Dice probability paradox

Factual Questions
1

There are 36 equally probable ways two dice can land:

1 - 1, 1 - 2, 1 - 3, 1 - 4, 1 - 5, 1 - 6
2 - 1, 2 - 2, 2 - 3, 2 - 4, 2 - 5, 2 - 6
3 - 1, 3 - 2, 3 - 3, 3 - 4, 3 - 5, 3 - 6
4 - 1, 4 - 2, 4 - 3, 4 - 4, 4 - 5, 4 - 6
5 - 1, 5 - 2, 5 - 3, 5 - 4, 5 - 5, 5 - 6
6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6
A magician blindfolds seven people, then for each of them he rolled a
pair of fair dice and asked them the probability of their dice totaling 7.

He said to the first, here’s a hint: truthfully, at least one of your dice shows a 6.
The subject counted 11 cases of at least one 6, two of which, 1-6 and 6-1, total 7.
So he answered, my chances of having 7 are 2/11. Very good, said the magician.

He then said to the second, I’ve looked at your dice, and at least one of them is a 5.
This subject counted 11 cases of at least one 5, of which 2-5 and 5-2 made 7.
So he answered 2/11. Very good, said the magician.

He told the third subject, I see at least one 4 on your dice.
That subject also found 11 cases of 4, of which 3-4 and 4-3 made 7.
So he answered 2/11. Very good, said the magician.

The next subject was told at least one of his dice was a 3.
Like the others, he found 11 cases, and of them only 4-3 and 3-4 were favorable.
So he answered, my chances of having 7 are 2/11. Very good said the magician.

The next two were told their dice showed at least one 2 and one 1, respectively.
They found 5-2, 2-5 for one, and 6-1, 1-6 for the other, among 11 cases gave 7.
They both answered their chances of 7 were 2/11. Very good said the magician.

The seventh subject had been listening to all of this. And before the magician
could speak, he said, I don’t need a hint. I know that you’re going to tell me
some number appears on at least one of my dice. And you’ve already confirmed
what the right answer is in each case. So whatever you were going to say,
I know most certainly what the odds probability of my dice totaling 7 are is.
My answer is 2/11.

The seventh subject is wrong.

The first subject knew he must have had one of the following showing:

6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6
1 - 6
2 - 6
3 - 6
4 - 6
5 - 6

Eleven equally possible combinations, two of them add up to seven. 2/11 possibility that his dice add up to seven.

The same for the next five subjects.

The seventh didn’t wait to for the magician to speak, so he had thirty-six possible dice landings, and six of them add up to seven. 1/6 possibility that his dice add up to seven.

Now, the seventh man knew that the magician would make one of the statements he told the others, so why is his probability of having his dice add up to seven a little worse than the others?

I realize you can answer “you already showed why”. Even though I did show that it isn’t the same, I can’t wrap my mind around why the answer is different when we knew the magician would say one of the things he told the others.

2

If one of the dice is a six and the other is unknown, the odds of getting a seven are 1 in 6, just like with two unknown dice. Youre counting 2[sub]1[/sub]–6[sub]2[/sub] as different from 6[sub]1[/sub]–2[sub]2[/sub] but you’re not doing that for 6[sub]1[/sub]–6[sub]2[/sub] and 6[sub]2[/sub]–6[sub]1[/sub]

3

That’s not true. If one of the dice is a 6 (and we’re not told which one) we can extract the following eleven equal possibilities out of the thirty-six that I listed:

6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6
1 - 6
2 - 6
3 - 6
4 - 6
5 - 6

There are elven equally possible combinations we can have and two of them add up to seven.

As you can see form the list of thirty-six possible rolls, 6-6 can only happen once.

6,6 is not counted twice as there’s only one possible way to roll a 6, 6. The first die must be a 6 (1 out of 6) and so must the second (1 out of 6). 6x6=36

There are two possible ways to roll a 6 and a 1. The first die must be a 1 or a 6 (1 out of 3) and the second must be a 1 if a 6 has been revealed or a 6 if a 1 has been revealed ((1 out of 6). 3x6=18. That’s why there are two possible ways to roll a 6 and a 1 on the list of 36 equal possibilities.

4

Why is the order in which you roll the die important? If you know one of the dice is a six, you have six possibilities for the other. Even if you decide to make order important you have twelve possibilities. The first die is a six with six possibilities for the second or the second is a six with six possibilities for the first. The odds are 1 in 6 that you have a seven, just as it would be with two unknown dice.

5

The 7th guy is wrong because everyone else before him was also wrong.

Consider the possible ways that you can roll two dice, at least one of which is a 6:

First die is a 6:
6-1, 6-2, 6-3, 6-4, 6-5, 6-6

Second die is a 6:
1-6, 2-6, 3-6, 4-6, 5-6, 6-6

Notice that 6-6 shows up twice because there are two ways that it can happen. So there are not 11 ways of rolling two dice so that at least one is a 6, but rather 12 ways. The answer to all of the magician’s questions is 2/12 = 1/6.

