Craps - Other Bets
Larry from Detroit
Yes, you can make a pass line bet at any time. However, you give up the come out roll, which twice the chance of winning as losing. Making a late pass or come bet is called a "put" bet.
Gman from Lockport, New York
I hope it will answer your question if I say that place, buy, and odds on come bets are temporarily turned off for come out rolls, unless requested otherwise.
The average shooter throws the dice 8.53 times, including the seven-out roll. For more information on that, visit my odds table for the probability of every number of throws from 2 to 200.
SD from San Francisco, California
The player should always leave the odds on regardless of how many come bets are active. When considering the options, looking at the probability of winning alone is not enough. Yes, with one come bet the odds of losing the come bet are greater than winning, however the potential win is greater than potential loss. The reason the player should always leave the odds on are because it is a bet with zero house edge. By turning the odds off the player is making the overall game more heavily weighted towards the bets with a house edge, thus increasing the overall ratio of the expected loss to the total amount bet.
This is similar to a question I got last week. Yes, it is true that there are ten ways to roll a 6 or 8, and six ways to roll a 7. However, one must not look at the probabilities alone, but weight them against the payoffs. The place bet on the 6 and 8 pays 7 to 6 odds when fair odds would pay 6 to 5. By making six unit place bets on the 6 and 8, and taking the other down if one wins, the probability of winning 7 units is 62.5% and the probability of losing 12 units is 37.5%. If the player must cover both the 6 and 8, then the place bet is the way to go. This rate of return isn't bad but could be better. For the player who puts a priority on minimizing the overall house edge, the best strategy is to make combinations of pass, don't pass, come, and don't come bets, and always take the maximum allowable odds.
Bry from Chesterton, Indiana
For other readers, let me explain that a put bet is making a pass or come bet after a point has already been established. The player may choose the point to be established on the put bet. While the player can make an odds bet immediately on top of the put bet, the opportunity to win on the initial roll is lost. The effect is the same as making a place bet or buying odds, but the house edge is different depending on the multiple of odds allowed. In the case of 20 times odds the house edge of the put bet on the 4 and 10 is 1.59%, on the 5 and 9 is 0.95%, and on the 6 and 8 is 0.43%. At this high level of odds allowed, which is much greater than the norm, all put bets are better than the corresponding place or buy bets. This option should never be taken at a casino that offers less than 5 times odds. At 5 times odds exactly, the put bet on the 6 and 8 is slightly better than the place bet. At 10 times odds or greater, all put bets become better than their corresponding place or buy bets. I shall add something to my craps section about the put bets, thanks for the idea.
A put bet is like a come bet made after a point has already been established. I don't recommend put bets because the value from come bets comes from the first roll, before any point is established. On a the first roll of a come bet the probability of winning is 22.22%, and the probability of losing is 11.11%.
However, if you must cover other points quickly, the odds are generally equal or better on place bets and buy bets on the 4 and 10 (assuming commission paid only on a win). Only if you can get 10x odds, and take them, do put bets become a better value than the better of place and buy bets. Since so few places offer 10x or better odds, it is safe to say to just avoid put bets completely.
Ray from Plainfield, USA
The better system is to bet on the don't pass only and take full odds. Yes, betting on both does increase you chances of winning on any one bet. However you are suffering a higher combined house edge by betting on both the pass and don't pass and it will cost you in the long run.
Daniel from Geneseo, USA
Most casinos in Las Vegas do, except those owned by Caesars Entertainment (formerly known as Harrah's). In Reno and Lake Tahoe they pay 2-1 on the 12, but 3-1 on the 2.
Chris from Tyler, USA
Yes, it was luck. It helped that you stuck to the low house edge bets. However, next time, make the line bets with odds only, and don't bet the field, especially if it pays 2 to 1 only on both the 2 and 12.
Since you asked I added information about the put bet in my craps section. That should tell you what you want to know.
Rolland from Buffalo, USA
I've never heard of a Las Vegas casino paying 3-1 on both the 2 and 12, but the Santa Ana Star in New Mexico does (at least as of this writing in 2013). This rule results in a house edge of exactly 0.00%.
Bry from Chesterton, USA
Thanks for the compliment. The formula for the house edge in buy and lay bets is the commission divided by the bet plus commission. In this case, the best bet is to bet $39 for the $1 commission. On the buy bet the house edge would be 1/40 = 2.5%. Assuming you can lay $78 to win $39 on the 4 and 10, and still only pay $1, the house edge would be 1/79=1.27%. I'll leave the other situations as an exercise for the reader (I hated it when my math books would say that).
