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Consider a set of 3 6-sided die divided into 3 groups, with each group having 2 dice and every die being in exactly 2 groups. Let $p_e$ be the probability that, after rolling all the die, there is at least one odd number rolled in at least one of the groups. Calculate $p_e$ using the expression for $P(A \cup B \cup C)$, and using another (easier) method.

My solution so far:

Let $X,Y,Z$ be the three different dies and the three groups are:

$XY, XZ, YZ$

We want to find

$$P(XY \cup XZ \cup YZ) = P(XY)+P(XZ)+P(YZ)-P(XY\cap XZ)-P(XY \cap YZ) - P(XZ \cap YZ) + P(XY\cap XZ \cap YZ)$$

$P(XY) = 1-P(no.odd)=1-1/4=3/4$

$P(XY)=P(YZ)=P(XZ)=3/4$

$P(XY\cap XZ)=(3/4)(3/4)=9/16$

$P(XY\cap XZ)=P(XY\cap YZ)=P(XZ\cap YZ)=9/16$

$ P(XY\cap XZ \cap YZ)=3/4(3/4)(3/4)=27/64$

Therefore:

$$P(XY \cup XZ \cup YZ) = 3/4+3/4+3/4-9/16-9/16-9/16+27/64=63/64$$

Easier Method:

$$P(XY \cup XZ \cup YZ) = 1- P(XY \cup XZ \cup YZ)^c = 1-P(XY^c\cap XZ^c \cap YZ^c)$$

$P(XY^c)=P(no.odd)=1/4=P(XZ^c)=P(YZ^c)$

So we have $1-(1/4)(1/4)(1/4)=1-1/64=63/64$

Am I correct in my solution?

user130306
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    If you roll 3 dice then the probability of getting all even numbers is ${1 \over 2^3}$, so surely the probability that at least one odd number is rolled is $1-{1 \over 2^3}$? – copper.hat Sep 28 '22 at 05:02
  • It should just be the probability that at least one odd number is rolled, $1-\frac{1}{2}^3$. So that reasoning is not valid. – Nij Sep 28 '22 at 05:03
  • so then what would the easier solution be? @copper.hat – user130306 Sep 28 '22 at 05:05
  • I'm not sure it gets any easier than I wrote. – copper.hat Sep 28 '22 at 05:05
  • But does this account for "6-sided die divided into 3 groups, with each group having 2 dice and every die being in exactly 2 groups". with your solution are the dice being grouped? @copper.hat – user130306 Sep 28 '22 at 05:07
  • As far as I can tell, that has nothing to do with the computation. The probability that all evens are rolled is $1-p_e$. – copper.hat Sep 28 '22 at 05:09
  • @copper.hat hmm im not sure I fully understand. I understand how you got $1-\frac{1}{2^3}$ but the question asks to solve it two different ways. I suppose you showed me the 'easier' way. but how would I solve it by using the "harder" way i.e. by doing $P(A\cup B \cup C)$ – user130306 Sep 28 '22 at 05:13
  • What are $A,B,C$? Presumably you can use the exclusion inclusion formula. – copper.hat Sep 28 '22 at 05:13
  • @copper.hat well I just used $A,B,C$ as examples. What I mean is that you used the complement rule but I don't know how to solve the question by using the formula for the probability of the union of three events. – user130306 Sep 28 '22 at 05:15
  • @copper.hat since the question originally asked to solve the problem by using that formula, and then solving it again using an easier method – user130306 Sep 28 '22 at 05:16
  • @copper.hat Well in my original solution I had that $A=XY, B= YZ, C = XZ$ but I'm not sure what $A,B,C$ would need to be for me to get $7/8$ as the answer – user130306 Sep 28 '22 at 05:20
  • It is not at all clear to me what the groups have to do with the computation. Rolling at least one odd number has nothing to do with the grouping in my interpretation. – copper.hat Sep 28 '22 at 15:47
  • @copper.hat hmm I guess what I'm saying is that if we have 3 die, $X,Y,Z$, there need to be three groups (as stated in the problem). The three groups need to be $XY, XZ, YZ$ because each dice appears in exactly two groups (which is what the problem says). so I'm not sure why exactly my solution is incorrect? – user130306 Sep 28 '22 at 18:20
  • @user130306 I understand that, but as far as I can see the probability of at least one group rolling a die and getting an odd roll is exactly the same as rolling 3 dice and any of them being odd. – copper.hat Sep 28 '22 at 18:46
  • @copper.hat When you roll the three dice you're trying to find P( at least one odd in first group OR at least one odd in second group OR at least one odd in third group). I denoted the first group as $XY$, second group as $YZ$ and third group as $XZ$ – user130306 Sep 28 '22 at 18:50
  • @user130306 Yes, I got that as I said. But how is that different from any of the 3 rolls being odd? – copper.hat Sep 28 '22 at 19:04
  • @copper.hat hmm I see what you mean. but then shouldn't we have gotten the same answer? instead you got $7/8$ and I got $63/64$ – user130306 Sep 28 '22 at 19:14
  • @copper.hat actually what you're saying makes sense. If you have at least one odd then obviously at least one group will have an odd. So I get what you mean now by the groupings not being relevant. – user130306 Sep 28 '22 at 19:15
  • I could easily be wrong, but that is how I interpret the question. – copper.hat Sep 28 '22 at 19:17
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    @copper.hat yeah I'm confused on which interpretation is correct now. I solved $P(X.is.odd \cup Y.is.odd \cup Z.is.odd)$ and I got 7/8, which is what you got using the complement rule $1-P(none.are.odd)$ right? – user130306 Sep 28 '22 at 19:37

