If a die is rolled $10$ times, what is the probability that the number of even results differs from the number of odd results by at most $2$?

Question

Here is the problem:

A fair die is thrown independently $10$ times. What is the probability that difference between the amount of obtained even results and amount of odd results is lower than or equal to $2$?

I understand that problem on this way:

If the difference between the amount of obtained even results and the amount of odd results has to be lower than or equal to $2$ and we roll the die $10$ times we are looking for situation where $5$ results of rolls are even and $5$ are odd (difference is $0$, so lower than $2$) or we have $4$ even results and $6$ odd results (difference $2$) or we have $6$ even results and $4$ odd results (also difference $2$).

If we roll the symmetrical die, the chance for odd result and even result is $1/2$.

When we roll the die $10$ times we have $1/2^{10}$ possibilities which means that the chance for result we want is $1/1024$.

This is where I am stuck. Do you have any ideas how to solve it or do you see some mistakes in what I have written?

I forgot to mention that is it closed question and possible answers are $A=0,6563; B=0,2461; C=0,4512; D=0,1719$.

Answer 1

If we simplify the outcomes to even, odd, we get a binomial $(10,\frac12)$, yielding

$$\operatorname{Pr} = \dfrac{2\binom{10}{4}+\binom{10} 5}{2^{10}} = \frac{21}{32} \approx 0.6563$$

Added

The first term in the numerator takes care of both $6$ even and $4$ even, and the second term is for $5$ each of even and odd.

Answer 2

Here's another very simple solution that might help you understand:

There are three possible outcomes where the difference between the # of odd and even rolls is <= 2:

• 5 odds & 5 evens,
• 6 odds & 4 evens,
• 4 odds & 6 evens

For simplicity we can look at just the number of evens. Let E be the binomial random variable representing the number of even #s we have rolled.

Therefore, the total probability of the difference between the # of odd and even rolls being at most 3 is the sum of the probabilities of these 3 events:

P(E=5) = (10 choose 5) * 0.5^5 * 0.5^5
P(E=4) = (10 choose 4) * 0.5^4 * 0.5^6
P(E=6) = (10 choose 6) * 0.5^6 * 0.5^4   

P(E=5) + P(E=4) + P(E=6) = 0.656