Jim from Yakima, USA
The house edge is the same regardless of how many come bets you make assuming you always take the maximum allowable odds and leave the odds turned on during a come out roll. How many come bets you make should be a matter of personal preference.
Richard from Binghampton, USA
As long as you are backing up your pass and come bets with full odds, it doesn't make any difference how many come bets you make. However, it does reduce the overall house edge to keep the odds on your come bets working on the come out roll.
Jon Moriarty from Danville, New Hampshire
You should never remove a don't pass bet after a point is made! Once a point is made of 6 or 8 the don't pass has equity of 9.09% of the bet amount, which you would be throwing away by taking the bet down. The equity of a don't pass bet on a point of 5 or 9 is 20%, and on a 4 or 10 is 33.33%.
Mitch from Hopkins, USA
A put bet is like a come bet on a particular point. In other words the put bet does not get the benefit of winning on a 7 or 11 on the first roll, but the bet immediately gets "put" on the point of the bettors choice. The bettor may also immediately take odds on the put bet. Put bets are generally not a good idea because the player is twice as likely to win as lose on the first roll of a come bet, and the player is voluntarily forfeiting that roll with a put bet. However if the odds allowed are high enough a put bet, backed up with maximum odds, can have a lower house edge than a place or buy bet. My craps section states how high the odds need to get for this to happen. My craps advice in general is to stick to the line bets (pass, don’t pass, come, and don’t come) and backed up the odds.
John from Overland Park, Kansas
There are 4 ways to roll a 5 (1+4, 2+3, 3+2, and 4+1) and 6 ways to roll a 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). So the probability of rolling a 5, given that a 5 or 7 was rolled, is 4/(4+6) = 0.4. The expected value of the place bet on 5 is (0.4*$4.50 + 0.6*-$3.00)/3 = 0. So you're right there is no house edge on the 5, or the 9. On the 4 and 10 the expected value is ((1/3)*$5.50 + (2/3)*-3)/3 = -0.0556. In other words the house edge is 5.56%. I take it this is only true of $3 bets, the casino rounding the winnings up from $4.20 to $4.50. As I reported in the last issue the player can gain an advantage in blackjack if the dealer pays $4.00 for a blackjack on a $2.50 bet.
Dave from Northfield, Minneapolis
Don’t be so hard on yourself. Very few Internet casinos implement the buy bet odds correctly. When I say the odds are 39-21 I mean that if you bet $21 (including the commission) you will get back $39 plus the original $21 if you win. So if you win you get back $60. Consider making a $20 buy bet, plus the $1 commission. The $1 is gone forever, but if you win you get back $40 in winnings plus the original $20, for a total of $60. Either way you risk $21 and get back $60 if you win.
Charline from Las Vegas
It depends on how the games are played. If optimal strategy is compared to optimal strategy then craps is better. By betting only the line bets and taking maximum odds the combined house edge in craps is well under 1%. The best you can do is baccarat is bet on the banker at a house edge of 1.06%. However it wouldn’t surprise me if the actual house edge in craps is higher, due to all the sucker bets players make.
That is a good question. It is obviously more fun to go with the crowd than against it. The question is why does the crowd favor the pass line? Perhaps it is just tradition. Maybe when people first started playing craps in private games the don’t pass wasn’t even an option.
I understand that taking the odds on the pass side reduces the overall house edge, however I don’t understand how laying the odds can reduce the house edge on the don’t side. I’m very curious? By the way, I discussed this with several casino bosses and dealers yesterday and they all had opinions, but not reasons for these opinions. Thanks for your time.
Let’s say you have a $10 don’t pass bet and the point is a 4. You have a 2/3 chance of winning the bet, so the expected value is (2/3)*$10 + (1/3)*-$10 =$ 10/3 = $3.33. Now consider adding a $40 odds bet on top of it. Now you have a 2/3 chance of winning $30 and a 1/3 chance of losing $50. The expected value of both bets combined is (2/3)*$30 + (1/3)*-$50 = $10/3 = $3.33. So either way your expected gain is 3 dollars and 33 cents. With the don’t pass alone the player edge is $3.33/$10 = 33.33%. With the don’t pass and odds the player edge is $3.33/$50 = 6.67%. So, yes, the player edge as a percentage drops by making the odds bet. However that player edge is effective over more money. The way I think gamblers should view the house edge is as the price to pay for entertainment. If you want to pay as little as possible then taking or laying the odds is getting entertainment for free.
