Craps Odds - What You Should Know About Them and the House Edge - Sorry, the Dice Aren't Talking
There are many things to consider when deciding on the issue - the odds craps. Experts tend to ... well, I agree with most of them, you must first understand the craps likely to be carefully mounted on the game of craps game.
In fact, some will point out that you need to know the odds before making a bet, to know that betting the house (casino) to give a small edge on you.
Why is the house edge a role? It can be argued that theCraps can not be beaten. If you look at the odds craps returns to the mathematical proof of this statement. This is true, does not make sense to reduce the house edge to lose hope, the amount you will eventually lose?
There is a possibility that you might think - Craps can not be beaten? Heck, if I walked away from a winner, so it does not. This argument, if it were to be dismissed at odds craps and the house advantage,keep the water under certain conditions.
However, if a probability craps, do not think, is not that a particular session or a series of rollers can not be beaten. The idea is that the odds craps and the house edge is designed to ensure that the house will not be beaten for a prolonged period.
Let us examine for a moment.
We can begin to understand the craps a chance to look at the probability (or opposition case) to obtain a specific number. The first thingto do is calculate the number of possible combinations with a pair of dice.
You can see that there are six sides of a cube. Each page represents a specific number. The numbers - 1, 2, 3, 4, 5 and 6
There are two dice, so you multiply six by six to determine the number of possible combinations. In this case the number 36 (6 x 6 = 36).
So, treat each cube individually (to die on the left side and right side die B), determine how many ways there are rolesThe following numbers - 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12
Here are the results - two (one way), 3 (2 species), 4 (3 species), 5 (4 possibilities), 6 (5 species), 7 (6 options), 8 (5 species), 9 (4 stroke ), 10 (3 types), 11 (2 species), 12 (1 on).
Now you can calculate the probability by rolling the number of possibilities, a number to the number of possible combinations with a pair of dice (36). For example, you pull one way, the number 2, so you have a 1 in 36 chance of rolling two. The probability is 1 / 36or 2.78%.
Here are the odds of rolling each number - 2 (1 / 36, 2.78%), 3 (2 / 36, 5.56%), 4 (3 / 36, 8.33%), 5 (4 / 36, 11.11%), 6 (5 / 36, 13.89%), 7 (6 / 36, 16.67%), 8 (5 / 36, 13.89%), 9 (4 / 36, 11, 11% ), 10 (3 / 36, 8.33%), 11 (2 / 36, 5.56%), 12 (1 / 36, 2.78%).
The odds above, as is likely, or likely to arise in any autonomous role of the dice. Independent, because whatever the outcome of the next roll of the dice, is not dependent on or influenced by his previous rolesnuts.
Dice have no memory - - and given the fact that they are objects, without the ability to think or to perform calculations, in other words, the nuts have no brains - you may have heard the saying, it's safe to say that nuts can not remember anything so irrelevant previous roles.
By the same reasoning we can say that the dice do not know the probabilities, so that is not influenced by chance. But if this is true, we can not say that white cubescraps odds, so they cannot be influenced by craps odds? Ooops! Don't answer that just yet.
Now that you know the probabilities, your next step is to understand how this relates to craps odds.
First off, you cannot establish true craps odds without knowing the probability of rolling a specific number. One definition of odds, according to Merriam-Webster's Online Dictionary, is as follows -- the ratio of the probability of one event to that of an alternative event.
In other words, you need to know the probability of rolling a number in a specific situation, in order to determine the true craps odds.
Here is a simple formula for true craps odds on rolling any number before a 7 on the next roll: P7 divided by PN = true craps odds. The letter P stands for probability, and the letter N stands for the number to roll before seven.
Using this formula you can calculate the true craps odds of rolling a 2 before the 7. P7/P2 = true craps odds, so 16.67% (.1667)/2.78% (.0278) = 6.00. The true craps odds of rolling a 2 before the 7 -- is 6 to 1.
This same concept, not necessarily the same formula, is used to mathematically determine the true craps odds of all the bets in the game of craps. However, the house edge is calculated to favor the house, and this is what gives the house the advantage.
For example, the true craps odds of rolling a 6 before a 7 is - P7/P6 =.1667/.1389 = 1.2, or 6/5, or 6 to 5, or 6:5. However, the house pays 7:6 (7 to 6) when you make a place bet on the number 6. The difference between the true craps odds of 6:5 and the actual payout of 7:6 is the house edge, which is 1.52%.
With this in mind, what happens if you bet $12 to place the 6 (make a bet that the 6 shows before a 7), and the shooter rolls a 6?
The true craps odds would be a payout of 6:5 or 6 dollars profit for every 5 dollars you bet, which is about $14.40 profit. However, the house pays you 7:6, instead of the true craps odds, so you only get $14 profit...the difference being 40 cents.
