Reason #3 why the Wizard likes Bovada: Excellent Odds
In my opinion many online casinos are too stingy when setting the odds on their games. They think they will make more money that way but I believe they are misguided, because when players lose too quickly it's not fun, and those players might not come back.
Bovada is one of the few casinos that understands this. They offer generous odds to let you play longer and get you a better chance of winning. Among their generous offerings are Full-Pay Jacks or Better returning 99.54%, five other video poker games paying over 99%, single-zero roulette, and my favorite, Double Jackpot Poker, returning 99.63%!
Kudos to Bovada for not being afraid to give their players a good gamble.
Let me begin by saying that of all the gambling related sites, the Wizard of Odds is by far the best. My question has to do with a betting strategy for craps. My question has to do with what some people have termed variance. As you state in your Ten commandments, the house has an edge in the long term, but there are short term fluctuations.
A casino I played at had the 3,4,5 odds system where you were allowed 3x on the 4 and 10, 4x on the 5 and 9 and 5 x on the 6 and 8. I feel that with this "system" of placing odds, you reduce the fluctuations (with respect to standard 5x odds on all numbers) in your bankroll, and change the distribution of net gain/loss per session, i.e. you would produce a sharper peak located slightly more to the loss side than with 5x odds. Is this so, and could you put some numbers to it?
That is known as 3-4-5X odds, and is now pretty common. The following table shows all the possible outcomes, for the pass and odds combined, with full odds.
Return Table with 3-4-5X Odds
|Pass line win||1||0.222222||0.222222|
|Pass line loss||-1||0.111111||-0.111111|
|Point of 4 or 10 & win||7||0.055556||0.388889|
|Point of 4 or 10 & lose||-4||0.111111||-0.444444|
|Point of 5 or 9 & win||7||0.088889||0.622222|
|Point of 5 or 9 & lose||-5||0.133333||-0.666667|
|Point of 6 or 8 & win||7||0.126263||0.883838|
|Point of 6 or 8 & lose||-6||0.151515||-0.909091|
The standard deviation per pass line bet is 4.915632.
What is the average number of rolls until a shooter "sevens out"? I know that a 7 will appear every 6 rolls, but with come-out 7-11s and craps, plus the possibility of shooters making multiple points, I think the average number of rolls may be higher than expected. Is there any mathematical reference material on this?
Grshooter from Kansas City, Missouri
The average number of rolls per shooter is 8.525510. For the probability of exactly 2 to 200 rolls, please see my craps probability of survival page.
On average, during the course of 100 points being established in craps: (1) How many of those would be 4/10, 5/9 or 6/8, (2) During the 100 how many times would each point (4/10, 5/9, 6/8) vs. a 7 be made?
Jon from Danville, New Hampshire
Of those 100 points established, on average 41.67 would be on a 6 or 8, 33.33 would be on a 5 or 9, and 25.00 would be on a 4 or 10. You could expect on average 18.94 points made on a 6 or 8, 13.33 on a 5 or 9, and 8.33 on a 4 or 10.
Are the craps probability numbers with the odds taken 100% reliable. Also is the gaming industry your full time profession, and do you visit Atlantic City often? Also, how do you simulate billions and billions of hands, spins, and rolls. Is it computer generated and if so with which software?
DB from New York, USA
Well, anyone can make a mistake, but craps is an easy game to analyze mathematically so I would be very confident my odds on craps are right. Yes, gambling in one way or another is my full time self-employed profession. I have been to Atlantic City many times in the last few years but two months ago I moved to Las Vegas. So, I'm afraid I wouldn't be gracing Atlantic City with my presence much any longer. I prefer a combinatorial approach as opposed to random simulations whenever I can. Either way, I roll my own software with Visual C++. For random numbers I use a Mersenne Twister.
Before I ask my questions I just want to say your site is phenomenal! I have two craps questions I was hoping you could answer:
1) Your preference is to count the come out roll of 12 in the calculation of the house edge on the don’t pass. If one was to choose NOT to count it, would the house edge on the pass line combined with full double odds be exactly equal to that of the house edge on the don’t pass line combined with full double odds?
2) Does the overall house edge against player x go up if player x places come bets (which will be backed up with full double odds) after betting the pass line with full double odds? i.e. player x with just a pass line with full double odds = house edge .572%, player x with same bet but places two come bets with full double odds = house edge (.572%)x(3)?