Dan from Ottawa, Canada
Most player probably don’t know the don’t pass is a slightly better bet. Even those that do probably bet the pass bet because it feels better to be part of a group than to go against it.
I get challenged a lot on my support for come bets with odds. Those against come bets are always quick to say a number has to hit twice with a come bet and only once with a place or put bet. That is not a good way to look at it. First of all with come bets you have a 8/36=22.22% chance of winning on the first roll as opposed to only 11.11% of losing. You get no such advantage on the first roll of a place or bet bet. Even if you do roll a point number it can be any number. In other words there are six point numbers that can win on a come bet, and only one on a place or put bet. Ultimately the reason the come bet with odds beats place or put bets is the house edge is less.
The following table shows what each number of points pays, the probability, and contribution to the total return. The probabilities were determined by random simulation. The exact probability of making all six points is 0.000162.
The lower right cell shows an expected loss, or house edge, of 24.70%. It is my understanding the only allowed bet amount is $2.50, so the expected loss per bet would be about 62 cents.
Just a coincidence I assure you. The exact house edge in craps is 7/495, which by definition must be a rational number. In fact I would argue the house edge in all casino games must be a rational number because there are a limited number of possible outcomes in all games, resulting in a house edge of a perfect fraction. 2 is not a perfect square thus the square root of 2 must be irrational by definition. Therefore the two numbers can not be equal. To be specific the house edge on a $100 pass line bet would be $1.41414141... The square root of 2 is 1.4142135623731...
Thanks for the compliments. As you know I strictly look at the expected value of a bet. However, if taken to an extreme it doesn’t make sense at some point. If you made the proposition to me in your example I would only bet about $200 on it, although it has a 100% player advantage. The reason is if I won I would win $2,000,000 and I don’t really need more than that. However in craps a 2 to 1 win is not going to change my lifestyle if I win. 2 to 1 in craps is not a big long shot so I say get aggressive on the odds. If you’re uncomfortable with bets that have a low probability of winning you might take up betting the don’t pass and laying odds.
Wally from Houston
Thanks for the compliment. I recommend taking the match play. I’m sure the $100 in slot play was on specially designated machines. From anecdotal evidence I believe these free play slots are extremely stingy, set to pay back about 25%. That match play is worth about 48 cents on the dollar. I recommend betting in on the don’t pass in craps. The reason I favor that over blackjack is that blackjack has a lower probability of winning, thus reducing the value of the match play. For further explanation please see my October 30 2001 column.
Yes, you are. You are forgetting that a come bet wins on the first roll 22.22% of the time and loses 11.11% of the time. So you are missing the extra value of the first roll of a come bet. However if you had a crystal ball that told you that the first roll would result in a point number then you would be right.
I prefer to calculate the house edge as 1-(pr(win)*payout - pr(lose)). In this case it would be 1-((1/3)*1.8 - (2/3)) = 6.67%. However if you know the fair payout and the actual payout a convenient formula for the house edge is (f-a)/(f+1), where f=fair payout and a=actual payout. In this case (2-1.8)/(2+1) = 0.2/3 = 6.67%.
You’re welcome, thank you. The dice do not have a memory so at four throws you do not get any closer to sevening out. You could roll 1000 non-sevens and still be no closer or further away from a seven than you were the first throw. There is no optimal number of come bets, just make as many as you find the most fun.
No! The probability of winning is 15/36 = 41.67%, for a house edge of 16.67%. Even if the 2 and 12 both pay 2 to 1 on the field the house edge is only 5.56%. The over/under 7 bet ties for the any 7 bet as the worst bets on the table.
Blake Haas from Thousand Oaks
Yes. I can just imagine the follow up question to be why I recommend taking the odds if doing so doesn’t help to win more. What I suggest is betting less on the pass so that your need for action is mostly met by a full odds bet. For example if you are comfortable betting about $90 per bet, and the casino allows 5x odds, then I would drop the pass line bet to $15 and bet $75 on the odds. That will lower the overall house edge from 1.414% to 0.326%.
B.L. from NYC
The following table shows the house edge is 5.56%.
Haig from Englewood
The fewer the sevens the greater the odds favor the pass line bet. The following table shows the house edge according to the percentage of sevens, assuming the probability of all other numbers is proportional to the fair probability.
