Odds problem - roll a pair of dice

[Thudlow_Boink]

@Haldurson

Knowledge changes everything.

Yes; and the reason that people in this thread (at least the ones who understand probability) are getting different answers is because of different assumptions/interpretations about what knowledge you have in the situation described in the OP.

[x-ray_vision]

Cliffy, I made a similar point as Haldurson did. Can you answer the question I specifically asked you in post 325?

[Telemark]

@Grestarian

Haldurson said it best: The red herring doesn’t change the fact that the probability of any single 6-sided die roll being a 6 is 1/6

This is true, but it’s not the question at hand. Some of the possible rolls have been removed from the choice set, and what we’re left with has a 1/11 chance of being a six. The manner in which the die that we’re asking about is chosen isn’t the same as if you threw a single die and asked if it was a 6.

Again, if you doubt it then write a small computer simulation following the instructions
[ol]
[li]Roll two dice[/li][li]If neither is a 6 then ignore roll and go back to step 1[/li][li]If AT LEAST one die is a 6 then increment your throw counter, select a die that shows six as your primary die and call the other the remaining die[/li][li]If the remaining die is a 6, increment your success counter[/li][li]Go back to step 1[/li][/ol]

After running this simulation 10,000 times do you think this test will produce a success/throw ratio of 1/6 or 1/11?

If you remove step 2 and the first part of step 3 then your answer will be 1/6, but that’s not the setup.

[ftg]

@The_Great_Unwashed

For instance you seem to have said that the sequential rolling of two dice (one after the other) is an essentially different experiment to the “simultaneous” rolling of two dice. It isn’t, I’m wondering why you think it is.

You left out something very key about what I said: Rolling one die until you get a 6 and then rolling the second die is what I was objecting to. This conditional rolling of the second die changes things quite a bit.

You can roll one die on Monday and the second die on Friday. As long as the second roll happens regardless of what the first roll was. It suffices to not look at the first roll until the second roll is finished. Note that not saying anything to the other person until the 2nd roll is finished is implied by all this.

OTOH, rolling the first die. Looking at it and if it’s a 6 telling the other person before rolling the 2nd die is a separate problem.

So any situation with the events in this order:

1. Roll the dice. One at a time, together, doesn’t matter.
2. Look at both dice.
3. Tell the other person “At least one is a 6.”, if that is indeed the case.

satisfies the conditions for the 1/11 answer.

Some variations of the above also give 1/11. But if you push it a bit too far, you get something else.

The OP’s statement is clear enough that the 1/11 answer is the only one that fits unless you get into: What if the dice are 23 sided? What if the roller is lying? and other ultra-pedantic questions.

[Irishman]

Musicat, jtur88, you are both interpreting the question to be same as the following:

“You roll two fair six-sided dice. You then ignore one of them. What are the odds that the remaining die is a six?”

Going back the the original problem as stated and given that the list of answers excludes “1/11” but includes “1/6”, I am inclined to agree that the test writer meant the statement “a six is removed” to be a meaningful severing of the combined information you were given - that at least one die was a six.

This appears to be a trick question trying to get the statisticians to assume it is a different problem than it is. You are given irrelevant information - at at least one is a 6. It is irrelevant because that 6 is then removed from consideration, and you are told to look only at the remaining die.

The problem it resembles would be equivalent to:

“You roll two fair six-sided dice. If one of them is a six, what are the odds they are both sixes?”

For those attempting to solve that problem, then:

@chrisk

But the real sleight of hand, as it were, is that I’m possibly not deciding which die I’m calling “the other die” until after I’ve rolled both dice, and THAT changes the odds.

Agreed. The selection of a red and green die is irrelevant, because this version of the question does not specify the order. There are 11 equally likely possible outcomes. Of those, only one outcome has double sixes.

You can’t conflate the action of selecting the red 6 versus the green 6 as a separate condition. That is a separate step in the sorting process, and is another term in the probability calculation.

In this problem, (6,6) has the same likilihood as (1,6) or (6,1). If you pick red 6 half the time and green 6 half the time, then you have to split the probability of (6,6) in two for show Red (6,6) and show Green (6,6).

Alternately, you have to lump (1,6) and (6,1) as one option.

Think of it this way, if the dice are Red/Green (6,6), and half the time the selector picks Red 6 and half the time Green 6, you have to compare against the condition that if the dice are Red/Green (1,6), then half the time the selector will pick Green 6 and the other half the time Green 6. That is the level of equivalent likelihood options.

Analogy: think of an estate of $100,000 to be divided among 4 siblings equally. Everyone should get $25,000. But what if one of the siblings died, but left two children of his/her own? They get 1 share equal to 1/4 the whole pot, and they have to split that share between the two of them.