Jay from Hamilton, Ontario
Thanks for your kind words. Here are my answers.
1. If we define the house edge as the expected loss per unresolved bet (not counting ties) then the house edge on the don’t pass would be 1.40%, just barely less than the 1.41% on the pass line bet. If the player can bet more money on the don’t pass side, which is the case in real but not Internet casinos, then the combined house edge favors the don’t side more the greater the multiple of odds allowed.
2. Assuming the player keeps his odds on during a come out roll then the overall house edge does not change if the player adds come bets, backed up with the odds. However if the player keeps the odds off, which is the default rule, then the overall house edge will actually go up slightly by adding come bets.
First let me say I think your web site is absolutely outstanding. Thanks. I watched a new craps game being played at Grand Casino, Biloxi, MS. called "Four The Money". To win the shooter must throw the dice 4 times without a 7 coming up. What are the odds of throwing the dice:
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks
Stan Abadie from Harahan, Louisiana
You’re welcome, thanks for the kind words. The probability of throwing the dice n times without a 7, and then throwing a 7, is (5/6)n*(1/6). The probability of throwing n non-sevens, without specifying the next throw would be (5/6)n. So the probability of throwing the dice at least four times without a seven would be (5/6)4=625/1296=0.4823.
Based on approximately 150 rolls per hour in dice, how many decisions with regard to the point will be made. I was told by someone that there is a decision every 3.6 rolls. Is this correct?
Jeff from Las Vegas, US
The following are the possible outcomes of the pass/come bet and their associated probabilities:
- Player wins on come out roll: 22.22%
- Player loses on come out roll: 11.11%
- Player wins on a point: 27.07%
- Player loses on a point: 39.60%
So the player will win on a point about 1 in 3.7 rolls.
I’ve just starting learning the game of craps. In craps, the Don’t Pass is a better bet for the player than the Pass Line bet. But the few times I’ve played in the casinos, most people seem to be betting the Pass Line and not the Don’t pass. I’m either not correct on the odds b/t the two bets or is there some reason most players are taking the Pass Line bet over the Don’t pass line bet?
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 have a craps question. If I make a $100 pass line bet and then a $100 come bet every roll, what is my average action per roll? For example, I bet $100 on the come-out. The dice show a 4. I bet $100 come bet ($200 total on the layout). A five is rolled. I bet another $100 come bet ($300 on the layout). A seven is rolled. My total action was $100+$200+$300=$600, or average of $200 per roll. What is this number for the long run using this betting pattern? Essentially I am looking for my average bet. Thanks.
Good question. Let’s think of this in units as opposed to $100 bets. You will always have a bet on the pass or come. On any given roll the probability there is an old pass or come bet on the 4 is 3/9. This is the probability that by looking back at old rolls you will find a 4 before a 7. Likewise the probability of having a bet on 5 is 4/10 and on 6 is 5/11. So the average overall bet is 1+pr(4)+pr(5)+pr(6)+pr(8)+pr(9)+pr(10) = 1+3/9 + 4/10 + 5/11 + 5/11 + 4/10 + 3/9 = 3.3758 units. This average will not true at the beginning, while you are getting in to the game. It will only apply after all point numbers and the 7 have already been rolled at least once.
I rolled four hard 4’s without rolling a 7 or an easy 4. Any idea what the odds on doing that is? Can it be calculated?
The probability of winning the hard 4 bet is 1/9. So the probability of winning four times in a row is (1/9)4 = 1 in 6561.
How does the casino practice of calling established come bet odds "off" during the "come out" roll affect the house advantage, how is that computed, and how is the house advantage affected by leaving the odds on come bets turned on during come out rolls?
Good question. For those who don’t understand the question, unless otherwise requested, odds on come out bets are not active on come out rolls. So if the player rolls a seven on a come out roll any come bets will lose and odds on come bets will be returned. Likewise if the player’s point on the come bet is rolled on the come out roll the come bet will win but the odds will push. The answer depends on how we define the house edge. If we define it as expected loss to total bets made then turning the odds off would not matter. This is because the player is still betting the odds and it still counts as a bet even if it is returned as a push. However if you define the house edge as expected loss to bets resolved then turning the odds off on a come out roll does indeed increase the house edge. I wrote a computer simulation to determine this effect. Assuming the player takes fives times odds then turning the odds off on come out rolls increases the ratio of losses to total bets resolved from 0.326% to 0.377%, or an increase of 0.051%. So if you want to maximize your return on bets resolved then leave those come odds turned on.