House Edge in Craps According to Seven Probability
|Seven Probability||Pass House Edge||Don’t Pass House Edge|
Todd from Chicago
First, it doesn’t make any difference that the shooter is making lots of money. Your odds are the same on an ice-cold table. The past does not matter. However if you are going to play then wait patiently for a come out roll. Never make a pass bet after a point has been established.
Gary from Milwaukee, Wisconsin
You’re right. A come bet would have been just as good and require no waiting. I should have added that.
Tom from Culver City
First I would take a taxi to the Hard Rock, the closest major casino to the airport. I’m not sure how much odds the Hard Rock allows in craps but I would guess 3-4-5. If that is the case then bet 1/7 of your bankroll on the don’t pass bet, or $275 to round down. If a point is established then lay the maximum on the odds, or $1650. If you win you’ll be a lot closer to your goal, the amount will depend on the point. Win or lose bet the lesser of 1/7 of your bankroll and 1/7 of how far you are from $4000. If you get close to either extreme just get it over with and bet everything if you’re low, or whatever you need to close the gap on $4000 if you’re high, and forget about the odds. Four hours should be enough time. However don’t dilly dally. The lines at security can get pretty bad. If your outbound flight is in terminal C be sure to ask an agent about the secret entrance.
Mike from Westfield, MA
Thanks. If your goal is to win just one unit I agree craps is the best place to start. The don’t pass is slightly better than the pass. However if you lose your first bet I would switch to blackjack. Only when exactly where you started would I go back to craps. This is because you won’t need to double or split to win just one unit, and a hit/stand only blackjack game has a house edge of about 2.5%.
Al from Calgary
Yes. Assuming the commission is paid only on a win then it should be applied to the bet amount, not the win amount.
Alex from Englewood, FL
A $10 buy bet on the 3 or 11 would pay $29.50, and on the 2 or 12 it would pay $59.50. The house edge on the 3 or 11 is 1.25%, and on the 2 or 12 it is 0.71%.
Charles from Buffalo
As if the normal odds were not bad enough. Shame on the IP. Lowering the 2-number horn bets from 15 to 14 increases the house edge from 11.11% to 16.67%. Lowering the 1-number horn bets from 30 to 29 increases the house edge from 13.89% to 16.67%. No, I don’t know of any other casinos that do this, but I don’t look for this sort of thing either.
I asked the Bone Man of Next Shooter.com. He said it is all the hard ways and either the 7 or 11. It is referred to either "buffalo -- seven" or "buffalo -- eleven."
Jay from Whitestone
I agree. I don’t recommend making put or place bets, because as you said, the odds are better making a line bet and then maximum odds. However, some people absolutely must bet on the points directly. If that is a given, I advocate making the best bet between the place, buy, and put, which I explain in greater depth in my craps section.
Donald from Las Vegas
That is very tight to limit the dealers like that. On a $2 bet the house edge goes up to 29.02%, and a $5 bet it is 41.94%.
Eric from Boston, MA
Assuming the commission is always payable, then the house edge would be 4% at a $3 commission (3/75), compared to 5.33% at a $4 commission.
Allan from San Diego, CA
With 3-45-x odds, you don’t have to bet as much on a point of four as a six, yet the reward is the same. With all gambling, you shouldn’t just look at the reward, you also have to look at how much you are risking.
Jason from Murrieta
According to my calculations, the “anything but the point” strategy will have a loss of 0.11988 units per pass bet resolved. The expected amount bet, counting only resolved bets, is 5.09865 units. So, defining the house edge as the ratio of expected loss to total bets resolved, it is 2.351%. Under the “anything but the point” strategy, the house edge is 9.19394/5.09865 = 2.341%, which includes the push on 12 as a resolved bet.
A.B. from Zion, IL
After a pass bet, on a point of 4 or 10, the house edge swings to 33.33%. On a 5 or 9, it is 20%, and on a 6 or 8, it is 9.09%. The player has these same advantages on a don’t pass bet, once a point has been established.
Ron from Collinsville, IL
Crapless Craps offers those two bets too. There is one way to roll a 2, and six ways to roll a 7, so the probability of winning a place bet on the 2 is 1/7. Same probability is the same for the 12. As explained in the baccarat question, if the probability of something is p, then fair odds are (1/p)-1 to 1. In this case fair odds would be 6 to 1. The house edge can be expressed as (t-a)/(t+1), where t is the true odds, and a is the actual odds. In Crapless Craps the place bet on the 2 and 12 pays 11 to 2. Using this formula, the house edge on the 2 and 12 is (6-5.5)/(6+1) = 0.5/7 = 7.14%.
