Rolling dice and conditional probability
Task number: 3542
Vašek has made three rolls of a regular 6-sided die. The sum of his rolls equals \(7\).
Variant
Is it more probable that his first roll was a one or a two?
Hint
Transform this question into a problem of rolling two dice.
Solution
If Vašek first rolls a one, it remains to roll a total of six on two dice.
If he first rolls a two, it remains to roll a total of five on two dice.
On two dice a seven is the most likely total; successively smaller values are decreasingly likely.
Answer
It is more likely that his first roll was a one.
Variant
What is the probability that his first roll was a two?
Solution
The number of ways to form a sum of 7 from three positive integers is \(\binom{6}{2}\) – we distribute 7 balls into three pigeonholes such that no pigeonhole is empty, which is the same as placing one ball into each hole and distributing the remaining four in any manner.
(Notice that in the given problem it will never happen that one pigeonhole contains more than six balls. To generalize this approach for larger sums it would be necessary to use e.g. generating functions.)
If the first roll is a two, there are \(\binom{4}{1}\) cases.
The probability we are seeking is the ratio of these two values.
Comment
Notice that unlike the previous variant the probability is not the same as the probability of throwing a five on two dice – that is \(\frac{4}{36}\).
That would correspond to the probability of rolling a total of seven under the assumption that the first roll was a two, not the probability that the first roll was a two under the assumption that the total roll was seven.
However the solution to the previous exercise is correct, because \(P(\text{1st is two}|\text{sum is 7}) =\frac{P(\text{1st is two and sum is 7})}{P(\text{sum is 7})} =\frac{P(\text{1st is two})P(\text{remaining sum is 5})}{P(\text{sum is 7})}=\)
\( =\frac{\frac16P(\text{remaining sum is 5})}{\frac{15}{216}} =\frac{36}{15}P(\text{remaining sum is 5})\).
The fact that in both cases the probability differs by a multiplicative factor of \(\frac{36}{15}\) follows from the fact that throwing a one or a two in the first roll are equally likely.
