The expected value if you roll a die three times is 3.5.
pr(six) = 1/6 → expected 6s in 90 rolls = 1/6 × 90 = 15
When rolling a fair six-sided die, the probability of rolling a three on any single roll is 1/6. If you roll the die 100 times, you can expect to see the number three about ( \frac{1}{6} \times 100 ), which is approximately 16.67 times. Therefore, you should expect to see the number three about 17 times in 100 rolls.
The expected value is 20 times.
fifty percent
The expected number of rolls under 3 is the number of rolls times the probability of rolling less than 3. So: E(rolls less than 3 out of 30) = 30 * 1/3 = 30/3 = 10
pr(six) = 1/6 → expected 6s in 90 rolls = 1/6 × 90 = 15
Three times.
The expected value is 20 times.
The expected number of times is the probability x number of throws. Since you have a prob of 1/6 for a seven, then (1/6) * 160 = 26.67 times you would have success. We generally would round up the expected number of successes to the whole number 27.
fifty percent
The expected number of rolls under 3 is the number of rolls times the probability of rolling less than 3. So: E(rolls less than 3 out of 30) = 30 * 1/3 = 30/3 = 10
100/6 = 16.6 times
The expectation is 50 times.
If a DIE (not dice) is rolled 90 times, the expected value of the sum of the first and second rolls is 7 if you assume that the die is fair. It does not matter how many times you roll the die, as long as it is at least 2.
Type your answer here... 15
Never. If you roll it 1 time, there can be no "then".
16.66... % of the time - if the die is fair. And the percentage is not affected by the number of times you roll it.