In Crapless Craps the 3 and 11 pay 11 to 4. Using the same formula, t=3, and a=2.75, so the house edge is 0.25/4 = 6.25%.
Ron from St. Louis
Let’s find the breakeven point. The expected value of placing the 6 or 8 is [(5/11)*7 + (6/11)*-6]/6 = -(1/11)/6 = -1.52%.
Let b be the buy bet. The expected value is [(1/3)*(2b-1) + (2/3)*-b] / b = (-1/3)/b
Equating the two bets:
-1/66 = (-1/3)/b
3b = 66
b = 22
So, at a bet of $22 the odds are the same. The odds are better on the buy bet for bets of $23 to $39.
Rick from Poway
I assume the come bets were $5 each. Under this assumption, you were actually making put bets, which you can do at any time, at any casino that allows them. Put bets are generally allowed in Nevada, and generally not allowed in Atlantic City. I normally don’t recommend put bets, because you skip the come out roll on the come bet, which has a 22.2% chance of winning, and only a 11.11% chance of losing. My advice is to start over with new come bets.
Robert M from Summit, NJ
Here you go:
Average winning pass line bets = 1.244898
Average total pass line bets = 2.52510
Expected points made = 0.683673
Expected rolls = 6.841837
Also, if you want to make a bet on the table, you’ll have to pay a dollar ante to the casino. You pay only $1 per come out roll. Once the point is established you can bet as much/little as you’d like without another payment of ante. The table limits are from 5 dollars to 300 dollars.
If the dealer went through 39 cards (out of 54) before re-shuffling the deck, you can count/see 26 of those cards. Previously, you’ve said that if there are a lot of 5s and 6s left in the deck, you would bet the "yo 11" bet. Can you develop a more effective strategy and way for betting in this casino? I truly feel that this game is beatable. Would a count of high/low, like counting cards in blackjack, work? Thanks.
Chuck from New York
I still say hop bets, like the yo-11, are the way to go. Using chips, you could keep track of how many cards of each face are left in the decks.
With 26 unseen cards, if any one face had 6 left in the deck, you would have a 43.1% advantage on a hard hop bet (two of the same face), assuming it paid 30 to 1. With only 5 left, the house would have a 4.6% advantage.
The easy hops are even more exploitable. If the two dice sides in majority have at least 10 left combined, both with a minimum of 3 left, out of 26 unseen cards, then make an easy hop bet on those two numbers. If two numbers have 5 left, you will have a 23.1% advantage. If one has 4 and one has 6, you will have an 18.2% advantage. If one has 3 and one has 7, you will have a 3.4% advantage. All this assumes easy hop bets pay 15 to 1.
None of the above takes into consideration the $1 fee. As long as you are making large bets, it won’t make much difference.
Tim from Grimsby, ON
Unless life changing amounts of money are involved, I disapprove of hedging, per my seventh commandment of gambling.
I'm going to ignore the fact that if you hit the 5 you could hedge more to lock in an even larger win, and just look at this as if it ended after a 5 or 7. At this point your net will will be $785 or $50. You should start by taking down the odds bet. That will change the scenario to winning $755 or $70. Then you should lay the odds on the 5. Let b represent your lay bet against the 5. If you lose the bet, you’ll have $755-$b. If you win the bet, you’ll have $70 + (19/31)×$b. So, equate the two sides, and solve for b:
755-b = 70 +(19/31)×b
685 = (50/31)×b
That will lock in a win of $330.30. So, if rounding were not an issue, then lay $424.7 against the 5. However, rounding always is an issue, so I would lay $403 against the 5 ($390, plus $13 commission on possible win of $260).
Leo from Fort Worth
Comparing buy bets to place bets, on points of 6 and 8, the place bet always has the lower house edge. On points of 4 and 10, the buy bet always has the lower house edge. On points of 5 and 9, it depends on whether the commission is always paid or only paid on wins. If the commission is paid up front, as is usually the case, then place bets are better; otherwise, buy bets are. To be more specific, the house edge placing the 5 and 9 is 4.00%. The house edge buying the 5 and 9, when the commission is paid up front, is 4.76%. When it is only paid on wins, the house edge is 2.00%.
Rob R. from Berkley, CA
The average number of bets such a player will have on the table for any one throw is 3.6. If the player is betting $10 at a time, for example, then a fair average bet would be $36.
