Calculate the conditional probability that the sum of two dice tosses is even given that at least one of the tosses gives a five.

I'm a bit confused by this. Shouldn't the probability just be 1/2, since we know that at least one of the dice tosses gave us a five, thus the other must give us an odd number?

  • $\begingroup$ The issue with this simplified argument is the case that both dices show a $5$. Did you calculate the probability with the theorem of Bayes' ? $\endgroup$ – Peter May 12 at 7:15
  • $\begingroup$ @Peter No I didn't. I'm not sure how we can calculate the reverse conditional probability? (probability that one of the die toss gives us a five given that the sum is even, that seems to be even more difficult to compute directly) $\endgroup$ – Gummy bears May 12 at 7:18
  • $\begingroup$ The possible events are $15,25,35,45,55,65,51,52,53,54,56$ , $11$ elementary events. Hence $\frac{1}{2}$ cannot be the right result. $\endgroup$ – Peter May 12 at 7:18
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    $\begingroup$ @Henry I'm quite sure 1/2 isn't the right answer. Also, can you give some justification for your 13/27 answer? I'm confused as to how you can get that? $\endgroup$ – Gummy bears May 12 at 7:21
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    $\begingroup$ The clue that the probability is not just $\frac12$ is the fact that your condition is "at least one is a five." In such cases, the number on "the other" die is not an independent variable. In fact the whole concept of "the other" die is dubious; which die is the "other" one when you roll double fives? $\endgroup$ – David K May 12 at 12:42

A = event when one of the tosses gives a five. (Sample space for the conditional probability) Let (n1, n2) represent the outcomes of die1 and die2 A = { (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6) } // (5,5) must be counted once only

Thus n(A) = 11

B = sum of two dice tosses is even n(B|A) = { (1,5), (3,5), (5,5), (5,1), (5,3)|

P(B|A) = n(B|A)/n(A) = 5/11

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    $\begingroup$ This is exactly the way I know that 1/2 is not the correct answer. I was just wondering if there was a more intuitive way of arriving at this? Something that can be thought of rather than listed and calculated? $\endgroup$ – Gummy bears May 12 at 7:25
  • $\begingroup$ I don't think we can come up with a generic formula around arriving at this. The key to this problem is around carefully counting the sample space and the favorable outcomes without double counting. For example, you can extend this question to a n dice being cast, where each die can have m outcomes. There again while writing down the sample space and favorable outcomes, one needs to identify and eliminated double counting of possible outcomes. $\endgroup$ – Don Bosco May 12 at 7:40
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    $\begingroup$ @Gummybears, one reason the chance is lesser than $1/2$, is your sample space is larger. If the dies were labelled red and blue, the chance that the sum is even, given the red die rolls $5$ is $1/2$. However, that is more restrictive than saying atleast one die rolls a 5. $\endgroup$ – Quasar May 12 at 8:13
  • $\begingroup$ @Gummybears, if your question was how to ensure the sample space should be counted correctly. A = event when one of the tosses gives a five. n(A) = n(first die gives "5" as outcome ∪ 2nd die gives "5" as outcome) = n(first die gives "5" as outcome) + n(2nd die gives "5" as outcome) = n(first die gives "5" as outcome ∩ 2nd die gives "5" as outcome) = 6 + 6 - 1= 11 Apply same principle while counting number of favorable outcomes for event B|A. Hope this helps! $\endgroup$ – Don Bosco May 12 at 11:13
  • $\begingroup$ @Gummybears , "I was just wondering if there was a more intuitive way of arriving at this" yes - I explained it in my answer. The other folks here are too smart to explain it :) :) $\endgroup$ – Fattie May 12 at 17:57

By the definition of conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Let $A$ be the event of the two rolls adding to an even number and let $B$ be the event of rolling at least one five.

Note that $A \cap B$ = the event of the two rolls adding to an even number AND rolling at least one five. Now we need to find $P(A \cap B)$ and $P(B)$:

There are three ways we can roll at least one five.

1) You roll a five on the first roll but not the second. The probability of this is $(1/6)*(5/6) = 5/36.$

2) Not rolling a five on the first roll, but rolling a five on the second: Probability of this is $(5/6)*(1/6) = 5/36.$

3) Or you can roll two fives: Probability $= (1/6)*(1/6) = 1/36.$

These three events are disjoint, so to find the probability of getting $\textit{at least}$ one five, we add them: $5/36 + 5/36 + 1/36 = 11/36$.