1 Answers1

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"Let $p_c$ be the probability that, after rolling all the die, there is at least one odd number rolled in at least one of the groups."

It never occurred to me that MathSE reviewers would interpret this to mean that the number on each die was to be examined to see if it was even or odd.

I interpreted the problem differently. That is, I interpret the problem to intend that the Math student is supposed to examine the three sums Die-1 + Die-2, Die-1 + Die-3, and Die-2 + Die-3. My interpretation is based on the idea that if the problem composer intended the original interpretation instead, why did the problem composer create the three groups, where each group represented examining the two dice in that group.

What would be the point?

First of all, the probability of at least one odd number is equal to $(1)$ minus the probability that you don't have $(3)$ even numbers.

You will have $(3)$ even numbers, if and only if, all $(3)$ dice have the same parity, odd or even. Without loss of generality, you can assume that the first die has parity even.

Then, you will have all $(3)$ dice with the same parity if and only if the 2nd and 3rd dice are also of even parity.

The probability of this happening is $\displaystyle ~\left[\frac{1}{2}\right]^2 = \frac{1}{4}.$

Therefore, the easy approach is to compute

$$1 - \frac{1}{4} = \frac{3}{4}. \tag1 $$

So, (1) above is the probability that you have at least one odd sum, among the three groups.

For the hard way, I recommend approaching the problem combinatorically, where the probability will be expressed as

$$\frac{N}{D},$$

with $D = 2^3.$ Here, you should resist the idea of the simpler approach of $D = 6^3$, because there is no reason to differentiate each of the $(6)$ numbers that may show on a specific die. The nature of the problem is such that it is sufficient to only focus on the odd/even parity of each die.

Let $S_1$ denote all of the ways that Die-1,Die-2 can have different parities.

Similarly, define $S_2, S_3$ to cover Die-1,Die-3 and Die-2,Die-3, respectively.

