3 over 4 , or 8th quared
If we are thinking of getting a '6', here are the odds. Wth one dice, its 1 in 6. So,with two dice its 1 in 216 with three dice its 1 in 7776 with four dice its 1 in 279936 with five dice its a huge 1 in 10077696
To roll a three on any one dice, the odds are 1/6. To roll a three on any one of a pair of dice, the odds are 1/6 x 1/6 which is 1/36 or 1 in 36 chance.
If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.
2 out of 12
(1/6)(1/6)(1/6) = 1/216 Tha'ts the total number of outcomes. Now have a look at the possible combinations, describes as [die1] [die2] [die3] 344 434 443 That means the odds are (216/3=) 1/72 chance of rolling 3, 4, 4. with three dice.
6
Assuming fair dice: With 3 dice there are 6 x 6 = 36 possible permutations (the two dice can be identified and are independent of each other). Of these there are 3 ways (1+3, 2+2, 3+1) of getting four. Probability of getting a four with 2 dice on one roll = 4/36 = 1/9 As rolls are independent, the probability of getting 4 with 2 dice 3 times in a row is 1/9 x 1/9 x 1/9 = 1/729 So the odds of it are 728 to 1 against, ie 1 in 729 or about 0.14 %.
i am not sure
-3
The odds are 1:3. The probability is 1/4 or 25%.
There is a one out of four chance of having both dice even numbers.
The odds of getting four of a kind in Texas Hold'em are approximately 1 in 4,165.