Now to calculate $P(A \cap B)$: If you roll two fives, the sum is even. Otherwise, the roll that is not a five must be an odd number (ODD + ODD = EVEN). In each of cases 2) and 3), the only other rolls that give you an even sum are 1 and 3. That's a total of 5 scenarios out of a possible 36. So $P(A \cap B) = 5/36.$ Finally,

$$P(A|B) = \frac{5/36}{11/36} = \frac{5}{11}.$$

  • $\begingroup$ Now that's intuitive ;-) $\endgroup$ – Fattie May 12 at 17:51

If the dies were colored, say red and blue, we asked the question, what's chance that sum is even, given that the red die rolls a five, that's $1/2$. But, you see, either red or the blue die could roll a five. So, your sample space is slightly larger.

Fix $X_{i}$ to be the random variable representing the number on die $i$, $i=1,2$. You could draw a nice tree to think of the possible outcomes of this experiment and count them.

(1) When the first die is rolled, there are three possibilities - $X_{1}=5$, $X_{1}=1 \cup X_{1}=3$, $X_{1}\text{ is even}$.


$P(X_{1}=1 \cup X_{1}=3)=1/3$.

$P(X_{1} \text{ is even})=1/2$.

(2) The second die is now rolled, again having three possibilities each - $X_{2}=5$, $X_{2}=1 \cup X_{2}=3$, $X_{2}\text{ is even}$. So, we have $3 \times 3=9$ branches.

$P(X_{1}=5 \cap X_{2}=5)=1/6 \times 1/6=1/36$.

$P(X_{1}=5 \cap (X_{2}=1 \cup X_{2}=3))=1/6 \times 1/3=2/36$.

$P(X_{1}=5 \cap X_{2}\text{ is even})=1/6 \times 1/2=3/36$.

$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}=5)=1/3 \times 1/6 = 2/36$.

$P((X_{1}=1 \cup X_{1}=3) \cap (X_{2}=1 \cup X_{2}=3))=1/3 \times 1/3 = 4/36$.

$P((X_{1}=1 \cup X_{1}=3) \cap X_{2}\text{ is even})=1/3 \times 1/2 = 6/36$.

$P(X_{1} \text{ is even} \cap X_{2}=5)=1/2 \times 1/6 = 3/36$.

$P(X_{1} \text{ is even} \cap (X_{2}=1 \cup X_{2}=3))=1/2 \times 1/3 = 6/36$.

$P(X_{1} \text{ is even} \cap X_{2}\text{ is even})=1/2 \times 1/2 = 9/36$.

Counting the outcomes where the sum is even and atleast one roll is 5, the numerator $1/36+2/36+2/36=5/36$.

Restricting our attention to events, where atleast one roll is 5, the denominator = $1/36+2/36+3/36+2/36+3/36=11/36$.

So, the chance is $5/11 < 1/2$.


As suggested above, the Bayes th. can help too:

Let $S$ be the event that the sum is even, and $A5$ the event that at least one $5$ shows up, and $S^c$ is the event that the sum is odd.

Then, $$P(S|A5) = \frac{P(A5|S)P(S)}{P(A5|S)P(S)+P(A5|S^c)P(S^c)} $$

$P(S)=P(S^c)=0.5$, $P(A5|S)= 5/18$, as there are $5$ in $A5$ for even sum. Also, $P(A5|S^c)= 6/18$ when the sum is odd.

Thus, $P(S|A5) = 5/11$.


It depends upon how you found out that at least one of the dice is 5.

Are you familiar with the children’s game Guess Who? Given a set of 24 characters, you have to guess which one is mine by asking yes/no questions. Suppose you ask me “Does your person have glasses?” And I answer yes. Since most of the characters do not have glasses, there’s a lot of information in my answer. Out of 24 characters, only 6 remain.

It’s a similar principle with the dice. If you ask, “is at least one of them 5”, and I answer yes, you eliminate 25 possibilities and leave 11. Whereas the original 36 possibilities have an equal distribution of odds and evens, the remaining 11 do not.

By asking a question, you learned something about the result of rolling the dice. You’ve narrowed the possibilities, and gotten a little closer to knowing for sure whether the sum is even.

If you didn’t ask about a particular number, but I just looked at one of the dice and said “at least one of them is X”, then I haven’t given you any information. I don’t even know the answer myself, since I only looked at one of the dice. In that case, the answer is 1/2.

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    $\begingroup$ Indeed, this answer goes for an "intuitive explanation." (Which I believe is what the OP is after, based on comments.) $\endgroup$ – Fattie May 12 at 18:02

A visual explanation. Given that a pair is in the right half, what’s the probability that it’s in the top half?

11    13                           15      
   22    24    26        
31    33                           35
   42    44    46
                       51    53    55
   62    64    66

   12    14    16      
21    23                           25
   32    34    36
41    43                           45
                          52    54    56
61    63                           65

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    $\begingroup$ Now that's clever! :) $\endgroup$ – Fattie May 13 at 11:35

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