Dice, conditional probability - where am I wrong?

Question

There is a problem which I solved wrong. I know the correct solution from the book, but I cannot find the gap in my reasoning.

The problem is: three fair dice are rolled at the same time. What is the probability of getting at least one "1" upon condition that at least one die shows a "6"?

The solution in the book is quite complicated, with a result of 30/91.

My idea - before checking the book solution - was: let's just throw out the die that rolled a "6", now we have two dice and are looking for the probability of getting at least one "1" from them. Now this is fairly easy, it's 1-(5/6)^2 = 11/36. WRONG.

Yes I know this is wrong. I went as far as write a program that simulated 10,000,000 rollings of three dice, and the relevant percentage worked out very close to 30/91, the result in the book, not my result.

But where did I go wrong?

Answer 1

You computed the probability of at least one '1' rolling 2 dice, and as such it is correct. But the book problem is different. You don't know which die falls '6' and that is why the conditional information affects the solution.

I am not sure that it is possible to answer your question more clearly. You assume that the conditional information is not useful and does not affect the answer, but it is not so. This situation reminds me the Monty Hall problem, where it is also not obvious that conditional information is important, and you would better "shut up and calculate" as physicists say.

Answer 2

This is subtle. I went back and forth trying to delve into your analysis, so that I could find the key that would expand your intuition. I am unable to find the right key. Therefore, the best that I can do, is give the inferior, somewhat off-point, explanation of Conditional Probability, as discussed in this Wikipedia article.

Suppose that you have two events $E_1$ and $E_2$, that may or may not be independent of each other.

The Wikipedia article linked to above indicates that

$$p(E_2 | E_1) = \frac{p(E_1,E_2)}{p(E_1)}. \tag1 $$

In (1) above, the LHS represents the probability of event $E_2$ occurring, given that event $E_1$ has occurred.

In (1) above the numerator on the RHS represents the probability that events $E_1$ and $E_2$ both occur.

Before proceeding further, I recommend that you draw a Venn Diagram, representing events $E_1$ and $E_2$ as two circles that intersect each other.

The $p(E_2)$ computation may be intuited as the ratio between the area of the $E_2$ circle and the area of the whole universe.

The $p(E_2|E_1)$ computation may be intuited as the ratio between the area where circles $E_1,E_2$ intersect and the area of circle $E_1$.

With this sort of visual intuition, the formula given in (1) above should $\color{red}{\text{make sense}}.$ If it doesn't (yet) seem intuitive, then I recommend stopping, focusing on the Venn diagram, and trying to understand why the formula in (1) above is correct.

If all else fails, print out this answer, show it to your Math teacher, and ask him to help you expand your intuition. Probability depends on intuition. You will get into trouble if you prematurely try to attack Probability problems without first expanding your intuition.

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Assuming that you accept the formula given in (1) above, the Math is not that bad.

Let $E_1$ denote the event that at least one $6$ was rolled.

Let $E_2$ denote the event that at least one $1$ was rolled.

You are being asked to compute $p(E_2|E_1)$.

Here, $p(E_1) = 1 - (5/6)^3 = \dfrac{91}{216}.$

So, the problem reduces to computing $p(E_1,E_2).$

The easiest way to do this is to consider the $91$ rolls where at least one $6$ occurred. How many of these $91$ rolls do not include a $1$.

Let $S$ denote the set $\{2,3,4,5\}.$

There is one way that you could roll three 6's.

There are $3 \times 4$ ways that you could roll two 6's and a third number from the set $S$. That is, there are $3$ ways of choosing which of D-1,D_2,D_3 will not be a $6$, and then $4$ ways of choosing which element from set $S$ will appear.

Then, there are $3 \times 4^2$ ways that you could roll one 6 and 2 elements from set $S$. That is, there are $3$ ways of choosing which of D-1,D_2,D_3 will be the lone $6$, and then $4$ choices for each of the other two dice.

Putting this all together, there are $1 + 12 + 48 = 61$ ways of not rolling a 1, contained within the $91$ ways of rolling a $6$, out of the $216$ rolls.

Therefore, there are $91 - 61 = 30$ ways of rolling at least one $1$, contained in the $91$ ways of rolling at least one $6$.

So, as your posting indicated, the correct computation, based on the formula in (1) above is

$$\frac{\frac{30}{216}}{\frac{91}{216}} = \frac{30}{91}.$$

Answer 3

Well, the first problem is that you can't throw out the die that rolled a 6, because there could be more than one of them. That's easy enough to fix: just always throw out the first die that rolled a 6.

Now the problem is that, of the 36 possibilities for the remaining two dice, some of them are more likely than others. For example, the result 11 can be obtained in three ways (the original roll could be 611, 161, or 116), but the result 66 can only be obtained in one way (the original roll has to be 666).

Define a "success" as rolling at least one 1. We can solve the problem by considering three cases:

• A 6 was rolled first: 36 possibilities, 11 successes, 25 failures.
• A 6 was rolled second but not first: 30 possibilities, 10 successes, 20 failures.
• A 6 was rolled third but not second or first: 25 possibilities, 9 successes, 16 failures.

Each individual possibility is equally likely, regardless of which case it falls under. Therefore, we can find the answer by just adding the numbers from all three cases together. Overall, there are 91 possibilities, of which 30 are successes and 61 are failures. So the probability of success is 30/91.

Answer 4

To get at least one 6 you can roll:

{6xx,x6x,xx6,6x6,66x,x66,666}

where x is not 6.

6xx for example can occur in 25 ways,

6x6 in 5 ways,

so we have 91 ways of rolling at least one 6.

How many ways can we get at least one 1 from these?

6xx: 61x = 4 ways, 6x1 = 4 ways, 611 = 1 way so 9 ways for 6xx.

Same for x6x and xx6, so these give 27 ways.

66x: Only one way, same for 6x6 and x66, so these give 3 ways.

So we have 30 ways of getting at least one 1.

So 30/91.

Answer 5

Your solution essentially said "just look at the first die and if it is six then see that the probability at least one of the other dice is one is $\frac{11}{36}$". But you ignored the possibility that it might not be six while one of the other dice is six. You cannot just say this applies equally to looking at the second die and the third die, as more than one might be a six.

There are several ways of doing this correctly. One is a counting argument saying there are $6^3=216$ equally likely ways of rolling three fair dice and the number which have $a$ sixes and $b$ ones is $\frac{3!}{a!\, b!\, (3-a-b)!}1^a1^b4^{3-a-b}$ when $a+b\le 3$, giving this table

    a:   0   1   2   3
        --  --  --  --
b: 0 |  64  48  12   1 
   1 |  48  24   3   0  
   2 |  12   3   0   0
   3 |   1   0   0   0

and counting the cases with $a\ge 1$ and $b>1$ and dividing it by the number of cases with $a\ge 1$ gives a result of $\frac{24+3+3}{48 + 12+ 1+24+3+3} =\frac{30}{91}$.