W-2G if:
The winnings (not reduced by the wager) are $1,200 or more from a bingo game or slot machine,
The winnings (reduced by the wager) are $1,500 or more from a keno game,
The winnings (reduced by the wager or buy-in) are more than $5,000 from a poker tournament,
The winnings (except winnings from bingo, slot machines, keno, and poker tournaments) reduced, at the option of the payer, by the wager are:
$600 or more, and
At least 300 times the amount of the wager, or
The winnings are subject to federal income tax withholding (either regular gambling withholding or backup withholding).
Tax-Exempt Organizations
A tax-exempt organization conducting gaming activities may be required to withhold income tax and report on Form W-2G. See Pub. 3079, Tax-Exempt Organizations And Gaming.
Withholding
There are two types of withholding on gambling winnings: (a) regular gambling withholding at 25% (33.33% for certain noncash payments) and (b) backup withholding at 28%. If a payment is already subject to regular gambling withholding, it is not subject to backup withholding.
Regular Gambling Withholding
You may be required to withhold 25% of gambling winnings for federal income tax. This is referred to as regular gambling withholding. Withhold at the 25% rate if the winnings minus the wager are more than $5,000 and are from:
Sweepstakes,
Wagering pools,
Lotteries, or
Other wagering transactions if the winnings are at least 300 times the amount wagered.
Do not withhold at the 25% rate on winnings from bingo, keno, slot machines, or any other wagering transaction if the winnings are $5,000 or less. However, see Backup Withholding below.
Regular gambling withholding is figured on the total amount of gross proceeds (the amount of winnings minus the amount wagered), not merely on the amount in excess of $5,000.
Report the amount you withheld in box 2 of Form W-2G. Also file Form 945, Annual Return of Withheld Federal Income Tax, to report all your gambling withholding.
Noncash payments. A noncash payment, such as a car, must be taken into account at its fair market value (FMV) for purposes of reporting and withholding. If the FMV exceeds $5,000, after deducting the price of the wager, the winnings are subject to 25% regular gambling withholding. The tax you must withhold is computed and paid under either of the following two methods.
The winner pays the withholding tax to the payer. In this case, the withholding is 25% of the FMV of the noncash payment minus the amount of the wager.
The payer pays the withholding tax. In this case, the withholding is 33.33% of the FMV of the noncash payment minus the amount of the wager.
If you use method 2, enter the sum of the noncash payment and the withholding tax in box 1 of Form W-2G and the withholding tax paid by the payer in box 2.
Backup Withholding
You may be required to withhold 28% of gambling winnings (including winnings from bingo, keno, slot machines, and poker tournaments) for federal income tax. This is referred to as backup withholding. You should backup withhold at the 28% rate if:
The winner does not furnish a correct taxpayer identification number (TIN),
25% has not been withheld, and
The winnings are at least $600 and at least 300 times the wager (or the winnings are at least $1,200 from bingo or slot machines or $1,500 from keno or more than $5,000 from a poker tournament).
Figure any backup withholding on the total amount of the winnings reduced, at the option of the payer, by the amount wagered. This means the total amount, not just the payments in excess of $600, $1,200, $1,500, or $5,000, is subject to backup withholding at 28%.
Reportable Gambling Winnings
Report gambling winnings on Form W-2G if:
The winnings (not reduced by the wager) are $1,200 or more from a bingo game or slot machine,
The winnings (reduced by the wager) are $1,500 or more from a keno game,
The winnings (reduced by the wager or buy-in) are more than $5,000 from a poker tournament,
The winnings (except winnings from bingo, slot machines, keno, and poker tournaments) reduced, at the option of the payer, by the wager are:
$600 or more, and
At least 300 times the amount of the wager, or
The winnings are subject to federal income tax withholding (either regular gambling withholding or backup withholding).
Tax-Exempt Organizations
A tax-exempt organization conducting gaming activities may be required to withhold income tax and report on Form W-2G. See Pub. 3079, Tax-Exempt Organizations And Gaming.
Withholding
There are two types of withholding on gambling winnings: (a) regular gambling withholding at 25% (33.33% for certain noncash payments) and (b) backup withholding at 28%. If a payment is already subject to regular gambling withholding, it is not subject to backup withholding.
Regular Gambling Withholding
You may be required to withhold 25% of gambling winnings for federal income tax. This is referred to as regular gambling withholding. Withhold at the 25% rate if the winnings minus the wager are more than $5,000 and are from:
Sweepstakes,
Wagering pools,
Lotteries, or
Other wagering transactions if the winnings are at least 300 times the amount wagered.
