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Two dice are rolled, what is the probability that die 1 shows a 5 given that a 10 has been rolled?

I'm a little confused as to how to solve this exercise. I get two different results depending on how I go about solving it.

Method 1 (logic): The probability of having die 1 show a 5 when two dice are rolled is 1/36. The probability that a 10 is rolled is 3/36. Now I put this in the basic probability format and it gives me (1/36)/(3/36)=1/3.

Method 2 (baye's theorem): P(A/B) = P(A)xP(B/A) / P(B) = 1/36 x 1/6 (bc if die 1 shows a five and I need to get a 10, then die 2 needs to show a five also, hence 1/6) / (3/36) = 1/18

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  • $\begingroup$ P(1st die is 5)=1/6 your text was a bit unclear on that. We get P(1st die is 5 | sum is 10) = P( both dice 5) / P(sum is 10) = (1/36)/(3/36) = 1/3. Your Baye's them calculation must contain an error. $\endgroup$ – jdods Jul 9 '16 at 2:14
  • $\begingroup$ $P(B|A)$ is not $\frac 1{36}$ you get to assume that one die is a 5. It is not the same as the chance of rolling double 5's outright. $\endgroup$ – Doug M Jul 9 '16 at 3:07
  • $\begingroup$ Another approach to is to discard all rolls that do not sum to 10, at the first step. Leaving (6,4) and (5,5), and (6,4) is twice as likely as doubles. $\endgroup$ – Doug M Jul 9 '16 at 3:10
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Observe that our probability space is $(\Omega, \mathbb P)$ where

$$\text{sample space: }\Omega = \{\{1,1\}, \{1,2\}, ..., \{6,5\}, \{6,6\}\}$$

$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{36}$$

We want to compute

$$P(\text{5 is rolled} | \text{sum is 10}) \tag{*}$$

Observe that the event $\text{5 is rolled}$ corresponds to the sample outcomes $\{5,1\}, ..., \{5,6\}$ and the event $\text{sum is 10}$ corresponds to the sample outcomes $\{5,5\}, \{4,6\}, \{6,4\}$.

Hence,

$$(*) = P((\{5,1\} \cup ... \cup \{5,6\}) | \{5,5\} \cup \{4,6\} \cup \{6,4\})$$

$$ = \frac{P((\{5,1\} \cup ... \cup \{5,6\}) \cap (\{5,5\} \cup \{4,6\} \cup \{6,4\}))}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$

$$ = \frac{P(\{5,5\})}{P(\{5,5\} \cup \{4,6\} \cup \{6,4\})}$$

$$ = \frac{P(\{5,5\})}{P(\{5,5\}) + P(\{4,6\}) + P(\{6,4\})}$$

$$ = \frac{1/36}{3/36} = 1/3$$


Another way to look at it is to change our probability space

$$\text{sample space: }\Omega = \{\{5,5\}, \{4,6\}, \{6,4\}\}$$

$$\text{probability measure: }\forall \omega \in \Omega, \mathbb P(\omega) = \frac{1}{3}$$

So if we want to compute $P(\{5,5\})$, by definition of our probability measure, we have 1/3.

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Presumably $B$ is the event a $10$ has been rolled, and $A$ is the event Die 1 shows a $5$. Then $\Pr(A)=1/6$, or, if you prefer, $6/36$, $\Pr(B\mid A)=1/6$, and $\Pr(B)=3/36$.

Substituting and simplifying, we get that $\Pr(A\mid B)=1/3$. The two methods give the same answer.

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  • $\begingroup$ Thank you. However, both methods give the same answer when method 1 considers the probability of getting die 1 to show 5 as 1/36; and method 2 considers the probability of getting die 1 to show 5 as 1/6. The problem itself is a bit unclear, as jdods pointed out, but even if we consider the same probability of getting die 1 to show a 5 for both methods, we still get a different answer, why is that? What am I not reasoning correctly? $\endgroup$ – Jose Jul 9 '16 at 2:47
  • $\begingroup$ Method 1 is in essence uses the definition of conditional probability, $\Pr(A\mid B)=\Pr(A\cap B)/\Pr(B)$. It would be better to make $A$ and $B$ explicit, since informal arguments can distressingly often lead to error. Method 2 is essentially the same, except that it uses explicitly $\Pr(B\mid A)\Pr(A)$ to compute $\Pr(A\cap B)$. $\endgroup$ – André Nicolas Jul 9 '16 at 2:57

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