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.


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.

  • $\begingroup$ Thanks for the answer, Im reading it now, to be clear: I do not care which dice is 1 or 4 in event one, as long as one is either. That's why I thought I should double count. $\endgroup$ – user929304 Aug 25 '16 at 13:31
  • $\begingroup$ By double counted I mean you counted the cases (1, 1), (4, 4) four times. there are only 2 such cases. Your calculation double counts this because (2 * 6) already includes both these cases and then you add 2 * 6 together, giving you a count of 4 for those cases. $\endgroup$ – user3284549 Aug 25 '16 at 13:33
  • $\begingroup$ ah yes, by the way, in the second line of Event 1:, you meant to write: 4 cases where dice 1 has value 4 and .... , right? $\endgroup$ – user929304 Aug 25 '16 at 13:37
  • $\begingroup$ Yes, oops. Fixed that now. $\endgroup$ – user3284549 Aug 25 '16 at 13:39
  • $\begingroup$ With two dice with 6 sides, you have 6 possible matching pairs (1, 1), (2, 2), (3, 3), (4, 4), (5,5), (6,6). The answer is thus simply 36-6 = 30. $\endgroup$ – user3284549 Aug 25 '16 at 13:44

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.

  • $\begingroup$ +1. Moreover, you double counted several cases when calculating $p_1$ (namely (1,1) and (4,4)) $\endgroup$ – H. Potter Aug 25 '16 at 12:57
  • $\begingroup$ Write alle possible outcomes in both events and find alle outcomes which are in both. This will be the intersection of the two events. $\endgroup$ – New_to_this Aug 25 '16 at 13:09

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