Question

A four sided, balanced die has the values 1, 2, 3, and 4. There are three successive throws. Find the probability of throwing the same value on exactly two throws

Answer 1

Total possible permutations = \(4*4*4 = 4^3\)

Next observe that you can fix same number on two dice in \(4\) ways,
After that, the third die can be any of remaining \(3\) numbers
So there are \(4\times 3 = 12\) ways for this to happen and accounting for permutations we have \(12\times 3 = 36\) total favorable outcomes.

Answer 2

Take the ratio of favorable and total for the probability :
\[\dfrac{36}{4^3}\]

Answer 3

You could also work it by considering a 3 bit string.

you can choose first digit to be anything so probability is 1
after that, you may choose the second digit to be same as first digit : 1/4
Finally the third digit must be different : 3/4

So the probability for having first two same digits and last digit different would be
\[1\cdot \dfrac{1}{4}\cdot \dfrac{3}{4}\]

Multiply that by \(3\) to account for order

Answer 4

Another way is to subtract from 1 the probabilities having all different numbers (4*3*2/64) and all identical (4/64).
1-(4*3*2)/64-4/64=36/64

Answer 5

Ahhh okay, I understand it now! It was driving me crazy lol. Thanks for the help!