Let us say that we make n rolls, where n is an even integer. Since we do not care about the number returned by a roll besides whether it is even or not, we could simplify this a bit by saying every roll has 2 equal likely possible outcomes, that is even or odd. This also means that you should not assume I am correct without first checking if I did not made any mistakes. Please note that I do not guarantee that my answer is correct, so if anyone could verify it then that would be great. The probability of rolling an even number on a 2 sided dice, on a 10 sided dice, or even on a 1,000,000 sided dice are all the same!īut if the number of odd and even sides are not equal (if the dice has an odd number of sides) then there is not the same probability! The more sides, the closer the fraction will get to 1/2, but it will never be 1/2! So as long as the dice has the same number of even and odd sides (if the dice has an even number of sides) it makes no difference how many sides it has. The probability of getting an even number on the first dice is different than on the second dice! With a 7 sided dice there are 3 even numbers (2,4,6), but 4 odd numbers! (1,3,5,7)Īnd on a 101 sided dice there are 50 even numbers, and 51 odd numbers. HOWEVER, if the dice has an uneven number of even and odd sides (like a 7 or 101 sided dice) the probabilities will be DIFFERENT. On a 100 sided dice there are 50 possible even numbers (I won't write them all, but I encourage you to if you don't get it)ģ/6 and 50/100 are both the same fraction, 1/2! So the probability of rolling an even number on both of those dice is the same! The number of sides would not change the answer. On a 6 sided dice there are 3 possible even numbers (2,4,6) Noticing these patterns can make counting much easier. It is the sum of the first 6 perfect squares. For example, for a sum less than 14 it just so happens that the number of favourable outcomes is 1 + 4 + 9 + 16 + 25 + 36 = 91. It is important to find some sort of pattern for organized counting. It can be a hassle to list out and count the individual favourable outcomes. If my calculations are correct, there are 91 ways to get a sum less than 14. Count how many results are favourable (produce a number less than 14) and divide that by the total number of outcomes to get the probability. The same applies for an addition which results in a number less than 14. There are a total of 91 ways to get 6 in at least one of the three dice. You can calculate the probability of another event just by finding the total number of outcomes, in this case 216, and counting how many results involve getting one or more 6. Therefore, the probability is still 1/8 after reducing the fraction, as mentioned in the video. And yes, the number of possible events is six times six times six (216) while the number of favourable outcomes is 3 times 3 times 3. Rolling three dice one time each is like rolling one die 3 times. Your first question is basically the same as the one described in the video.
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