The “paradox” is misleading because the magician responds “very good” to each of the first 6 people even though their answer was wrong.

6

No, you don’t separate probabilities that way. People have made the same mistake regarding the Girl-Boy problem.

If there are two children and we know at least one is a girl, and we want to know what the probability is that they are both girls, you don’t separate the probabilities like this:

First child: Girl
GG GB

Second child: Girl
BG GG

and come to the conclusion that the probability of having two girls is 1/2.

What you do is look at all of the equal possibilities of having a child:

GG
GB
BB
BG

and remove the one we know isn’t a possibility (BB)

GG
GB
BG

and come to the conclusion that the probability of having two girls is 1/3.

If you don’t believe me, read through the thread I linked to.

There’s one way it can happen. One die must turn up 6 and so must the other.

7

It’s not; it’s the information you are given that is important just as in the Girl-Boy puzzle I linked to above.

If we are told two dice are rolled and the first is a 6, the following are the possible combinations:

6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6

If instead we’re told that at least one of the dice is a 6, then the following are the possible combinations:

6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6
1 - 6
2 - 6
3 - 6
4 - 6
5 - 6
Again, this is similar to the Girl-Boy problem. If we’re told that there are two children and the first is a girl, the following are the equally likely possibilities:

G-B
G-G

1/2 probability of having two girls.

If instead we’re told that there are two children and at least one is a girl, the following are the equally likely possibilities:

G-B
G-G
B-G

1/3 probability of having two girls.

8

Fantome: The difference between this problem and “normal” problems is that the degenerate case (6, 6) has two ways of showing up in the magician’s sentence (assuming he isn’t doing anything funny when picking which number to report). He could choose the (say) blue die and report its number or he could choose the red die and report its number. Thus, if he says a “6” is present, (6, 6) has twice the chance of being reported by the magician than does (1, 6).

(Another way to say this: If the roll is (6, 6), he will definitely say “6”, but if it is (1, 6), he has only a 50% chance of saying “6”.)
On preview: In the boy-girl problem, you have two possible games. Game 1: You have a family with two kids, you pick one at random, and you ask whether the other kid has opposite gender. Game 2: You have a family with two kids, you check if either is a girl. If not, pick a new family and repeat. If yes, you ask whether there is a boy.

When Game 1 starts with a girl, the other kid will be a boy 50% of the time. However, it will sometimes start with a boy.

Your dice game is like (1), as there is no “re-rolling” to satisfy any particular starting condition. Your magician would as happily say “at least one child is a boy” as he would “at least one child is a girl”. To make the game like (2), you could have the magician operate thus: If there is a six anywhere, I will report “6”. If there is not a six anywhere, I will pick one of the dice at random and report its number. In this version, a report of “6” does indeed lead to the 11 equally likely cases you enumerated. But in the version where the die he chooses to report from is randomly chosen, the probability of sum=7 remains 1/6.

9

The calculation done by the first six people is essentially the same as the famous problem “Given that at least one of my two children is a boy, what is the probability that I have one boy and one girl?” The usual interpretation of this problem is exactly as Fantome explained: of the universe of four equiprobable cases (BB,BG,GB,GG) for my ordered children, exactly one (GG) is ruled out by my statement, leaving a probability of 2/3 that I have one boy and one girl. I don’t get to count the BB case twice just because it has two Bs. But just as with the OP, the probability that I have one boy and one girl, unconditioned on my statement (but given only that I have exactly two children) is 1/2, not 2/3. [On preview I see Fantome has already given this example.]

Write a quick program to test it if you don’t believe it; Fantome correctly enumerated the 36 equally-likely cases for two dice, of which precisely 11 have at least one 6 showing.

The problem is with the reasoning of the seventh person. He is forgetting that the magician’s statements are not all mutually exclusive. In standard probability language, the seventh person is (mis)using the conditional-probability formula
P(A) = sum_B P(A|B)P(B) .
To be used correctly, this sum must be taken over a complete set of mutually-exclusive events B. The seventh person takes as event A “the dice sum to 7” and as the six events B “at least one 1 is showing”, …, “at least one 6 is showing”, so that all P(A|1)=…=P(A|6)=2/11. Now he implicitly assumes (wrongly) that these six events are mutually exclusive and (also falsely) that all six P(B) are equal to 1/6, and therefore gets P(A)=2/11.

If instead I define my events B as “the smallest number showing is a 1”, …, “the smallest number showing is a 1” then these events are mutually exclusive and complete, and so the formula above can be used to get the 1/6 value we expect. However, in this case the six values P(A|B) are not all equal to 2/11 (and nor are the values of P(B) equal to 1/6). Alternately, I may use an inclusion-exclusion argument to compensate for my overcounting.

