Rolling a die 6 times!

Probability Level 3

A blind man rolls a fair die marked $1$ , $2$ , $3$ , $4$ , $5$ and $6$ on it's six faces. After the first rolling, he erases the number on one of it's six faces turning up and marks it with $1$ . He rolls the die again and erases the number on the face turning up and marks it with $2$ . He repeats this exercise four more times and marks $3$ , $4$ , $5$ and $6$ on faces turning up in subsequent trials.

If the probability that the face initially marked with $4$ still has $4$ marked on it after the sixth trial is $\dfrac{a}{b}$ , where $a,b$ are coprime positive integers, then find the value of $a+b$ .

The answer is 67681.

1 solution

Gediminas Sadzius
Oct 13, 2019

There are two ways of having 4 still marked on the face initially marked with 4 after six rolls:

The man does not roll the face 4 in the six rolls. The probability of this happening is $(\frac{5}{6})^{6}$ .
The man rolls the face 4 on the 4th roll, and then does not roll the face 4 in the last two rolls. The probability of this happening is $\frac{1}{6}\times(\frac{5}{6})^{2}$ . Note that it does not matter what the man rolls on the first three rolls as long as he rolls the face 4 on the 4th trial.

So the total probability is $(\frac{5}{6})^{6}$ + $\frac{1}{6}\times(\frac{5}{6})^{2}$ = $\frac{21025}{46656}$ . The answer is 21025+46656= $\boxed{67681}$ .

Rolling a die 6 times!

The answer is 67681.

1 solution

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