Do not withhold at the 25% rate on winnings from bingo, keno, slot machines, or any other wagering transaction if the winnings are $5,000 or less. However, see Backup Withholding below.
Regular gambling withholding is figured on the total amount of gross proceeds (the amount of winnings minus the amount wagered), not merely on the amount in excess of $5,000.
Report the amount you withheld in box 2 of Form W-2G. Also file Form 945, Annual Return of Withheld Federal Income Tax, to report all your gambling withholding.
Noncash payments. A noncash payment, such as a car, must be taken into account at its fair market value (FMV) for purposes of reporting and withholding. If the FMV exceeds $5,000, after deducting the price of the wager, the winnings are subject to 25% regular gambling withholding. The tax you must withhold is computed and paid under either of the following two methods.
The winner pays the withholding tax to the payer. In this case, the withholding is 25% of the FMV of the noncash payment minus the amount of the wager.
The payer pays the withholding tax. In this case, the withholding is 33.33% of the FMV of the noncash payment minus the amount of the wager.
If you use method 2, enter the sum of the noncash payment and the withholding tax in box 1 of Form W-2G and the withholding tax paid by the payer in box 2.
Backup Withholding
You may be required to withhold 28% of gambling winnings (including winnings from bingo, keno, slot machines, and poker tournaments) for federal income tax. This is referred to as backup withholding. You should backup withhold at the 28% rate if:
The winner does not furnish a correct taxpayer identification number (TIN),
25% has not been withheld, and
The winnings are at least $600 and at least 300 times the wager (or the winnings are at least $1,200 from bingo or slot machines or $1,500 from keno or more than $5,000 from a poker tournament).
Figure any backup withholding on the total amount of the winnings reduced, at the option of the payer, by the amount wagered. This means the total amount, not just the payments in excess of $600, $1,200, $1,500, or $5,000, is subject to backup withholding at 28%.
November 6th, 2012 at 12:48:10 AM
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Sorry for digging this up, but IMHO the FIRE BET suffers from too high a payout. Now that the Wizard of Odds (thanks to miplet, et al for the answers) has the math done, its easy to create a more structured ramp. Waiting for 4 numbers to hit, and paying high-odds for all 6, just doesen't seem right. I mean imagine pai gow poker bonus side-bets configured this way... quads or better to win and 20G for 5-Aces Paired for example. I think the bet-ramp is poorly made/greedy, but back-fired. At $2000 for all 6, with 10 Players up for a $1, is it any wonder???
As an example I offer this structure with an 8.93% HA... if the 1st 3 bets pay 2:1, 6:1, 20:1 the HA is about 15%. $1 ONLY... Good Luck & Good Game!
2 Points........ 3:1
3 Points........ 6:1
4 Points...... 15:1
5 Points...... 60:1
6 Points.... 250:1
As an example I offer this structure with an 8.93% HA... if the 1st 3 bets pay 2:1, 6:1, 20:1 the HA is about 15%. $1 ONLY... Good Luck & Good Game!
2 Points........ 3:1
3 Points........ 6:1
4 Points...... 15:1
5 Points...... 60:1
6 Points.... 250:1
Some people need to reimagine their thinking.
November 6th, 2012 at 12:46:06 PM
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Unfortunately, I believe the high payout of the Fire Bet is part of its appeal. I would never lower the payout of 6 points below $500 for a $1 wager. But I agree that 3 points should be paid out. I'm more indifferent about 2 points being paid.
This below gives the house a 16.7% edge and a hit frequency of 4.4% (which is okay for a bet like this!)...these are payouts on a "for 1" basis:
0-2...0
3...5
4...20
5...200
6...1000
This below gives the house a 16.7% edge and a hit frequency of 4.4% (which is okay for a bet like this!)...these are payouts on a "for 1" basis:
0-2...0
3...5
4...20
5...200
6...1000
November 6th, 2012 at 7:55:02 PM
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I partly agree, but really, how many ppl know the odds... if you open this bet in an E.Coast casino chances are it lives. In Vegas, where the high payouts exists, epic fail
Some people need to reimagine their thinking.