10

This is incorrect. If he says a 6 is present, the following are all equally likely:

6 - 1, 6 - 2, 6 - 3, 6 - 4, 6 - 5, 6 - 6
1 - 6
2 - 6
3 - 6
4 - 6
5 - 6

1 chance in 11 that a 6, 6 has been rolled and 1 in 11 that a 1, 6 has been rolled (and a two chance in 11 that a 1 and a 6 has showed up in any order).

I appreciate the explanation, but do you think you could explain that without using formulas (way over my head)?

Obviously the magician will say one number, and if we wait for him to speak, the probability of having a 7 will be 2/11. If that will definitely be the probability we can calculate if we wait for him to speak, why not take it for granted that the probability is 2/11 now?

Again, I know how to calculate the answer; I just can’t understand why it works the way it does.

11

The truth is, if we were to analyze the case a little differently, the probabilities might be different (as hinted at by Omphaloskeptic). There are the following 66 possibilities:

Roll: 1, 1. Told: At least one is a 1
Roll: 1, 2. Told: At least one is a 1
Roll: 1, 2. Told: At least one is a 2.
Roll: 1, 3. Told: At least one is a 1
Roll: 1, 3. Told: At least one is a 3.
…
Roll: 5, 6. Told: At least one is a 5.
Roll: 5, 6. Told: At least one is a 6.
Roll: 6, 6. Told: At least one is a 6.

By convention, we assume each possible die roll outcome is equally probable; thus, the first row above has probability 1/36, the sum of the second and third rows’ probabilities is 1/36, etc.

However, beyond this, we haven’t been given any information about, and have no clear convention regarding, the particular probabilities of row 2 and row 3. They could be 1/36 and 0, or 1/72 and 1/72, or all sorts of other things. Much depends on this.

[Being called away. More to come later]

12

This is actually an extrapolated case of the Boy-Girl problem.

Let me simplify your problem:

Suppose you have a coin, one side of which is marked “1” and the other side “2”.

There are 4 equally probable ways two coins can land:

1 - 1, 1 - 2,
2 - 1, 2 - 2

A magician blindfolds three people, then for each of them he flips a pair of fair coins and asked them the probability of their coins totaling 3.

He said to the first, here’s a hint: truthfully, at least one of your coins shows a 2. The subject counted 3 cases of at least one 2, two of which, 1-2 and 2-1, total 3. So he answered, my chances of having 3 are 2/3. Very good, said the magician.

He then said to the second, I’ve looked at your coins, and at least one of them is a 1. This subject counted 3 cases of at least one 1, of which 2-1 and 1-2 made 3. So he answered 2/3. Very good, said the magician.

The third subject had been listening to all of this. And before the magician could speak, he said, I don’t need a hint. I know that you’re going to tell me some number appears on at least one of my coins. And you’ve already confirmed what the right answer is in each case. So whatever you were going to say, I know most certainly what the odds probability of my coins totaling 3 are is. My answer is 2/3.

The third subject is wrong.

This now, is exactly the same as the Boy-Girl Problem. And because of that, it’s (hopefully) easier to see that the two cases covered by “person one” and “person 2” are *not *mutually exclusive.

The magician essentially tells person 1, “the two coins are NOT 1 - 1.” Then he tells person 2, “the two coins are NOT 2 - 2.” Person 3’s error here is to assume that these two statements fully describe all possibilities.

If that makes sense, your original problem is just an extrapoltion of this simplified one.

13

I think this is another problem where it’s going to depend on exactly how the problem is set up, and where the OP doesn’t precisely and clearly specify it.

For any roll, you can always pick one of the two numbers and say “you have at least one X”. If you roll the dice first, then pick the number X from the numbers you get, the chance of rolling 7 is always 1/6.

If you decide beforehand that the first number to be specified is a 6, then roll the dice, what do you do if no 6 shows up? You roll the dice again until you get at least one 6. In this case, the probability of rolling 7 is indeed 2/11.

So the OP needs to specify precisely how the numbers told to the blindfolded are chosen.

On preview, many more posts have shown up, so I’m just going to post this, then go read the thread.

14

Whoa! I didn’t think of that angle.

Cool! Looking forward to see where this is going.

Yeah, I see that. Still having a hard time wrapping my head around this though. I’ll look at it again tomorrow when I’m a little less tired. Thanks all.

15

Ok, I’m just going to complete the post I was making earlier and post it again.

