Two dice throw probability

Question

We throw two conventional dice, and ask the probablity of the following event:

• The main probability $p_m$ to compute is composed of two sub-events: first sub-event, having either of the two dice with a value of 1 or 4, and second sub-event: the two taking values 1 and 4 at the same time.

So I'm trying to compute the probability of either of these two sub-events occuring. My attempt:

• First subevent: for a single die, the probability of either 1 or 4 is 1/6. For two dice, we have 36 outcomes, 6 ways (1,x) can occur, x can be any value, similarly 6 ways (4,x), and because we have two dice, both are multiplied by two, so finally the probablity of subevent one is: $$p_1=\frac{2*6+2*6}{36}\approx 0.66$$
• For the second subevent: there are two ways 1 and 4 can occur at the same time, that is, (1,4) or (4,1) so $p_2 = 2/36 \approx 0.05$
• Finally for the main event, $p_m = p_1+p_2 \approx 0.71.$ Is the estimation correct? Have I gone wrong somewhere with the composition of probabilities?

Note: we assume the two dice are indistinguishable.

Answer 1

You're counting event 2 in your event 1. You are also double counting the cases where the 2 dice have the same value.

In your calculation for event 1 you said dice 2 can have any value (6 possible) after choosing 1 value for dice 1, this includes the cases where they are both equal. By adding these probabilities twice (for 2 dice), you've doubly counted the probability that they are equal.

Note that letting dice 2 have any value in your calculation for event 1 means you also include the cases where one dice is 1 and the other is 4.

You can either subtract the intersection, or you can make up new independent events. Or you can split the events up into the possible cases.

I'll show you a solution using the classic definition of probability: satisfactory cases/total amount of possible cases

Event 1:

4 cases where dice 1 has value 1 and dice 2 has a value that isn't 1 nor 4.

4 cases where dice 1 has value 4 and dice 2 has a value that isn't 1 nor 4.

These 8 cases go both ways (swap the values of the 2) thus you have 2*8 = 16 cases like this.

There's also 2 cases where these dice share values. (both are 4 or both are 1)

In total 18 cases to satisfy the condition for event 1.

Note that none of these cases include the event that both values 1,4 are present at the same time.

Event 2:

1 case where dice 1 has value 1 and dice 2 has value 4

1 case where dice 1 has value 4 and dice 2 has value 1.

So you have 2 cases for event 2.

Note that these 2 cases can't possibly happen at the same time as the previous counted ones, they are disjoint cases, so now to get the total probability, you add all the cases that satisfy these conditions and divide by total amount of possible cases.

Divide by total amount of possible cases (6^2) and you get (18+2)/36 = 20/36.

Answer 2

The two events are not independent, and hence you cannot just add the probabilities of the two events; you also need to subtract the probability of the intersection of the two events.