Does this mean you lost $.40? Hmmm...You put $12 on the table, won $14 profit, plus you get to keep your $12 bet...would you feel like you lost money at this point?
Do you think the dice know just how much the house edge cost you?
Okay, that's quite a bit to think about, so let's dig a little deeper.
You know that the number 6 will be rolled five times in 36 rolls...in theory. You also know that the number 7 will be rolled six times in 36 rolls...in theory.
Let's alternate the 6 and 7 such that 6 is rolled before 7, then 7 is rolled before 6. Further, let's do this to reflect the theory that 6 will be rolled five times and 7 will be rolled 6 times. Additionally, we will make a $12 place bet on 6 for each time we alternate the 6 and 7.
By the way, this will represent a total of eleven bets. Five of the bets will be a win for 6, and six of the bets will be a loss due to the 7. This will make more sense as the example progresses.
You start with a $12 place bet on 6 and it wins. This gives you a profit of $14.
Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled before a 6. You lose the $12 place bet, and now have a total profit of $2 ($14 previous profit minus the $12 loss).
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $16 (the previous total profit of $2 plus the $14 profit on this bet).
Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $4 ($16 previous profit minus the $12 loss).
So far you have rolled 6 twice and 7 twice.
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $18 (the previous total profit of $4 plus the $14 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $6 ($18 previous profit minus the $12 loss).
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $20 (the previous total profit of $6 plus the $14 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $8 ($20 previous profit minus the $12 loss).
You have rolled 6 a total of four times and 7 a total of four times. This means you have one more roll of 6 and two more rolls of 7 to go.
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14 for this bet, and an overall profit of $22 (the previous total profit of $8 plus the $14 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $10 ($22 previous profit minus the $12 loss).
Since you have exhausted the rolls of 6 in our hypothetical scenario, you still have one more roll of 7 to go. This means making one more place bet on 6.
You make your final $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of -$2 ($10 previous profit minus the $12 loss).
Based on the information above, if your bankroll was only the $12 you began with, you just lost 17% of your bankroll. If your bankroll was $100, you just lost 2% of your bankroll.
Here is the real question -- Was the loss due to the probability of rolling 6 before 7, or due to the house edge?
By checking out the same scenario, using the true craps odds, we can get a better idea of the impact of the house edge.
You start with a $12 place bet on 6 and it wins. This gives you a profit of $14.40.
Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled before a 6. You lose the $12 place bet, and now have a total profit of $2.40 ($14.40 previous profit minus the $12 loss).
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $16.80 (the previous total profit of $2.40 plus the $14.40 profit on this bet).
Next, you make another $12 place bet on 6, but, since we are alternating results, the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $4.80 ($16.80 previous profit minus the $12 loss).
So far you have rolled 6 twice and 7 twice.
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $19.20 (the previous total profit of $4.80 plus the $14.40 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $7.20 ($19.20 previous profit minus the $12 loss).
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $21.60 (the previous total profit of $7.20 plus the $14.40 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $9.60 ($21.60 previous profit minus the $12 loss).
You have rolled 6 a total of four times and 7 a total of four times. This means you have one more roll of 6 and two more rolls of 7 to go.
Next, another $12 place bet on 6 and it wins. This gives you a profit of $14.40 for this bet, and an overall profit of $24 (the previous total profit of $9.60 plus the $14.40 profit on this bet).
Next, you make another $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $12 ($24 previous profit minus the $12 loss).
Since you have exhausted the rolls of 6 in our hypothetical scenario, you still have one more roll of 7 to go. This means making one more place bet on 6.
You make your final $12 place bet on 6, but the 7 is rolled again before a 6. You lose the $12 place bet, and now have a total profit of $0 ($12 previous profit minus the $12 loss).
Based on the information above, if your bankroll was only the $12 you began with, you just broke even. If your bankroll was $100, you just broke even.
By examining the two hypothetical scenarios above, it should be plain to see that the house edge is not solely responsible for your losses.
The probability of making a number before 7, and the house edge combined, led to the loss. What would have happened if we disregarded the probabilities, and rolled 6 and 7 five times each?
Looking at the first scenario, with the house edge factored in, you would be ahead, with a profit of $10. Looking at the second scenario, with the true craps odds factored in, you would be ahead, with a profit of $12.
What does this mean? Craps odds are not solely responsible for the long term loss expected in the game of craps.
It takes a combination of the probabilities (the number combinations that will be produced over the long run), plus the odds (actual payouts that factor in the house edge), and in certain cases, the rules of the game (for example, the rule that bars 12 on the come out roll when betting Don't Pass).
Does this mean that you can make a profit in the short run? Yes! How do you determine what the long run is?
Great question! Maybe you should ask the dice.;-)
