Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?

Attempt: I feel I'm having trouble formalizing the expression for what I believe to be happening. This is a question concerning conditional probabilities so as such I will need first the probability of obtaining a sum of $6$:

$$P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)$$

So in terms of notation, the ordered pairs represent the way in which a sum of $6$ could be attained (ex: $(4,2)$, where you roll a $4$ on the first die and $2$ on the second). With that being taken into account:

$$\left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{1}{5}\right)\left(\frac{1}{36}\right) = 0.0495$$

Now what was requested was the probability of a $3$ coming up given the sum of 6:

\begin{align}P((3,3)|6) &= \frac{P(6|(3,3))P(3,3)}{P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)}\\\\ &= \frac{0.0056}{0.0495}\\\\ &= 0.113\end{align}

But I still feel I'm not treating the condition on the sum of $6$ properly.. Is this correct?

  • $\begingroup$ Hint: $P(X_1+X_2=6)=P(X_1=4,X_2=2)+P(X_1=2,X_2=4)+P(X_1=5,X_2=1)+P(X_1=1,X_2=5)+P(X_1=3,X_2=3)=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{5}{36}$ $\endgroup$ – callculus Jun 10 '16 at 18:30
  • $\begingroup$ Perhaps I've looking at this a bit too simply, but the problem seems trivial... The only way for two dice to sum up to $6$ is if each side comes up a $3$, so the probability is 100%? $\endgroup$ – Mortified Through Math Jun 10 '16 at 18:32
  • $\begingroup$ @Alb But there are 4 other ways (in total 5) to get a sum of 6. $\endgroup$ – callculus Jun 10 '16 at 18:34
  • $\begingroup$ @callculus crap, right. Please excuse the post lunch dumb comment... $\endgroup$ – Mortified Through Math Jun 10 '16 at 18:35
  • 1
    $\begingroup$ $P(6|(4,2))$ is the probability you have rolled a $6$ given that the dice show $(4,2), P(6|(4,2)) = 1$ Correct that and you will get $P(6) = \frac 5{36}$ and $P((3,3)|6) = \frac {P((3,3))}{P(6)}$ $\endgroup$ – Doug M Jun 10 '16 at 19:04

$P(1,5)= P(2,4)=P(3,3)=P(4,2)=P(5,1) = 1/36$

Thus, you need $P(3,3)/P(6) = 1/5$. ($P(sum= 6|(3,3))=1$, and sum can be 6 in those 5 ways above with total probability 5/36)


I would use a table to first see this intuiutively

$$\begin{array}{|cc|cccccc|}\hline &&&&&D_1\\ &&1&2&3&4&5&6\\ \hline &1&2&3&4&5&\color{red}6&7\\ &2&3&4&5&\color{red}6&7&8\\ D_2&3&4&5&\color{blue}6&7&8&9\\ &4&5&\color{red}6&7&8&9&10\\ &5&\color{red}6&7&8&9&10&11\\ &6&7&8&9&10&11&12\\ \hline\end{array}$$

The red numbers show all combinations where the dice sum to $6$ and the blue shows the one event we are interested in, where at least one dice shows a $3$

So we have $\dfrac 15$ elements which satisfy the given constraints

Now consider it mathematically

We know that $P(\text{sum}=6)=\dfrac 5{36}$

We also know that $P((3,3))=\dfrac 1{36}$

We can therefore say that \begin{align}P(\text{sum}=6\text{ AND one dice shows }3)&=\frac{P((3,3))}{P(\text{sum}=6)} \\\\ &=\frac{\frac1{36}}{\frac5{36}}\\\\ &=\frac1{36}\cdot\frac{36}5\\\\ &=\frac 15\end{align}


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