You say the house edge on the pass line bet in craps is 1.414%. Is there any coincidence that this number is the square root of 2?
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...
First, great site. During a recent visit to Harrah’s, they gave me an option of either $100 match play or $50 in slot play. In your opinion with which is the best to take. (I took the match play). Also, for the match play would it be better to play all $100 on one hand, or multiple smaller hands (10 x $10 hands). Thanks
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.
The American Mensa Guide to Casino Gambling has the following "anything but seven" combination of craps bets that shows a net win on any number except 7. Here's how much MENSA advises to bet in the "Anything but 7" system:
- 5- place $5
- 6- place $6
- 8- place $6
- field- $5
- total= $22
They claim the house edge is 1.136%. How is that possible if every individual bet made has a higher house edge?
Good question. To confirm their math I made the following table, based on a field bet paying 3 to 1 on a 12. The lower right cell does shows an expected loss of 25 cents over $22 bet. So the house edge is indeed .25/22 = 1.136%.
Mensa Anything but Seven Combo
|Number||Probability||Field||Place 5||Place 6||Place 8||Win||Return|
The reason the overall house edge appears to be less than the house edge of each individual bet is because the house edge on place bets is generally measured as expected player loss per bet resolved.
However, in this case the player is only keeping the place bets up for one roll. This significantly reduces the house edge on the place bets from 4.00% to 1.11% on the 5 and 9, and from 1.52% to 0.46% on the 6 and 8.
For you purists who think I am inconsistent in measuring the house edge on place bets as per bet resolved (or ignoring ties) then I invite you to visit my craps appendix 2 where all craps bets are measured per roll (including ties).
In one of your answers you state that the average number of rolls for a shooter in craps is 8.522551. How is that number obtained?
Steve S. from Long Island, NY
First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = 1/4. Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758. Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...
x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551
Is the combined house edge in craps of 0.014% (taken from your chart) on don’t pass and laying 100x odds the lowest house edge of any casino game? And, does 0.014% casino edge mean that for every $100 you wager you will lose 1.4 cents?
There are still video poker games that with proper strategy pay over 100%. I’ve also seen a blackjack game at the Fiesta Rancho and Slots-a-Fun in Las Vegas that had a basic strategy advantage. As I argue in my sports betting section betting NFL underdogs at home against the point spread also has resulted in a historical advantage. So 100x odds in craps is still one of the best bets out there, but not the very best. Yes, 0.014% means that per $100 bet you lose 1.4 cents on average.
I’ve noticed a small disturbing pattern at the craps table that I thought might be worth mentioning on your site. Players will bet the don’t come bar, but if a 6 or 8 is rolled as the point they say "no action" and they keep their money on the don’t come bar. The Luxor even had a boxman ENCOURAGE me to do it saying it’s what "smart people who know the odds are better on the don’t tend to do" or something to that effect. Not sure how you could incorporate this into your site but I’ve seen players doing it and casinos encouraging it and it’s really stooopid.
I agree that this is a very bad decision and poor advice from the dealers. Once a point of 6 or 8 has been rolled the player edge on a don’t pass or don’t come bet is (6/11)*1 + (5/11)*-1 = 1/11 = 9.09%. Taking "no action" is the same as trading it for a bet with a 1.36% house edge. So this decision costs the player 10.45%. To any dealers encouraging this I say shame on you.
At the showboat in Atlantic City there’s a new bet on the layout where the big 6/8 was. I’m wondering what the odds were on this one roll bet. 6-7-8 pay even money, hard 6/8 pay double. Thanks.
B.L. from NYC
The following table shows the house edge is 5.56%.
If you have reason to believe that the seven is weighted and is coming up more than it should, does that favor the don’t or the pass side of craps?
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|
Hello oh great and powerful Wizard. Love your site and the great education it has given me. Today I am asking a question regarding the math for determining the odds of certain "groups" of wagers. For instance, the groups of 2 bets wagering on both the 6 and 8 in craps, or the group of 4 bets wagering as an "inside" bet in craps. We know that for the 6 OR the 8, ((5/11)*7 + (6/11)*(-6))/6 = 1.515 %. BUT what if we wager on both the 6 and the 8 at the same time? Using a formula similar to that above: (((10/36)/(10/36+6/36))*7+(((6/36)/(6/36+10/36))*-12))/12 = -1.04167%. - 10 chances to win 7, and 6 chances to lose 12. No? Am I out to lunch?! Thanks for considering this problem.
