# Dice Probability

Roxy Palace Welcome offer: 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. Play craps online at Cherry Jackpot. About all that can be said is that you won't lose your money quite as fast as you will in other casino games of pure chance. The probability of exactly 14 sevens is 7. By getting a sum of 11?

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By getting a sum of 11? ChrisJWelly, if the game isn't won in the first roll, the game is won by rolling in a consecutive roll the same number rolled in the first roll, or lost by rolling a 7.

If you want a seven to come out, there are 30 ways to lose and 6 ways to win. This translates into odds of 30 to 6 or 5 to 1. The difference between this and the above answer is the way odds are stated. They can be stated either using "in" or "to. When these are reduced, they become 1 way to win "in" 6 ways to lose and 5 ways to lose "to" 1 way to win.

Thus, you get "1 in 6" which is equivalent to "5 to 1". What are the odds of rolling 's or 8's before rolling a 7 in craps? You have a To find total probability in this situation, you multiply all of your chances together..

You may just be finding this out now, but casinos have known it for a long time. The casinos simply make their payouts less than the probability and they can't lose.. While I recognize that your question deals with the probablity of an event recurring, for the average gambler, there is another way to look at this.

Dice have no memory. Therefore, each time you roll them, the probability of rolling a given number remains the same. Each time you roll the dice, the chances of rolling a 6 before a 7 is 6 to 5 against you. The same odds apply to rolling an 8 before a 7. 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.

Is this so, and could you put some numbers to it? That is known as X odds, and is now pretty common. The following table shows all the possible outcomes, for the pass and odds combined, with full odds.

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 s 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? The average number of rolls per shooter is 8. For the probability of exactly 2 to rolls, please see my craps probability of survival page.

On average, during the course of points being established in craps: Of those points established, on average You could expect on average 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? 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.

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: Thanks for your kind words. Here are my answers. 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. To win the shooter must throw the dice 4 times without a 7 coming up. What are the odds of throwing the dice: How does the math work for this? Based on approximately 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. Player wins on come out roll: 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. I have a craps question. The dice show a 4. A five is rolled. A seven is rolled. What is this number for the long run using this betting pattern? Essentially I am looking for my average bet. You will always have a bet on the pass or come. This is the probability that by looking back at old rolls you will find a 4 before a 7. 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.

Any idea what the odds on doing that is? Can it be calculated? 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?

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. 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. 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. Is there any coincidence that this number is the square root of 2? Just a coincidence I assure you. 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.

Therefore the two numbers can not be equal. The square root of 2 is 1. In your opinion with which is the best to take. I took the match play. Thanks for the compliment. I recommend taking the match play. That match play is worth about 48 cents on the dollar. 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 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.

How is that possible if every individual bet made has a higher house edge? To confirm their math I made the following table, based on a field bet paying 3 to 1 on a So the house edge is indeed. 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. 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. How is that number obtained?