The truth is, if we were to analyze the case a little differently, the probabilities might be different (as hinted at by Omphaloskeptic). There are the following 66 possibilities:

Roll: 1, 1. Told: At least one is a 1
Roll: 1, 2. Told: At least one is a 1
Roll: 1, 2. Told: At least one is a 2.
Roll: 1, 3. Told: At least one is a 1
Roll: 1, 3. Told: At least one is a 3.
…
Roll: 5, 6. Told: At least one is a 5.
Roll: 5, 6. Told: At least one is a 6.
Roll: 6, 6. Told: At least one is a 6.

Now, just because there are 66 possibilities, we shouldn’t assume that each has probability 1/66 (to assume “N possibilities” always means “N equally probable possibillities” is a common probabilistic mistake). Indeed, by implicit convention, we instead have that each possible die roll outcome is equally probable; thus, the first row above has probability 1/36, the sum of the second and third rows’ probabilities is 1/36, etc.

However, beyond this, we haven’t been given any information about, and have no clear convention regarding, the particular probabilities of row 2 and row 3. They could be 1/36 and 0, or 1/72 and 1/72, or all sorts of other things. Much depends on this.

If we go with the most natural “1/72 and 1/72” setup (e.g., after rolling the pair of dice, the magician flips a coin to decide which one to tell you about), then everyone who announces “2/11” is actually wrong. For example, if the magician says “You have at least one 1”, then this encompasses “Roll: 1, 1; Told 1” (prob. 1/36) as well as “Roll: 1, n; Told 1” (prob. 1/72) and “Roll: n, 1; Told 1” (prob. 1/72) for each of n = 2, 3, 4, 5, 6. The total probability of this is (1/36 + 5/72 + 5/72) = 12/72, out of which only 2 possibilities of probability 1/72 each have the dice adding up to 7; this gives a final conditional probability of (2/72)/(12/72) = 2/12 = 1/6. NOT 2/11. In this setup, everyone who announces “2/11” is wrong.

The point is, when the magician announces “At least one of your dice is a 4” or whatever, he is not merely giving you the information that at least one of your dice is a 4; you are also being provided with the further information that, also, out of the two dice, the one he chose to tell you about was a 4. This further information can cause the correct response to not be “2/11”, but something else (in the above setup, “1/6”; in other setups, perhaps something else). Taking this into account, there will be no paradox.

16

I can confirm this as true, as I did exactly that. I wrote a quick-and-dirty program in QuickBASIC (I can’t believe I still have that on my computer) which rolls a pair of dice 5,000,000 times and adds up all instances of at least one die showing 6. Each time running it comes up with a value very, very close to .30555… (11/36). And never once was the value anywhere near .33333… (12/36).

17

Expanding on my post:

Now, remember… the problem you (the OP) have given is not fully specified. After rolling the two dice, we have no idea what process the magician uses to decide what to tell you; perhaps he always tells you about the first die rolled, perhaps he flips a fair coin to decide, perhaps he flips an unfair coin, perhaps he generally prefers to announce the minimum number but doesn’t always, perhaps he has a much more complicated system.

So, “He flips a coin to decide what to tell you” is just one possible way to finish off your problem, not the only one. That warning having been emphasized, let’s look at that particular case another way, in case the above discussion was confusing. We can most naturally construe it as containing 72 possibilities (6 die faces * 6 die faces * 2 coin faces), and, particularly nicely, all of these 72 possibilities will be equally probable.

After being told “At least one of your coins is a 4”, there will be 2 possible ways to have a dice-total of 7 (“3, 4, tails” and “4, 3, heads”), while there will be 12 total possibilities still in play (“n, 4, tails” and “4, n, heads” for each of n = 1, 2, 3, 4, 5, 6); thus, in this setup, the conditional probability of having a dice-total of 7 will be 2/12, and NOT 2/11.

18

In this case, you decided beforehand what number you want to specify. This is equivalent to the magician deciding the first person would be told he has a 6, and rerolling until he got a 6. That may be what the OP intended, but he never explicitly says so.

19

Take it up with Omphaloskeptic. I was just running the experiment he suggested.

20

I was going to try to explain it in plain English, but I think Indistinguishable already did it better than I would have.

Part of the problem, here and with a lot of probability problems, is with the translation between fuzzy plain English and a well-defined probability space. A statement like “at least one of my children is a boy” is pretty unusual in plain English, so it wouldn’t be unreasonable to suspect a trick from the speaker or to wonder how this statement was chosen. In most word problems that statement is assumed to translate directly into the event-space restriction that you write out in your OP. But as Indistinguishable shows, there are a lot of reasonable ways to choose these statements that give different answers.

The events given in the OP, in which after each of six consecutive dice rolls the magician is apparently able to truthfully make a somewhat unlikely (probability 11/36 for each roll, with the interpretation above) statement about the dice, are indeed pretty improbable. I guess this is leading several people to suspect some sort of trickery on the magician’s part. It would be more plausible if the magician were allowed to re-roll until he saw his desired number, but of course that would remove the apparent symmetry with the seventh subject.