Andy from Hollywood
I get a lot of questions about combinations of craps bets. Normally I don’t answer them but when you address me as "the great and powerful Wizard" it greatly improves your odds of getting a reply. Your mistake is that both bets are not resolved all of the time. When you win either the 6 or 8 you are taking the other bet down, which brings down the expected loss because you are betting less. So your math is right but you are comparing apples to oranges.
Normal Craps are not allowed in California. Here many casinos are using cards to act as dice, using A,2,3,4,5,6 to act as the 6 sides of the dice. I would assume by using multiple decks it would alter the odds. (i.e 4 decks = 16 aces, 16 2’s, ect.) Does this favor the house as in blackjack... or does this favor the player? The player could bet at higher or lower numbers based on the half of the cards out of the shoe before a shuffle(assuming a mid shoe shuffle).
Joe from Eureka, CA
You’re correct, dice alone can not determine the outcome in craps. There are various ways of using cards in place of dice and still have the odds exactly the same. One way is to use two separate decks, thus there is no effect of removal. Another way is to have a 7-card deck, featuring the numbers 1 to 6, plus a seventh "double" card. The first card drawn can never be the double card. If it is then it is put back in and the process repeats from the beginning. If the double card is drawn second then it counts as whatever the first number drawn was. Regardless of how the casino does it I have never seen hard evidence of a case where the odds were different than if two dice were used. So I think you are omitting something from the rules.
You mentioned in one of your articles an upcoming appearance on "The Casino" (apparently, it’s been cancelled). I have searched and searched to no avail in finding some kind of link to his episode. I find the idea of a story involving his advice to some young gamblers and how to most likely turn $1,000 into $5,000 quite intriguing. Please respond with some insight/leads as to how I might go about finding a copy of this episode online or purchase a video recording of it, or at the very least come across a written transcript of the episode. Thank you for your time.
Yes, there was a story taped in which some frat boys at UNLV were trying to parlay $1,000 into $5,000 to buy a high end television. They sought out my advice on how to best achieve this goal quickly. I was limited to the games at the Golden Nugget. The Nugget has 10x odds in craps, which I felt offered the opportunity to achieve the goal. It was my strategy on each come out roll to bet min(bankroll/11, (5000-bankroll)/21), subject to convenient rounding, and take the maximum odds. This way we would never go over $5,000 after a 4 or 10 win, would always have enough to take full odds, and would risk the maximum amount if we didn’t have enough to get to $5,000.
For the first bet, this formula would call for a pass line bet of $90.91, but I rounded it up to $100. Then a point was rolled, I think a 6 or 8. On the second roll the shooter sevened out. So the entire grand was lost in two rolls. It apparently didn’t make for very entertaining television and that story never made the air.
Two questions I can anticipate being asked would be (1) why did I have them bet the pass as opposed to the don’t pass, and (2) why didn’t I bet $91 on the line and $910 on the odds, adding the extra dollar out of my own pocket. To answer the first question, I think that for purposes of going for a quick big win the pass line is better. While the overall house edge is less on the don’t pass, I felt it would have taken more rolls to achieve the $5,000 goal, thus exposing more money to the house edge. To answer the second question, there is not much difference between 9x odds and 10x odds and I thought it would look better on television to be betting only black chips, at least to start.
At a recent charity casino night (not real money) there were some unusual rules for both Blackjack and Craps, and I wasn’t sure which to play. In BJ, Dealer stand on Soft 17, Double after splitting alowed (except on aces), Doubling allowed on 3 cards, BJ pays 2:1, no insurance, no surrender. In craps, COME bets paid 2:1 on 4 and 10, but no odds allowed on COME bets. I played craps until the table just got so crowded it wasn’t any fun any more, but I suspect my pass line / always COME strategy was better odds than I got at the BJ table. Was I right?
Greg from Fairfax
As my blackjack section shows, the 2 to 1 on blackjacks is worth 2.27% and doubling on 3 cards is worth 0.23%. Otherwise the rules look standard. All things considered, the house edge in the blackjack game has a player advantage of 2.1%. The probability of winning on a 4 or 10 in craps is (6/36)×(3/9) = 5.56%. Every time this happens you get an extra unit, so it is worth 5.56%. Normally the house edge on the come bet is 1.41%, so overall the player edge under this rule is 4.15%. So I agree that craps was the better game to play.