Then, the overall computation is

$$\frac{|S_1 \cup S_2 \cup S_3|}{2^3}.$$

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Let $T_1$ denote $|S_1| + |S_2| + |S_3|.$

Let $T_2$ denote $|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|.$

Let $T_3$ denote $|S_1 \cap S_2 \cap S_3|.$

Then, in accordance with Inclusion-Exclusion theory, the overall computation will be

$$\frac{T_1 - T_2 + T_3}{2^3}.$$

To enumerate $|S_1|$, you can have Die-1, Die-2 be odd/even or even/odd. In each case, Die-3 is unconstrained.

Therefore, $|S_1| = 2 + 2 = 4.$

By symmetrical considerations, you have that $|S_2| = |S_3| = |S_1|.$

Therefore,

$$T_1 = 12. \tag2 $$

To enumerate $S_1 \cap S_2|$ you want the number of ways that Die-2 can have a different parity than both Die-1, Die-3.

Die-2 can be either odd or even. In each case, the parities of both Die-1 and Die-3 will then be set.

Therefore, $|S_1 \cap S_2| = 2.$ Again, by symmetrical considerations,
$|S_1 \cap S_2| = |S_1 \cap S_3| = |S_1 \cap S_3|$.

Therefore,

$$T_2 = 3 \times 2 = 6. \tag3 $$

The enumeration of $|S_1 \cap S_2 \cap S_3|$ must be $(0)$.

The reason for this is that if you take any two of the three subsets, (for example) $S_1$ and $S_2$, then the subset given by $S_1 \cap S_2$ demands that Die-1 and Die-3 both have a different parity than Die-2.

This requires that Die-1 and Die-3 have the same parity.

Similar considerations will apply throughout the attempt to enumerate $|S_1 \cap S_2 \cap S_3|$.

Therefore,

$$T_3 = 0. \tag4 $$

So, using Inclusion-Exclusion, the overall probability that at least one of the sums is odd is

$$\frac{12 - 6}{8}.$$

user2661923
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  • Hmm I'm sorry I'm confused. The previous comment said the answer was $7/8$ but you are saying that the answer is $3/4$. Is the previous justification incorrect? Isn't it $1-\frac{1}{2^3}$ – user8358234 Sep 28 '22 at 13:40
  • @user8358234 The answer is $~\dfrac{3}{4}.~$ One of the reasons that I posted was that there seemed to be confusion on this point. Further, my answer arrives at the $~\dfrac{3}{4}~$ computation in two independent ways. Perhaps what was overlooked was that if all three dice are showing an odd number, rather than an even number, then the sum of any two of the dice will be even. – user2661923 Sep 28 '22 at 13:58
  • @user2661923 hmm I'm not sure this is correct. As copper.hat suggested, the probability might just be the probability that either X is odd or Y is odd or Z is odd which is just 7/8 – user130306 Sep 28 '22 at 19:38
  • @user130306 "Let $p_c$ be the probability that, after rolling all the die, there is at least one odd number rolled in at least one of the groups." If the intent was to examine each die individually, then there would be no reason to discuss creating the three groups. So, the only interpretation that seems consistent with the problem composer's intent is to examine the sum of each pair of dice. – user2661923 Sep 28 '22 at 20:26
  • @user2661923 yes I see your point too. you're saying that if we roll the dice, if the sum of the pair of the dice is odd. But the question says 'there is at least one odd number rolled in at least one of the groups'. so I'm assuming that's why it's not a sum. because it says "at least one odd number". if you're looking at a sum of two things in each group, then how could you have "at least one odd number" in each group? do you see what I mean? – user8358234 Sep 28 '22 at 20:58
  • sorry, I'm not sure if I explained that too well – user8358234 Sep 28 '22 at 20:58
  • @user8358234 How do you know whether my original interpretation of the problem is not the one intended by the problem composer? Why pair up each die into groups and then, in each group question whether each die in the group is odd? This type of convolution is certainly not needed for Inclusion-Exclusion. Under the assumption that the problem composer intends that each die be examined, rather than the sum of each pair of dice, Inclusion-Exclusion could have been employed without the convolution of groups of pairs. – user2661923 Sep 28 '22 at 22:38