Determine which of the following assertions is true:

a) The probability that when throwing twenty dice at least ten of them will have a value of at least four equals one half.

b) The probability that when throwing nineteen dice at least ten of them will have a value of at least four equals one half.

Justify your reasoning. Can you determine the answer without a probability calculation?

• #### Solution

Even without a probability calculation, we can observe that if the number of dice $$n$$ is odd, we can pair the elementary events as follows: On all dice we replace the value rolled $$i$$ with $$7-i$$ (or, if values on opposite sides of the dice add to 7, we can turn all dice over so the value on the underside comes to the top).

This pairing gives us a bijection between the set of events $$A$$, containing the throws when at least ten of the dice have an above-average value, and $$B$$, where at most nine dice have an above-average value.

Because the elementary events have identical probabilities and the sets $$A$$ and $$B$$ have the same size, the second assertion is true.

In the first case, where the number is even, some cases would remain unpaired, i.e. those in which exactly 10 dice have an above-average value. The probability will exceed $$\frac12$$ by half the number of such cases, i.e. by $$\frac{\frac12 \binom{20}{10}}{2^{20}}$$.

By calculating we can determine that the probability in the first case comes out to $$\frac{616\,666}{1\,048\,576}$$.  