On a Crapless Craps table in Tunica, you can buy the 2, 3, 11, and 12. You listed the house edge when you place those numbers, but not when one is bought. What is the house edge on buying the 12 for $30 if you only pay the commission of $1 (rounded down from $1.50) when you win? According to my math, it’s about .47%, which would make it a VERY good bet. I got this by calculating the total money exchanged on all decision rolls ($211, including the vig) and the amount lost ($1). Am I doing this correctly? I want to make sure because this makes it a VERY appealing bet to make! Please detail how you arrived at the house edge as well, so I can make sure I am, in fact, doing it correctly. Thanks so much!
Will from Rector
I didn’t know they had a buy bet in Crapless Craps. The following table shows the house edge of place and buy bets, assuming there were no rounding of winnings. In your example of a $30 buy bet on 2 or 12 the winnings would be 6*$30-$1=$179. So the expected return is [(1/7)*$179 + (6/7)*-$30] / $30 = -0.0048, so we’re very close.
Place and Buy Bets in Crapless Craps pass and buying odds in Crapless Craps
|Bet||Pays||Prob. Win||House Edge|
|Place 2, 12||11 to 2||0.142857||0.071429|
|Place 3,11||11 to 4||0.25||0.0625|
|Buy 2, 12 (commision only on wins)||119 to 20||0.142857||0.007143|
|Buy 3,11 (commision only on wins)||59 to 20||0.25||0.0125|
|Buy 2, 12 (commision always)||119 to 21||0.142857||0.047619|
|Buy 3,11 (commision always)||59 to 21||0.25||0.047619|
I am a crap dealer in a casino that offers the fire bet (pay table A, 20.83% edge). The limits on the fire bet are $1-$5 (for players and dealers), but the dealers are limited to $1000 payout. What does that do to the house edge?
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%.
The Grand Victoria Casino in Elgin, Illinois offers a promotion called "Craps for Cash." A shooter wins a $4,000 cash bonus for making all six points on the same hand. All that's required is a $5 bet on the pass line. How does this affect the house edge on this particular game?
John B. from Riverside, Illinois
We can see from my analysis of the Fire Bet that the probability of a shooter making all six points is 0.000162435. So, the value of the promotion per shooter is $4,000 × 0.000162435 = 0.649739.
The next question to be asked is what is the expected loss per shooter. The house edge on the pass line bet is 7/495 = 1.414141%. The tricky part is how many pass line bets will a shooter make, on average.
There are four possible states the shooter can be in. Let's define each one as the expected number of future pass line bets for that shooter.
- A = Come out roll
- B = Point of 4 or 10 made
- C = Point of 5 or 9 made
- D = Point of 6 or 8 made
Here are the equations showing the probability of each state leading to the next state.
A = 1 + (12/36)*A + (6/36)*B + (8/36)*C + (10/36)*D
B = (1/3)*A
C = (2/5)*A
D = (5/11)*A
A little algebra results in A = 2.525510, the number of pass line bets made per shooter.
So, the expected loss per $5 shooter is $5*2.525510*0.0141414 = 0.178571.
The expected amount bet by the shooter is $5*2.525510=$12.627551.
Finally, the expected return is the expected win divided by the expected bet: (0.649739-0.178571)/12.627551 = 3.73127%. So the house edge is -3.73%.
Why are the odds of a hard four different from the odds of a hard six? Isn’t there just one way out of thirty-six possible combinations to hit doubles (double 1,2,3...)?
James from Santa Cruz
Yes, the probability of each double is 1/36. However you have to compare that to the probability of rolling a losing comination. For a hard four, there are 8 losing rolls (two each of 1-6, 2-5, 3-4, and 1-3), so the probability of winning is 1/9. For a hard six, there are ten losing rolls (two each of 1-6, 2-5, 3-4, 1-5 and 2-4), so the probability of winning is 1/11. The hard six pays more because the probability of winning is less.
I was playing craps at Harrah’s in St Louis, and noticed they have added place bet positions for the 2, 3, 11, and 12 to the table. I don’t remember what they paid. Do you know the odds for these bets? Thanks.
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%.
In a recent article, it was revealed that Ty Lawson, the starting point guard at UNC said, "The only time I lost was in Reno; that’s when everybody on the team lost," he said. "It’s the only place I lost. The other five or six times I did gamble, I won at least $500.”