George from the Jungle
I checked, and indeed they do have that bet there. The return table below shows a house edge of 24.8%.
|4 or 10 four or more times||1000||0.000037||0.036892|
|5 or 9 four or more times||500||0.000207||0.103497|
|4 or 10 three times||120||0.000524||0.062847|
|6 or 8 four or more times||100||0.000698||0.069815|
|5 or 9 three times||95||0.001799||0.170927|
|6 or 8 three times||70||0.004294||0.300609|
I get a probability of winning of 2/11, for a house edge of 9.09%.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
If the Fire Bet were offered in Crapless Craps, what would be the probability of winning?
As a reminder, in Crapless Craps the 2, 3, 11 and 12 do not immediately resolve a pass line bet, but are considered points, much like the 4, 5, 6, 8, 9, and 10.
The first step in my solution requires calculating the probability of any given outcome of the passline bet, as follows.
Crapless Craps Possible Outcomes
|Come out roll||1/6||0.166667||1/6|
|Point 2 win||(1/36)*(1/7)||0.003968||1/252|
|Point 3 win||(2/36)*(2/8)||0.013889||1/72|
|Point 4 win||(3/36)*(3/9)||0.027778||1/36|
|Point 5 win||(4/36)*(4/10)||0.044444||2/45|
|Point 6 win||(5/36)*(5/11)||0.063131||25/396|
|Point 8 win||(5/36)*(5/11)||0.063131||25/396|
|Point 9 win||(4/36)*(4/10)||0.044444||2/45|
|Point 10 win||(3/36)*(3/9)||0.027778||1/36|
|Point 11 win||(2/36)*(2/8)||0.013889||1/72|
|Point 12 win||(1/36)*(1/7)||0.003968||1/252|
|Point 2 loss||(1/36)*(6/7)||0.023810||1/42|
|Point 3 loss||(2/36)*(6/8)||0.041667||1/24|
|Point 4 loss||(3/36)*(6/9)||0.055556||1/18|
|Point 5 loss||(4/36)*(6/10)||0.066667||1/15|
|Point 6 loss||(5/36)*(6/11)||0.075758||5/66|
|Point 8 loss||(5/36)*(6/11)||0.075758||5/66|
|Point 9 loss||(4/36)*(6/10)||0.066667||1/15|
|Point 10 loss||(3/36)*(6/9)||0.055556||1/18|
|Point 11 loss||(2/36)*(6/8)||0.041667||1/24|
|Point 12 loss||(1/36)*(6/7)||0.023810||1/42|
If you add all the ways to lose, you get 7303/13860 = apx. 0.526912.
The next step in my solution to this problem uses calculus. It relies on the fact that the answer would be the same if there was a random period of time between pass line bets resolving. Let's call the mean time between bet resolutions 1 and distributed by the exponential distribution, meaning it has a memoryless property.
Let x represent the time since the shooter started his turn.
The probability the shooter did not get a point 2 win is exp(-x/252). Thus, the probability that get got at least one point-2 wins is 1-exp(-x/252).
The probability the shooter did not get a point 3 win is exp(-x/72). Thus, the probability that get got at least one point-3 wins is 1-exp(-x/72).
The probability the shooter did not get a point 4 win is exp(-x/36). Thus, the probability that get got at least one point-4 wins is 1-exp(-x/36).
The probability the shooter did not get a point 5 win is exp(-2x/45). Thus, the probability that get got at least one point-5 wins is 1-exp(-2x/45).
The probability the shooter did not get a point 6 win is exp(-2x/45). Thus, the probability that get got at least one point-6 wins is 1-exp(-x/72).
Note these probabilities are the same for 8 to 12, so we can square them to show they have been achieved twice each.
The probability the shooter did not lose is exp(-7303x/13860).
The probability of losing is 7303/13860.
We can solve for the problem by integrating from t = 0 to infinity of the probability of the product of all winning requirements have been met, the losing outcome has not been met, and the probability of losing given a bet has been resolved.
The function being integrated is exp(-7303x/13860)*(1-exp(-x/252))^2*(1-exp(-x/72))^2*(1-exp(-x/36))^2*(1-exp(-2x/45))^2*(1-exp(-25x/396))^2*(7303/13860).
Put that into an integral calculator like the one at integral-calculator.com. Remember to put in the limits from 0 to infinity. The answer will be what is expressed as the answer above.