November 6th, 2012 at 10:17:29 PM
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Quote: 98ClubsI partly agree, but really, how many ppl know the odds... if you open this bet in an E.Coast casino chances are it lives. In Vegas, where the high payouts exists, epic fail
?? The fire bet is a significant success, at least I thought? It even exists in Missouri.
November 6th, 2012 at 10:37:40 PM
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Previously in this thread (1 yr old) theres talk about certain well-known Vegas Casinos removing the game. Seemed odd to me that a bet with 20-25% HA would ever get pulled, but evidently, the Casinos were not bankrolled enough to back the bet. As I stated earlier, with 10 Customers betting $1, the risk is 20G's. Second, it looks like the required verite for the high payout delayed the game. It was even posted that a 5-minute dbl. check meant the House was losing $$$, and of course, the interruption may have cooled the Shooter.... not what you want both sides of the game. After seeing this just recently, I agreed, a bet that was bad bizzness. So it seemed to me that a low payment, and low payouts (well 250:1 ain't low on a Crap table) could keep the game moving and generate some promise. Maybe $500 for all 6 would be OK, but the lack of proper bankrolling in the first place seemed to doom the bet.
Some people need to reimagine their thinking.
November 7th, 2012 at 2:29:17 AM
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Smaller casinos that have a limited craps clientele probably can't risk the fire bet.
I sometimes stop at the Gold Strike at Jean on the I-15 on the way to Vegas. they have one table, $3 minimums and I think the max bet is $500. I asked them if they were going to get the Fire Bet and a supervisor told me they just couldnt afford it. If one player hit it for $5 -- that's five-thousand which is more than this little casino makes in a day at the craps table. they usually have only one side open, using two dealers.
I sometimes stop at the Gold Strike at Jean on the I-15 on the way to Vegas. they have one table, $3 minimums and I think the max bet is $500. I asked them if they were going to get the Fire Bet and a supervisor told me they just couldnt afford it. If one player hit it for $5 -- that's five-thousand which is more than this little casino makes in a day at the craps table. they usually have only one side open, using two dealers.
September 1st, 2016 at 8:35:49 AM
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how did you get that. 0 pts= 98/165 where does the 98 and 165 come from?
January 21st, 2018 at 6:07:43 PM
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I recently had a very long roll of about 50-60 shots...without making the fire bet. I made 8 or 9 points, but only 3 unique ones. That struck me as fairly rare. Anyone have an idea exactly how rare?
Let Pn,k be the probability that an active ‘n’ point roll has at least ‘k’ unique points. What is the smallest value of ‘n’ where Pn,k is greater than 50%? 90%? 99%? Compute for ‘k’ from 1-6. By definition, P2k-1,k is zero & P2k,k is non-zero.
Note: k = 1 is interesting as 1-(Pn,1) represents the probability that an ‘n’ point roll has no points.
Let Pn,k be the probability that an active ‘n’ point roll has at least ‘k’ unique points. What is the smallest value of ‘n’ where Pn,k is greater than 50%? 90%? 99%? Compute for ‘k’ from 1-6. By definition, P2k-1,k is zero & P2k,k is non-zero.
Note: k = 1 is interesting as 1-(Pn,1) represents the probability that an ‘n’ point roll has no points.
January 21st, 2018 at 10:23:50 PM
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My initial back of the envelope estimate tells me it’s not that rare. Like roughly 1 in 37 chance that you would hit less than 4 unique points in a run lasting 60 rolls. I’ll try to spend some more time on it tomorrow.
It’s all about making that GTA
September 6th, 2018 at 7:26:19 PM
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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.
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
December 11th, 2019 at 1:52:31 AM
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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.
December 11th, 2019 at 9:50:18 AM
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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.
March 23rd, 2020 at 2:47:51 PM
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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:
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.
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.
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:
Once again, I don't see why light bulbs are mentioned.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.
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.
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.Quote: WizardThis 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.
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.
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.Quote: WizardThe 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
Last edited by: Ace2 on Mar 23, 2020
It’s all about making that GTA
March 23rd, 2020 at 3:13:35 PM
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Thanks for your comments. I can easily fix some of them.
However, I have already struggled with how to explain this a lot. I changed the wording several times before you see what you see now.
Let me chew on this. Please keep in mind that I'm trying to dumb this down as much as I can. Just throwing that integral up there will help almost nobody. You know how many times we had to go back and forth for even me to get what's going on, at least I think I do.
However, I have already struggled with how to explain this a lot. I changed the wording several times before you see what you see now.