Ben from Austin, TX
If we ignore the house edge (which is very low in craps if played properly), the probability of winning $500, as opposed to losing $1,000, is 2/3. The probability of 4 out of 5 winning sessions would be 5×(2/3)4×(1/3) = 32.9%.
My question is based on dice odds. I know that there are six ways to get 7 and one way to get 12, but what are the chances of getting six 7’s before one 12? Are they even, and if not, how many twelves should be added to the equation to make it an even proposition?
The probability of rolling a 7 is 1/6, and the probability of rolling a 12 is 1/36. The probability of rolling a 7, given that a roll is a 7 or 12 is (1/6)/((1/6)+(1/36)) = 6/7. So the probability that the first six times a 6 or 12 is rolled it is a 6 every time is (6/7)6 = 39.66%.
If you rephrase the question to be what is the probability of rolling five 6’s before a 12, then the answer is (6/7)5 = 46.27%. With four rolls it is (6/7)4 = 53.98%. So there is no number of 7’s before a 12 that is exactly 50/50. If you’re looking for a good sucker bet, suggest you can either roll four 7’s before a 12, or a 12 before five 7’s.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
At a craps table with the Fire Bet, I saw a player make every point except 10, and still rolling. Assuming it was a come out roll, what is the probability of making a point of 10 at that point before sevening out?
On the come out roll there are three possible outcomes at this point.
- Sevening out.
- Repeating a point already made (4 to 9).
- Rolling a 10 on the come out roll, and then making it.
We need to quantify the second and third probabilities only. The shooter will eventually make a point, and then eventually make it or seven out. The probability that the point established and then made is 4 to 9 is:
(3/24)×(3/9) + (4/24)×(4/10) + (5/24)×(5/11) + (5/24)×(5/11) + (4/24)×(4/10) = 0.364394.
The probability of establishing a 10 point and then making it is (3/24)*(1/3) = 0.041667.
Let p be the probability of making a 10 point before sevening out. If the player makes any other point, he is right back to where he started from. So...
p = 0.364394 × p + 0.041667
p × (1-0.364394) = 0.041667
p = 0.041667/(1-0.364394)
p = 0.065554
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
What is the average number of points hit by a craps shooter before he sevens out?
Given that a point is established, the probability that the shooter makes the point is pr(point is 4 or 10) ×pr(making 4 or 10) + pr(point is 5 or 9) × pr(making 5 or 9) + pr(point is 6 or 8) ×pr(making 6 or 8) = (6/24) × (3/9) + (8/24) × (4/10) + (10/24) × (5/11) = 201/495 = 0.406061.
If the probability of an event is p, then the expected number of times it will happen before failure is p/(1-p). So, the expected number of points per shooter is 0.406061/(1-0.406061) = 0.683673.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
What is the expected number of rolls of two dice for every total from 2 to 12 to occur at least once?
This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.
First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.
Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.
The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.
So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.
Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.
What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:
pr(A or B) = pr(A) + pr(B) - pr(A and B)
You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,
pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3) - pr(4 before 2 and 3) = (3/4)+(3/5)-(3/6) = 0.85.
The probability of not getting the four along the way to the two and three is 1.0 - 0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15*12 = 43.8.
Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.
What is the probability of getting the five before achieving the two, three, or four? The general rule is:
pr (A or B or C) = pr(A) + pr(B) + pr(C) - pr(A and B) - pr(A and C) - pr(B and C) + pr(A and B and C)
So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)-pr(5 before 2 and 3)-pr(5 before 2 and 4)-pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)-(4/7)-(4/8)-(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1 - 83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)*9 = 44.5.
Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.
Here is the general rule for pr(A or B or C or ... or Z)
pr(A or B or C or ... or Z) =
pr(A) + pr(B) + ... + pr(Z)
- pr (A and B) - pr(A and C) - ... - pr(Y and Z) Subtract the probability of every combination of two events
+ pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
- pr (A and B and C and D) - pr(A and B and C and E) - ... - pr(W and X and Y and Z) Subtract the probability of every combination of four events
Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.
The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.
Expected Number of Rolls Problem
|Highest Number Needed||Probability||Expected Rolls if Needed||Probability not Needed||Probability Needed||Expected Total Rolls|
This question was raised and discussed in the forum of my companion site Wizard of Vegas.