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A sister has two fair six sided dice, one red one blue, the dice are rolled together. The sister announces to her brother who is sat in another room and unable to see the dice, that there is at least one six in the outcome. Then asks him what are the odds that the other die is a six. He replies $1/6$, she says no it has to be $1/11$. Who is right or most likely to be right? Please help settle this family feud!

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    $\begingroup$ The sister is right. $\endgroup$ – SirXYZ May 2 '17 at 10:12
  • $\begingroup$ The dice are independent, so it'll be $\frac16$. The sister mistakenly assumes that every number from $2$ through $12$ has the same probability to be thrown, but this is false (there is for example only one way to throw $2=1+1$, but a lot more ways to throw $7=1+6=2+5=3+4=4+3=5+2=6+1$) $\endgroup$ – vrugtehagel May 2 '17 at 10:13
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    $\begingroup$ The dice are independent, but the sister is correct since the event of "at least one die shows a six" does not specify the identity of the "other die". Of the eleven unbiased outcomes in that event, only one is "both are six", so $1/11$ is the probability that "the other is a six" when "at least one die shows a six" $\endgroup$ – Graham Kemp May 2 '17 at 11:07
  • $\begingroup$ @vrugtehagel How do the sums of the dice even come in to this? $\endgroup$ – Arthur May 2 '17 at 11:08
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    $\begingroup$ "At least one of the die is six. What is the probability both are six?" Answer: 1/11. "One of the die is six. What is the probability the other is six?" Answer: 1/6. "At least one die is six. What is the probability the other is six." Answer: This question semantically and logically makes no sense and is utterly meaningless. "At least one" is non-specific. "the other" is specific. $\endgroup$ – fleablood May 2 '17 at 17:11
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In two ways:

  1. We are computing the probability that both dice are a $6$ given all of the information he knows, not her. It is worth noting that she can see the other die, so she knows with certainty (probability $1$) whether both are $6$s or not.

    She rolls both dice at the same time. There are 11 pairs of dice for which her statement that "at least one of the dice are a $6$" is true, i.e., $$(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,1),(6,2),(6,3),(6,4),(6,5)$$ we are not told which of the dice was a $6$, so we cannot cancel any possibilities (which is how he incorrectly concluded $1/6$).

    Only one of these 11 are are both $6$s, hence the probability is $1/11$.

  2. In a similar vein, but more algebraic, let $X_2=I[\text{die 1 is $6$}]$ and $X_1=I[\text{die 2 is $6$}]$, then

    $$ \mathbb{P}[X_1+X_2=2 \,|\, X_1+X_2\geq 1] = \frac{\mathbb{P}[X_1+X_2 = 2]}{\mathbb{P}[X_1+X_2\geq 1]} = \frac{\mathbb{P}[X_1=1,X_2=1]}{1 - \mathbb{P}[X_1=0,X_2=0]}$$

    by independence and denoting $\mathbb{P}[X_1=1]=p_1$, $\mathbb{P}[X_2=1]=p_2$, \begin{align}= \frac{\mathbb{P}[X_1=1]\mathbb{P}[X_2=1]}{1 - \mathbb{P}[X_1=0]\mathbb{P}[X_2=0]}= \frac{p_1p_2}{1-(1-p_1)(1-p_2)} = \frac{1}{\tfrac{1}{p_1}+\tfrac{1}{p_2}-1}\end{align} Note that this allows for biased dice (e.g., what if the first die were weighted to land on $6$ 90% of the time?). But if we assume they are not biased, then $p_1=p_2=\tfrac{1}{6} \implies \tfrac{1}{p_1} = \tfrac{1}{p_2}=6$, giving the same result of $\frac{1}{(6)+(6)-1} = \frac{1}{11}$.

    We could then answer the same question if we replaced dice with coins ($p_1=p_2=\tfrac{1}{2}$), which gives $\tfrac{1}{(2)(2)-1}=\tfrac{1}{3}$.

Edit If the problem were different and instead she doesn't see the second die then the probability that she would guess correctly is $1/6$. If he knew that she had only seen one die then the probability he would be correct in his guess is $1/6$, but if he didn't have that information then it would still be $1/11$.

The confusion here is very similar to the Monty Hall problem.

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  • $\begingroup$ Thanks for your reply. I agree that you need to consider what the brother knows. From the sisters statement, the brother , in my view, can only be certain that the sister has seen one die which is showing a six, he is in the other room and is unable to ascertain whether she can see both dice. Given that, his response of 1/6 may be correct. $\endgroup$ – greengd May 2 '17 at 12:55
  • $\begingroup$ @greengd I mean it really depends on just how loosely you want to interpret the wording. Based on the fact the she states $1/11$, you could then argue that he should update his guess to $1/11$ because it is very likely that she has calculated it in the same way I have above, and hence should assume she has seen both dice. But at that point we have severely over-analysed the problem. I think I have presented the most logical and obvious interpretation, and the other response agreed with me. $\endgroup$ – adfriedman May 2 '17 at 13:08
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This question can't be properly answered from the information provided (which is why it's so good at provoking arguments). To give a proper answer, you would need to know on what basis the sister decided what fact about the dice to announce. Suppose she thought to herself "I will look at the red die, which will be showing a number, $n$. I'll shout to my brother there is at least one $n$ in the outcome". She does this and it happens to be a 6 that comes up on the red die. All this faffing around with the red die obviously doesn't affect what number the blue die has landed on (the sister might not even have looked at it), so the probability that the other die is a 6 too is clearly 1/6.