Let me chew on this. Please keep in mind that I'm trying to dumb this down as much as I can. Just throwing that integral up there will help almost nobody. You know how many times we had to go back and forth for even me to get what's going on, at least I think I do.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 17th, 2022 at 10:17:15 AM
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Here's another way to calculate the probability of winning the Fire Bet. This method uses inclusion-exclusion and, though not as slick as the Poisson integration posted above, it does not require an integral calculator.
There are seven possible outcomes relevant to the Fire Bet, weighted as follows:
Win 4...............55
Win 5...............88
Win 6.............125
Win 10.............55
Win 9...............88
Win 8.............125
Seven out…...784
Sum.............1320
Let a=55, b=88, c=125, v=784 and m=1320
The probability of winning is:
1
- [v/(v+a) + v/(v+b) + v/(v+c)]*2
+ [v/(v+2a) + v/(v+2b) + v/(v+2c) + {v/(v+a+b) + v/(v+a+c) + v/(v+b+c)}*4]
- [8v/(v+a+b+c) + {v/(v+2a+b) + v/(v+2a+c) + v/(v+2b+a) + v/(v+2b+c) + v/(v+2c+a) + v/(v+2c+b)}*2]
+ [v/(m-2a) + v/(m-2b) + v/(m-2c) + {v/(m-a-b) + v/(m-a-c) + v/(m-b-c)}*4]
- [v/(m-a) + v/(m-b) + v/(m-c)]*2
+ [v/m]
=~ .00016243 =~ 1 in 6,156
I can't think of a more efficient solution
There are seven possible outcomes relevant to the Fire Bet, weighted as follows:
Win 4...............55
Win 5...............88
Win 6.............125
Win 10.............55
Win 9...............88
Win 8.............125
Seven out…...784
Sum.............1320
Let a=55, b=88, c=125, v=784 and m=1320
The probability of winning is:
1
- [v/(v+a) + v/(v+b) + v/(v+c)]*2
+ [v/(v+2a) + v/(v+2b) + v/(v+2c) + {v/(v+a+b) + v/(v+a+c) + v/(v+b+c)}*4]
- [8v/(v+a+b+c) + {v/(v+2a+b) + v/(v+2a+c) + v/(v+2b+a) + v/(v+2b+c) + v/(v+2c+a) + v/(v+2c+b)}*2]
+ [v/(m-2a) + v/(m-2b) + v/(m-2c) + {v/(m-a-b) + v/(m-a-c) + v/(m-b-c)}*4]
- [v/(m-a) + v/(m-b) + v/(m-c)]*2
+ [v/m]
=~ .00016243 =~ 1 in 6,156
I can't think of a more efficient solution
Last edited by: Ace2 on Jan 17, 2022
It’s all about making that GTA
January 17th, 2022 at 10:24:45 AM
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Is the Firebet still around?
January 17th, 2022 at 1:09:59 PM
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There's a Firebet type A near me. I'd play a Firebet type C because it pays 299 to 1.
January 18th, 2022 at 7:58:25 AM
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I think it’s still popular in Vegas. Though I’ve never played itQuote: AlanMendelsonIs the Firebet still aground ?
link to original post
It’s all about making that GTA
January 18th, 2022 at 8:13:35 AM
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Quote: Ace2I think it’s still popular in Vegas. Though I’ve never played itQuote: AlanMendelsonIs the Firebet still aground ?
link to original post
link to original post
I think the All/Tall/Small and its progeny drive out the forever in Vegas.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
August 12th, 2022 at 3:45:42 PM
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Hey Wizard, I created a version of craps that uses d12 dice, i found calculating the odds for it not that bad except for the side bets/ the calculus equations. Can you explain how to calculate the fire bet for the d6 for 5 points-4 points and 3 points ect?
