First die is some number. Second die is anything but that number. Third die needs to be one of the first two. Fourth and fifth need to be whichever has doubled up already.
No, because we still get that one of the first two dice is the different one.
Since there are 5 places for the different die to be, and it can only be in one of these places, we can just sum up the probabilities of the different die being each of the 5 rolls.
Since all these probabilities are the same, we get 5*(probability of the different die roll being the last one).
Note that there is one more complication: we are also interested in cases where the different number is not the last one.
Fortunately, this is also easy to solve, since every other position has exactly the same probability.
1/1 * 5/6 * 2/6 * 1/6 * 1/6
Right?
First die is some number. Second die is anything but that number. Third die needs to be one of the first two. Fourth and fifth need to be whichever has doubled up already.
No, because we still get that one of the first two dice is the different one. Since there are 5 places for the different die to be, and it can only be in one of these places, we can just sum up the probabilities of the different die being each of the 5 rolls. Since all these probabilities are the same, we get 5*(probability of the different die roll being the last one).
Can you show the math please? I don’t follow the words.
511/61/61/65/6 Which corresponds to (Number of rolls)(any number)(same number)(same number)(same number)(different number)