To get an answer of 1/11 you need to make a similarly unjustified assumption about the sister's thought process (something like "If at least one die shows a 6 then I'll announce that fact, otherwise I won't say anything, and no maths question will be produced").

This is essentially the same as the "Tuesday Boy problem" as discussed by Rob Eastaway

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  • $\begingroup$ +1 Without really thinking about it I assumed that the sister was thinking like me. $\endgroup$ – drhab May 2 '17 at 13:36
  • $\begingroup$ ... or even if the sister decided to make an announcement versus staying quiet! In my family, the probably is probably near $1$, since my sister would likely only have done this as a trick question! This is basically the answer I was going to write myself. +1. $\endgroup$ – Hurkyl May 2 '17 at 14:54
  • $\begingroup$ I have a quibble with this answer (as I have a quibble with the analysis of the "Tuesday boy problem" you linked to). The statement "There is at least one six in the outcome" is a mathematically precise and well-defined event on the sample space, and this event has the 11 outcomes listed in the answer of @adfriedman. The problem could have been set up to describe the event ambiguously, hence inviting a linguistic discussion instead of a mathematical discussion (as in your answer and in Eastaway's analysis). But it wasn't, the event itself was described precisely. $\endgroup$ – Lee Mosher May 2 '17 at 15:25
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    $\begingroup$ @LeeMosher The event itself is well-defined. The circumstance under which the event is announced is not. $\endgroup$ – Thern May 2 '17 at 15:47
  • $\begingroup$ If she has computed the probability is 1/11 do you think it is more or less likely that she has seen the die? I think that makes it a justified assumption $\endgroup$ – adfriedman Jun 22 '17 at 20:12
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BLUF: If this happened in real life, I would expect the brother to more likely be right (but for the wrong reason). But I imagine the person who posed the question to you had in mind a scenario where the sister is more likely to be right.


This is a standard problem, whose main issue isn't even the formal mathematics — it's in the modeling of the problem. Specifically, the problem gives insufficient information to specify what's going on, leaving the listener to fill in their own ideas about what problem they want to solve.

It's further compounded by a grammar error — nowhere in the problem has a die been specified, so it makes no sense to speak of the "other" die. This error tends to have a very strong effect of tricking people into a particular interpretation of the problem.

I'm going to resolve the grammar issue by changing the problem slightly, my change in bold:

... The sister announces to her brother who is sat in another room and unable to see the dice, that there is at least one six in the outcome. Then asks him what are the odds that both dice are six. ...


That said, I expect the brother to most likely be right. I expect the typical sister that does this would have gone through something resembling the following process:

  • She thinks "I just learned this neat probability thing. I'll see if I can get one up on my brother."
  • She rolls two dice
  • She (uniformly randomly) picks one of the two dice, and announces its value to her brother
  • She then asks the probability question

The third bullet point is very important. The point being, for example, that the outcome of seeing two dice with $\{6, 3\}$ happens with probability $\frac{1}{18}$, but the outcome that the dice were $\{ 6, 3\}$ and the sister chooses to announce the $6$ rather than the $3$ happens with probability $\frac{1}{36}$.

So, from the information the sister announces, the relevant outcomes are:

  • With probability $1/36$, the dice read $\{ 6, 1 \}$ and the sister announces $6$
  • With probability $1/36$, the dice read $\{ 6, 2 \}$ and the sister announces $6$
  • With probability $1/36$, the dice read $\{ 6, 3 \}$ and the sister announces $6$
  • With probability $1/36$, the dice read $\{ 6, 4 \}$ and the sister announces $6$
  • With probability $1/36$, the dice read $\{ 6, 5 \}$ and the sister announces $6$
  • With probability $1/36$, the dice read $\{ 6, 6 \}$ and the sister announces $6$

So, given the information that the sister announced a $6$, you find that the probability that both dice read $6$ is, indeed, $1/6$.


Now, another example of how this could have gone — and probably the one that the question's originator had in mind, is that the sister did something more like

  • She repeatedly rolls two dice until at least one of the two dice read $6$
  • She announces there is a $6$ and asks the question

In this case, the outcome that the dice read $\{ 6,1 \}$ upon announcement really is $1/18$. The relevant outcomes are

  • With probability $1/18$, the dice read $\{ 6, 1 \}$ when the announcement is made
  • With probability $1/18$, the dice read $\{ 6, 2 \}$ when the announcement is made
  • With probability $1/18$, the dice read $\{ 6, 3 \}$ when the announcement is made
  • With probability $1/18$, the dice read $\{ 6, 4 \}$ when the announcement is made
  • With probability $1/18$, the dice read $\{ 6, 5 \}$ when the announcement is made
  • With probability $1/36$, the dice read $\{ 6, 6 \}$ when the announcement is made

and we see the probability is $\frac{1}{11}$.

