Odds problem - roll a pair of dice

Factual Questions
1

A math teacher friend of mine shared this problem on Facebook and we are disagreeing about the solution. Here is the problem verbatim:

“Two fair dice are rolled together, and you are told that ‘at least one of the dice is a 6.’ A 6 is removed, and you are ten shown the other die. What is the probability of the other die showing a six?”

My friend thinks the answer is 1/6, because the dice rolls are independent events and regardless of what one shows, the other has a 1/6 chance of being a six. I’m quite positive this is incorrect.

My reasoning is: if you roll a pair of dice, there are 36 possible combinations: (1,1), (1,2), (2,1), etc. Out of these 36 combinations, 11 contain at least one six: each 6 paired with 1-5 on the other die, and double sixes.

Since we are told that the pair contains at least one six, we have 11 possible combinations with equal chance of occurring. If a six is then removed, only 1 of the remaining 11 dice is another six. So I maintain the answer is 1/11.

However, the meme that she posted is multiple choice and 1/11 is not one of the available answers, which are: 1/18, 1/36, 2/11 and 1/6.

At first I jumped on the 2/11 thinking “there are two chances to pick a six from the pair of doubles!” But that logic falls apart I think. I’m back to 1/11. How am I wrong?

2

I agree with your analysis. I’ve seen that… meme? I don’t think that’s the right word… and I suspect that the correct answer was left out of the choices deliberately to make it more clickbaitey.

This is a variation of the old “…at least one is a boy” probability question that has been discussed here in numerous threads (e.g. here, here, here, here).

3

“A 6 is eliminated” isn’t enough information to give an answer. On what basis is the 6 eliminated? Does it mean that sixes have a half-life, and you measure the odds of one still remaining after the first one has gone kablooey? Or does it mean that you randomly choose one of the combinations including at least one 6, and eliminate a 6 from it?

4

Isn’t the question asking the probability of two sixes being rolled?

5

IMO the above part is right. Then you go wrong here:

How many different ways can we remove a six? Let’s assume we have a green and a red die so we can tell them apart. As you said, the possibilities expressed in green, red order are:

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

If the green die is a 6 we’re looking at one of the bottom 6 possibilities. 1/6th of which have another 6, the red die.

If the red die is a 6 we’re looking at one of the top 6 possibilities. 1/6th of which have another 6, the green die.

Result: it doesn’t matter which die is a 6. Red, green, or both. After you remove one of the sixes the chance of the other being a 6 is 1/6.

6

Late add:

You can restate the problem like this and it becomes really, really obvious:

A: Since the two dice don’t communicate, the odds of the second one being a 6 are *always *1/6. It doesn’t matter what the other die is doing.

This is one of those puzzles where the trick is to ignore most of the setup since that’s just there to distract you. “Now everybody watch my *left *hand very carefully as I fiddle with this magic wand…”

7

We have a condition here, IF one die shows a 6, THEN what are the odds the other die shows a 6. The answer is 1/6. Each valid trial requires that one of the die shows a 6, or else it’s not valid, rolling hard 10 or snake-eyes is dismissed.

Now, if the question is what are the odds of rolling a pair of 6’s, then that’s 1/36 … we have 36 valid trials of which only one result makes a positive.

ETA: nin’ja’ed by the great LSLGuy

8

The principle always holds in problems such as this:

**Entities **and their **outcomes **forever remain independent *with respect to each other *regardless of spatial, temporal, or any other separation.

9

Charma got it. The roll of each die is an independent process. You can completely ignore the fact that one of them shows a 6. The chance of any one 6-sided showing a 6 is always 1/6 no matter how many dice you roll and no matter what the other dice show.

10

I don’t think I agree… though as with many probabilities problems, it may be that I’m missing something or am making an assumption that others aren’t. Here’s my reasoning:

I roll the 2 dice. To begin with, there are 36 different possibilities, (counting the red die and the green die separately.)

I observe, “Hey, there’s a six here.” That eliminated 25 possibilities, all the ones composed only of 1 through 5 on both dice, and leaves 11 possibilities left, out of which only one is double-sixes.

Now, for each of these eleven possibilities, I can pick a six and take it off the table. Since I can always do that, doing it doesn’t change the original odds of rolling double sixes… the way the observation about having at least one six did. (That changed the odds, because there were many possible rolls in which I couldn’t make that observation.)

Thus, even after I’ve taken that die and removed it from the table, the odds that I originally rolled double sixes is still 1 in 11.

The remaining die still on the table will be a 6 if-and-only-if I rolled double sixes. Thus, the odds that the die on the table is a 6 is also 1 in 11.

No, the dice don’t communicate, and yes, the outcome of the dice is independent. But that observation of both dice together does change the answer, as does the process of “looking for” one six and selectively removing it from the table… assuming that I would announce my observation and remove one six every time I roll a result that has one or more sixes. That’s the assumption.

Edited to add: If, on the other hand, we assume that I removed one die from the table at random, and it happened to be a die showing a six… then I think the answer of 1 in 6 is correct.

11

The wording is distracting. If it had said,

then I would say the answer is 1/11, since there are 11 ways that two dice could show at least one six, but only one of them in which both dice are sixes.

On the other hand, if it said,

then the answer would be 1/6. After you remove one die, the odds that the other one is a six is the same as rolling that die by itself.

The odds where at least one is a 6 are different from where this one is a 6. The way the problem is worded, I’d say that the act of removing one of the dice makes the odds of the other one being a six 1/6.

12

Agreed.

13

(My emphasis)

Ah, but they do in a way communicate, since you’re told the pair contains “at least one 6”. So, if neither die showed a 6, they wouldn’t be on the table at all! So, if you are one of the dice, you can correctly infer: either myself, or that other die over there, is showing a 6!"

14

Ok, it’s not this because you know that one six was rolled. It does sound like the boy girl problem.

15

I think we’re confusing the odds of rolling at least one 6 with the OP. The OP gives us this one 6, we’re asked the odds of the other being a 6. We only have six possible trial results; (6,1),(6,2),(6,3),(6,4),(6,5),(6,6); all other combinations are ignored.

The operative phrase in the OP is “‘at least one of the dice is a 6.’”, it doesn’t matter which die is a 6.

16

Agreed.

17

But if we start not knowing if the red or green die, (for example,) is a 6, then we have 11 possible results at the early stage: (1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

18

I would say that the wording is not only distracting, it’s ambiguous. We have no idea whether there was intention or not in removing the 6, so we can’t tell if it changes the odds or how.

19

As in the case of the infamous Monty Hall problem, one way of settling this would be experimentally.

Roll two fair dice. At least one of the dice should be a 6, so if this is not the case, re-roll until it is.

Now, remove a 6.

Does the remaining die show a 6? Write down YES or NO.

Repeat this experiment many times, and then calculate the number of YESes you wrote down, dividing by the total number of trials.

20

Right, but how is it that a 6 being removed is decided?

We have:

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

Is on pair of dice randomly chosen to have a 6 removed?: 1 out of 11 pairs.

Or is one die out of ever die showing a 6 in the running to be chose?: 1 out of 12 dice.