If it's a six-sided die (like the kind used in craps and most dice games) the odds are 1 in 6.The probability of rolling a single number on any kind of die* is 1 ÷ (the number of sides).*Except, of course, for weighted, shaved, or otherwise illegal dice.
Answer 1:The odds are very easy to calculate. Simply divide the number of "valid" rolls against all possible rolls. For ease, you can write down all possible combination for the 2 dice.1-1; 1-2; 1-3; 1-4....and so on, remember 1-4 and 4-1 are different rollsThere are 36 unique possible combination, and 6 of them are doubles, so that's 6/36 chances (and since 6 goes into 36, 6 times, this reduces to 1/6) or about 17%Answer 2:Another way to look at this problem, generically, is to assume we have an 'n' face dice. In most cases, dice have 6 faces (1, 2, 3, 4, 5, 6). But why not create a solution that works for any number of sides? Well, if we are trying to calculate the probability of rolling two dice (dice-1 and dice-2) of 'n' sides at the same time and having them turn up as doubles, only one of the dice really matters. Here's why. Dice-1 is guaranteed to land on a number 1-n. This will happen every time (on a fair dice, disregarding freak incidents). What we are trying to calculate is the probability that dice-2 will land on the SAME number as dice-1. Dice-2 can only land on one of 'n' values: 1, 2, 3, 4, 5, 6, ... , 'n'. For you non math folks, this just means it must land on a number from 1 to 'n' where 'n' is the number of sides on your dice. Out of all of the sides that dice-2 can PHYSICALLY land on, one of the sides MUST necessarily have the same as the value that dice-1 landed on. That is to say, if dice-1 landed on the value 3, there must be some chance that dice-2 will also land on the value 3. The probability of this occurring on a fair die is 1 divided by the total number of possible outcomes, which would be 'n'. So, really, there is a 1/n chance that dice-2 will land on the same number as dice-1. Thus, our probability for rolling doubles is simple 1/n. For our 6 sided dice example, our dice-1 lands on some value between 1 and 6 and there is a 1/6 chance dice-2 will match it.
Assuming you are only rolling one die, then each of the six numbers (1,2,3,4,5,6) has an equal chance (1 sixth) of landing face up.
Ten-sided dice are often used in games that require a "percentage roll", such as Dungeons & Dragons. Players roll two 10-sided dice, and the numbers rolled are used as digits to form a number between 0 and 99. Using a "0" instead of "10" simplifies this process. For most other games and instances, zeroes on ten-sided dice represent tens. Just FYI - some ten-sided dice do actually have a "10" instead of a "0", but they're fairly rare. I got 4 of them in a board game called Filthy Rich, which - ironically enough - uses numerous references to rolling zeroes instead of tens. Go figure.
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The chance of rolling a certain number on a dice is 1/6, so the chance of rolling a 4 OR 5is 2/6. But if you have two dice you have the most chance of rolling a seven (6/36 or 1/6)
since there are 6 sides to it any of them could show up, 1,2,3,4,5, and 6 could all show up unlikely
2. This would be as a result of rolling a 1 on each of them.
If it's a six-sided die (like the kind used in craps and most dice games) the odds are 1 in 6.The probability of rolling a single number on any kind of die* is 1 ÷ (the number of sides).*Except, of course, for weighted, shaved, or otherwise illegal dice.
AnswerThe most likely number is seven.This can be 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.Remember that dice have no memory - if you have rolled 7 three times in a row the odds of rolling another 7 does not change.this is all wrong don't listen that
The most probable result of rolling two dice is a sum of seven. The probability of rolling a seven is 1 in 6 or about 0.167.All of the other possible sums have decreasing probability, all the way down to 1 in 36 or about 0.0278 for a sum of two or a sum of 12.
The problem can be split into two parts, rolling a 12, or rolling a 4 or less. This can be further broken down to rolling a 2, rolling a 3, rolling a 4, or rolling a 12. P(rolling 4 or less, or 12) = P(rolling 4 or less) + P(rolling 12) = P(rolling a 2) + P(rolling a 3) + P(rolling a 4) + P(rolling a 12) = 1/36 + 2/36 + 3/36 + 1/36 = 7/36
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Answer 1:The odds are very easy to calculate. Simply divide the number of "valid" rolls against all possible rolls. For ease, you can write down all possible combination for the 2 dice.1-1; 1-2; 1-3; 1-4....and so on, remember 1-4 and 4-1 are different rollsThere are 36 unique possible combination, and 6 of them are doubles, so that's 6/36 chances (and since 6 goes into 36, 6 times, this reduces to 1/6) or about 17%Answer 2:Another way to look at this problem, generically, is to assume we have an 'n' face dice. In most cases, dice have 6 faces (1, 2, 3, 4, 5, 6). But why not create a solution that works for any number of sides? Well, if we are trying to calculate the probability of rolling two dice (dice-1 and dice-2) of 'n' sides at the same time and having them turn up as doubles, only one of the dice really matters. Here's why. Dice-1 is guaranteed to land on a number 1-n. This will happen every time (on a fair dice, disregarding freak incidents). What we are trying to calculate is the probability that dice-2 will land on the SAME number as dice-1. Dice-2 can only land on one of 'n' values: 1, 2, 3, 4, 5, 6, ... , 'n'. For you non math folks, this just means it must land on a number from 1 to 'n' where 'n' is the number of sides on your dice. Out of all of the sides that dice-2 can PHYSICALLY land on, one of the sides MUST necessarily have the same as the value that dice-1 landed on. That is to say, if dice-1 landed on the value 3, there must be some chance that dice-2 will also land on the value 3. The probability of this occurring on a fair die is 1 divided by the total number of possible outcomes, which would be 'n'. So, really, there is a 1/n chance that dice-2 will land on the same number as dice-1. Thus, our probability for rolling doubles is simple 1/n. For our 6 sided dice example, our dice-1 lands on some value between 1 and 6 and there is a 1/6 chance dice-2 will match it.
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Assuming you are only rolling one die, then each of the six numbers (1,2,3,4,5,6) has an equal chance (1 sixth) of landing face up.