A Game Of Dice Continued

Recall from our previous example we obtained the impossible probability of rolling a single die a number of times and obtaining a value greater than one. To help understand the flaw in our reasoning let's make this even simpler. What is the probability of rolling at least one '4' after 2 tosses of a single die?

Intuitively we might say 1/6 probability of rolling a '4' on the first toss plus 1/6 probability of rolling a '4' on the second toss gives us 1/3. So is 1/3 the right answer?

What about the case where we roll a '4' on the first toss and a '4' on the second toss?

If we make a list of all possible combinations of the value of the first roll of the die and the value of the second roll of the die, we will end up with with 36 combinations:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

Recall our original question: What is the probability of rolling at least one '4' after two rolls of the die?

So all we have to do is to count the number of times we see at least one '4' show up in the 36 possible combinations. From the list we see eleven instances in which at least one '4' is present.

To find the probability we take the ratio of the number of combinations in which at least one '4' is present and the total possible combinations we can have.

Our answer is 11/36.

We have an 11/36 probability of rolling at least one '4' after 2 tosses of a die.

Mysterious? Counter intuitive. Absolutely! But with some straight forward logical reasoning we have come up with the true probability.

To be continued...