# Probability of two rolling dice

As I am taking some distance course on probability theory, I have no other option to ask my question than here. If they sound too basic, please bear me.

Q: I roll two dice. What's the probability I get at least one six?

A: P(first die is six) = $\frac{1}{6}$.

P(second die is six)= $\frac{1}{6}$.

P(both dice are six)= $\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.

P(at least one six)= $\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}=30.6\%$

This is what explained in the video. I think the question asked is "at least one six", so if I get six on both the dice, its ok for me, right? If that was the case, I need to add $\frac{1}{36}$ rather than subtracting, isn't it. Am I thinking right. Please clarify my doubt.

So, the solution will be

P(at least one six)= $\frac{1}{6}+\frac{1}{6}+\frac{1}{36}= 36.2 \%$

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No. $P(A\ \text{or}\ B)=P(A)+P(B)-P(A\ \text{and}\ B)$. That's what's being used here. You get at least one $6$ if the first die is $6$ or the second is $6$. – David Mitra Dec 19 '13 at 9:41
You already added 1/36 in each 1/6, so two times, so you need to remove one. – Jean-Claude Arbaut Dec 19 '13 at 9:42
Or said differently in the probabibily $1/6+1/6$ is contained following configurations: one six on dice 1and something else on dice 2, something else on dice 1 and one six on dice 2, one six on dice 1 and one six on dice 2 (two times). – Umberto Dec 19 '13 at 9:46
Yes, I agree, but the question is not "atmost one six" it is "atleast one six". So, six on first die is OK, six on second die is OK and also six on both dice is OK for me. Hence I need to add all the possibilities. If I am wrong, I need some intuition rather than looking into the formula P(A or B). – Brainy Dec 19 '13 at 9:51
A $6$ on the first die occurs if both show $6$. Same for a $6$ on the second. So when you add them, you counted a $6$ on both twice. Hence, the subtraction... – David Mitra Dec 19 '13 at 9:54

The first $\frac 1 6$ counts the number of solutions where the first dice rolls a 6: $$\frac{ |\{(6,1), (6,2), ... (6, 6)\}|} {36} = \frac 6 {36}$$

The second $\frac 1 6$ counts the number of solutions where the second dice rolls a 6: $$\frac{ |\{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac 6 {36}$$

You want to combine the solution sets:

$$\frac{ |\{(6,1), (6,2), ... (6, 6)\} \cup \{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac {\text{???}} {36}$$

The first solution set has 6 elements, the second has 6 elements, but they share 1 element.

$$\frac{ |\{(6,1), (6,2), ... (6, 6)\} \cup \{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac{ |\{(6,1), (6,2), ... (6, 5), (1,6), (2,6), ... (5, 6), (6,6)\}|} {36} = \frac {6 + 6 - 1} {36}$$

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Great answer! This has some superb intuition - the conventional answer to such questions is just $\frac{36}{36} - \left(\frac{6-1}{6}\right)^2$. This is a much more interesting way of looking at the problem. – Newb Dec 19 '13 at 10:49

In this answer, dice is the plural form of die. I assume that you are using regular 6-sided dice.

For one die, the probability of rolling 3 or lower is 12. For three dice, you want at least two out of three with 3 or lower.

P(all dice 3 or lower)=12×12×12=18 P(two dice 3 or lower)=12×12×12×3=38

Hope this helps........

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