For the d12 pass line i made the fronline winners for the passline "4, and "22" which both have 3/144=6/144 ways total, "5"=4/144, along with the "5", and "21" which both have 4/144 ways so 8/144 total, The crapout dice "3,23 have 2/144 so 4/144 total, and 2,24 have 1/144 so 2/144 total. 13 is the d12 version of the 7 out which makes 12/144
(yeah its kinda wierd how that works, i also had to make the hard 4 tie for the pass line, just like the 12 in the don't pass. By doing this i made the odds more match up to d6 craps,
So barring winning or losing on the passline is 144-20-12= 112
Point of 6 made and won =(5/112)*5/17=25/1904 1.313% 2.626% With 20
Point of 7 made and won = (6/112)*(1/3)=1/56 1.786% 3.572% with 19
Point of 8 made and won = (7/112)*(7/19)=7/304 2.303% 4.606% with 18
Point of 9 made and won = (8/112)*(2/5)=1/35 2.857% 5.714% with 17
Point of 10 Made and Won = (9/112)*(3/7)=27/784 3.444% 6.888% with 16
Point of 11 Made and won = (10/112)*(5/11)=25/616 4.058% 8.116% with 15
Point of 12 Made and Won = (11/112)*(11/23)=121/2576 4.697% 9.394% with 14
Any point made and 7-out = 2*((5/112)*(12/17))+2*((6/112)*(2/3))+2*((7/112)*(12/19))+2*((8/112)*(3/5))+2*((9/112)*4/7))+2*((10/112)*(6/11))+2*((11/112)*(12/23))=11829284/20021155 = 59.08%
I have no idea about calculus so the best i could do was plug the numbers in like you had on your fire bet page,
So i turned
(98*e^(-(98*x)/165)*(1-e^(-(25*x)/264))^2*(1-e^(-x/15))^2*(1-e^(-x/24))^2)/165=0.0001624347492698264
Into
(11829284*e^(-(11829284*x)/20021155)*(1-e^(-(121*x)/2576))^2*(1-e^(-(25*x)/616))^2*(1-e^(-(27*x)/784))^2*(1-e^(-x/35))^2*(1-e^(-(7*x)/304))^2*(1-e^(-x/56))^2*(1-e^(-(25*x)/1904))^2)/20021155= 2.04491E-10??
So the above problem was supposed to calculate making all 14 unique points, no idea if it is right. So how could i plug in the numbers to find all the different points? like 1 points made, and 3 points made? I understand how you calculated all 6 points for the d6 but i don't know how to do the rest.
For the d12 pass line i made the fronline winners for the passline "4, and "22" which both have 3/144=6/144 ways total, "5"=4/144, along with the "5", and "21" which both have 4/144 ways so 8/144 total, The crapout dice "3,23 have 2/144 so 4/144 total, and 2,24 have 1/144 so 2/144 total. 13 is the d12 version of the 7 out which makes 12/144
(yeah its kinda wierd how that works, i also had to make the hard 4 tie for the pass line, just like the 12 in the don't pass. By doing this i made the odds more match up to d6 craps,
So barring winning or losing on the passline is 144-20-12= 112
Point of 6 made and won =(5/112)*5/17=25/1904 1.313% 2.626% With 20
Point of 7 made and won = (6/112)*(1/3)=1/56 1.786% 3.572% with 19
Point of 8 made and won = (7/112)*(7/19)=7/304 2.303% 4.606% with 18
Point of 9 made and won = (8/112)*(2/5)=1/35 2.857% 5.714% with 17
Point of 10 Made and Won = (9/112)*(3/7)=27/784 3.444% 6.888% with 16
Point of 11 Made and won = (10/112)*(5/11)=25/616 4.058% 8.116% with 15
Point of 12 Made and Won = (11/112)*(11/23)=121/2576 4.697% 9.394% with 14
Any point made and 7-out = 2*((5/112)*(12/17))+2*((6/112)*(2/3))+2*((7/112)*(12/19))+2*((8/112)*(3/5))+2*((9/112)*4/7))+2*((10/112)*(6/11))+2*((11/112)*(12/23))=11829284/20021155 = 59.08%
I have no idea about calculus so the best i could do was plug the numbers in like you had on your fire bet page,
So i turned
(98*e^(-(98*x)/165)*(1-e^(-(25*x)/264))^2*(1-e^(-x/15))^2*(1-e^(-x/24))^2)/165=0.0001624347492698264
Into
(11829284*e^(-(11829284*x)/20021155)*(1-e^(-(121*x)/2576))^2*(1-e^(-(25*x)/616))^2*(1-e^(-(27*x)/784))^2*(1-e^(-x/35))^2*(1-e^(-(7*x)/304))^2*(1-e^(-x/56))^2*(1-e^(-(25*x)/1904))^2)/20021155= 2.04491E-10??
So the above problem was supposed to calculate making all 14 unique points, no idea if it is right. So how could i plug in the numbers to find all the different points? like 1 points made, and 3 points made? I understand how you calculated all 6 points for the d6 but i don't know how to do the rest.