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  • $\begingroup$ It's not a "grammar error" to omit which die was chosen, it actually gives some information, but intentionally not all. Using the notation I gave in my own answer, this is equivalent to $\mathbb{P}[X_1+X_2 = 2 | X_1+X_2\geq 1]$ which is completely different to (if for instance the red die were chosen) $\mathbb{P}[X_1+X_2 = 2 | X_1 = 1]$. $\endgroup$ – adfriedman May 4 '17 at 16:30
  • $\begingroup$ @adfriedman: I am not assuming they are indistinguishable -- I am merely not distinguishing between them. Listing both $(6,1)$ and $(1,6)$ would make the lists long enough that their presence would make the post awkward, and wouldn't do anything to improve the clarity anyways. $\endgroup$ – Hurkyl May 4 '17 at 16:36
  • $\begingroup$ "So, from the information the sister announces, the relevant outcomes are:" You are assuming that she announced based on the first die alone, which is not given information. So you need to include both $(6,1)$ and $(1,6)$, which do not represent identical outcomes. $\endgroup$ – adfriedman May 4 '17 at 16:39
  • $\begingroup$ @adfriedman: $\{ 6,1 \}$ is an unordered pair: there is no "first" die. If we add in order, the bullet you refer to covers both the 1/72 chance of "(6,1) and sister announces 6" and the 1/72 chance of "(1,6) and sister announces 6". $\endgroup$ – Hurkyl May 4 '17 at 16:40
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    $\begingroup$ @adfriedman: The grammar error is not omitting a choice of die, but making a later statement that refers to a choice which was not made. While "at least one die is 6" is a careful rephrasing of the usual problem statement, it really needs to continue with "... both dice are 6". $\endgroup$ – Hurkyl May 4 '17 at 16:43
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Let $X$ denote the number of sixes that have shown up.

Then: $$\Pr(X=2\mid X\geq1)=\frac{\Pr(X=2\wedge X\geq1)}{\Pr(X\geq1)}=\frac{\Pr(X=2)}{1-\Pr(X=0)}=\frac{\frac16\frac16}{1-\frac56\frac56}=\frac1{11}$$

Sister did not say something like: "the red (or blue) die shows a six". That would have made things different. In that case $\frac16$ is the correct answer.

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This is reminiscent of a "paradox" I saw in Martin Gardner's "Mathematical Games" column in the Scientific American 50 odd years ago. A bridge player, known to be truthful, looks at the 13 cards in his hand, and announces, "I have a Ace." What is the probability that he holds another Ace? Again, suppose he announces instead, "I have a black Ace." What is the probability that he holds another Ace? Finally, what is the probability that he holds a second Ace if he announces, "I have the Ace of Spades."

If you go to the trouble of computing these probabilities in the usual manner, for example dividing the probability that he holds at least two Aces by the probability that he holds at least one Ace for the first case, you find that as the specificity of his announcement increases, the probability that he holds a second Ace increases too. This seems paradoxical, since he can always announce the color or the suit; it doesn't seem like it should make any difference.

I can't remember how, or if, Gardner explained this. At the time it was a complete mystery to me. The explanation is similar to the answer given by aPaulT; we don't have enough information on how this person decides what announcement to make. For example, if he announces the color if and only if it is black, then the statement "I have an Ace," is tantamount to "I have a red Ace," and we have to recompute our answer.

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The brother is right.

Suppose you are the sister, and try to repeat the experiment. You roll a red 3 and a blue 5. What do you ask your brother - "one is a 3, what are the chances that both are 3's?" or "one is a 5, what are the chances that both are 5's?" ?

While a specific person may prefer 5's over 3's, and so always ask about 5's in this example, we have no information that would allow us to assume anything except that the sister chooses one of the two values at random, if there are two.

In the original question, there is one combination where the sister had to ask about 6's, and 10 where there was a 6 but only a 50% chance she would ask about it. This makes the answer 1/(1+10/2) = 1/6.

+++++

Note that the sister's logic is what makes people say the answer to the Monty Hall Problem is 1/2.

This problem, the Monty Hall Problem, and the Two Child Problem are all variations of the Bertrand Box Problem. Sometimes it is called Bertrand's Box Paradox, but the paradox is how Bertrand showed that the sister's answer can't be right.

Suppose the sister had said "I wrote one of the number's down on this hidden piece of paper. What are the chances that both dice landed on that number?" Then, whatever the chances are, they are the same no matter what is written. But if it is the same no matter what is written, then we don't need to see what is written to answer.

The chances of doubles, before she wrote down a number, were 1/6. She can always write down a number, so if it changes to 1/11 we have a paradox. The answer changed, but we gained no information that could make it change.

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