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1 in 6 The way you calculate this is as follows: There are 36 possible combinations you can get with two dice. Six of those combinations add up to 7 and 30 of those combinations add up to either 6, 5, 4, 3, or 2. Therefore, there are 30 ways you can roll a number other than seven and six ways you can roll a seven. 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." You can say the odds are either 6 ways to win "in" 36 possible combinations or you can say there are 30 ways to lose "to" 6 ways to win. 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".

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0Here are the odds in craps: 7 vs. 6 Odds: 6 to 5 7 vs. 5 Odds: 6 to 4 or 3 to 2 7 vs. 4 Odds: 6 to 3 or 2 to 1 7 vs. 3 Odds: 6 to 2 or 3 to 1 7 vs. 2 Odds: 6 to 1

The odds of rolling a 7 are 1/6. The odds of rolling two in a row are 1/36. The odds of rolling an 11 are 1/18. The odds of rolling two in a row are 1/324. The odds of rolling doubles are 1/6. The odds of rolling double twice in a row are 1/36.

In craps, the chance for rolling a 7 never changes. First roll, tenth roll - doesn't matter. The chance is always 6 out of 36 (or roughly 17%). The chance of rolling a 6 is always 5 out of 36 (or 20%). The chance of rolling an 8 is always 5 out of 36 (or 20%). The chance of hitting either a 6 or an 8 before hitting a 7 would be 10 to 6 (with the favor going towards hitting either the 6 or 8).

Odds of rolling seven once (2 dice) is 1:6, twice in a row, odds are 1:36. The reason is- the first die can be any number, but the second die must be a particular number. For example, if the first die is 6, then the second die has to be 1 to make 7. When an event occurs twice, and the two events are unrelated, then you can multiply the odds together= 1/6 * 1/6 = 1/36.

If you're asking the basic rules of craps, i.e. what bets win and lose based on the value of the dice, you should read a basic craps rules introduction; they're everywhere on the web. Basically, you or someone else rolls the dice. If you roll 7 or 11 you win; 2, 3, or 12 and you lose. If you roll anything else (4, 5, 6, 8, 9, 10), you now have to roll that number AGAIN before rolling a 7. That's the most basic bet, called the Pass line. The Don't Pass is basically the opposite, betting that the shooter will roll a 7 before rolling that number again - "making the point." There are a ton of other bets on the craps table, but many of them are self-explanatory. Again, find a basic introduction. If you're asking how you can maximize your odds, the answer is to play the Pass or Don't Pass with odds, and to buy numbers. Stay out of the field and off the horn (the big section of specific bets down the center of the table). If you're asking how you can gain an edge over the house, sorry, it can't be done. Pass line odds in most casinos pay true and correct odds, making them the only real even-money bet in the whole place. You can't gain an edge though, sorry.

well u can not roll a 7 on a dice

About 1.5% The odds of rolling a 7 on any particular throw is 1/6. (There are six ways you can roll a seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), out of 36 possible outcomes.) Therefore, the odds that you don't roll a 7 on a particular roll is 5/6. The odds that you don't roll a 7 in 23 rolls is: (5/6)^23, or approximately .015=1.5%

The odds of rolling a 7 with two dice is 6 in 36, or 1 in 6.Two six-sided dice will yield 36 different possible combinations in one roll. Note that rolling 1 and 6 is not the same as rolling 6 and 1. Yes, they both equal 7, but for the purposes of determining probability, each throw is unique. There are 6 possible ways to throw a 7, and they are 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. With 36 different possible outcomes, that means that 6 in 36 of those outcomes results in a 7. The odds of getting a 7 are 6 in 36, or 1 in 6 (reducing the two terms).We sometimes see the terms "odds" and "probability" interchanged, but it is important to distinguish them mathematically. If you are interested in the probability of rolling a 7 in a single roll of a pair of dice, use the link to the Related question.

No. The odds paid by the house are always less than the actual odds. The house advantage varies from 1.4% for a simple Pass/Come bet to 16.7% for an Any 7 bet. The only exceptions are the "Odds" bets, which pay at actual odds, but you can't place those independently of other bets.

The most likely number to be rolled in craps is the number 7

If the odds are 9 to 7, the chance of winning is 43.75%

Approximately 1/1120 * You have a 27.777% chance of rolling either a six or an eight on the first roll. * You have a 13.888% chance of rolling the same thing as the first roll. * You have a 13.888% chance of doing it again. * You have a 16.666% chance of rolling a seven. To find total probability in this situation, you multiply all of your chances together. .27777 x .13888 x .13888 x .16666 = 0.000892886 = .0892886% =1/1119.964 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.

"Big red" is a nickname for 7.

If the odds of a horse winning a horse race are 2 to 7 then the odds against that horse winning the race are 7 to 2.

To roll a 7, it doesn't matter what your first die rolls at, as every number on the die has a complement that will add up to 7. This means that your odds on rolling 7 are actually your odds on rolling the appropriate number on the second die. ie. 1/6. No matter what your first die is, the second can potentially give you a number that brings the total to 7. To roll an 11 however, your odds are not as high. This is because you must roll a five and a six to hit 11. Your first die can be either of the two, but your second die must be it's complement. In this case then, your odds are 2/6 * 1/6, or 1/18.

The odds against drawing a black 7 is 25 in 26.

If you mean casino craps, it depends on how you are betting, but a general answer to the question is, on the first roll, called the "come out" not "roll out," a 7 or 11 is considered a winning number.

If rolling 1 die it is 1; if rolling 7 dice it is 0.

It could be dice (craps) : 7 for a "seven natural" to be thrown (more rarely tossed)

Assuming a properly executed break, about 1/8 breaks. The odds increase the smaller the table - a good player will make the 8 ball 5% of the time on a none foot table. This will nearly double on a 7 foot table. The pockets are the same size on all pool tables in the US, so the smaller the table the greater open space and less distance to travel.

Your total possible number of combinations: 62 = 36 Combinations that can add up to 7: 1-6 2-5 3-4 4-3 5-2 6-1 So your odds of rolling a seven would be 6/36, or 1/6.

The probability of rolling a 7 with 2 dice is 6/36; probability of rolling an 11 is 2/36. Add the two together to find probability of rolling a 7 or 11 which is 8/36 or 2/9.

A natural number is any number that is an automatic winner or loser on the first roll. If you are playing regular craps, the natural winners are 7 and 11, and the natural losers are 2, 3, 11, and 12.

There are 7 periods in the periodic table.

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