Or tell the equivalent with the lottery. 6 winning lottery tickets, but 5 tickets are owned by 1 person apiece while the sixth ticket has two owners sharing the ticket. Each ticket gets an equal share of the winnings, and the two people sharing a ticket have to split their single share. They don’t get a full share apiece and force the pot to be divided seven ways.

@octopus

Sorry for the ugly formatting but padding spaces seem to be removed.

Try using [noparse]

[/noparse] tags.

Double 6	At least one 6	Odds	Inverse 
			
1        	9	0.111111111	9
2	      13	0.153846154	6.5
3	      41	0.073170732	13.66666667
4	      48	0.083333333	12
5	      66	0.075757576	13.2
6	      80	0.075	        13.33333333
7	      85	0.082352941	12.14285714
8	      86	0.093023256	10.75
9	      95	0.094736842	10.55555556
10	      97	0.103092784	9.7
11	     106	0.103773585	9.636363636
12	     135	0.088888889	11.25
13	     149	0.087248322	11.46153846
14	     150	0.093333333	10.71428571
15	     154	0.097402597	10.26666667
16	     181	0.08839779	11.3125
17	     182	0.093406593	10.70588235
18	     201	0.089552239	11.16666667
19	     207	0.09178744	10.89473684
20	     227	0.088105727	11.35

[Irishman]

I see now that I’m late to the party. When I posted that, I had eliminated the other option: that the test writer did not know what they were testing, so wrote the problem incorrectly. I now see that is actually the case. :smack:

@LSLGuy

@CHronos: As I said a bit ago, I wonder what the original puzzle maker was thinking when he/she asserted 2/11 was the right answer. To be sure there are *conceivable *rule sets that produce that result. What IMO there aren’t, is *simple & plausible *rule sets that do so.

Any ideas?

When throwing two fair six-sided dice, and specifying a 4 and a 6, but not specifying which die is which number, you have two dice to pick from and 11 possible pairs. This differs from the specifying 6 case by virtue that (6,6) is only one option, but (4,6) is different than (6,4).

@Thudlow_Boink

This is the interpretation I’d lean towards if the original wording had said “one of the dice is a 6” instead of “at least one of the dice is a 6.”

If it had simply said “One of the dice is a 6,” it would be natural and reasonable (IMHO) to interpret this as “I’ll tell you what one of the dice was: it happened to be a six. Now if I remove that one, what’s the probability that the other one is a 6?” And then your reasoning would be correct, and the answer would be 1/6.

But (again IMHO) the words “at least” don’t fit that interpretation. To me, “at least one of the dice is a 6” means “Out of all the possible ways the pair of dice could come up, the result was one of the ones in which at least one of the two dice shows a 6.” There are 11 such possibilities, and in only one of them would the remaining die be a 6 if a 6 were removed.

Ah, but someone else could just as easily say “you told me one of the dice is a 6, so the other dice is not a 6, QED.” Thus the verbage “at least one is a 6”, allowing that the other might be a 6 as well.

Language is [del]a bitch[/del] inherently ambiguous.

@septimus

Wrong.

I have two children and I tell you at least one of them is a girl. What is the chance both are girls?

Do the people getting OP’s problem wrong also get the two-children problem wrong? They are really the exact same problem. The correct answer is 1/(2N-1) not 1/N, where N=6 and N=2 for the respective problems.

Once you remove a specific die (the 6), you have narrowed the field just like specifying the older child.

@Cliffy_Biro

Does this help illustrate the 1/6 (2/12) argument? (Not that I agree with it.) The bold lines are the equally likely possibilities that are consistent with the event described in the problem statement. There are twelve of them, of which two are [6 6].

Dice	Die you learn about
====	===================
1 6	first
**1 6	second**
2 6	first
**2 6	second**
3 6	first
**3 6	second**
4 6	first
**4 6	second**
5 6	first
**5 6	second**
**6 1	first**
6 1	second
**6 2	first**
6 2	second
**6 3	first**
6 3	second
**6 4	first**
6 4	second
**6 5	first**
6 5	second
**6 6	first**
**6 6	second**

I think this post describes the flaw.

@Cliffy_Biro

When choosing a sample space, you must also consider the relative probabilities of the outcomes in it. It can make things simpler if you choose the sample space so that each outcome in it is equally likely. Otherwise you have to get into weighting things by their probabilities.
You can’t just say “there are four possible outcomes in the sample space, so the probability that either one of two particular outcomes will occur is 2/4.” That’s true if the four outcomes are equally likely, but not necessarily true otherwise.

**6 6	first**
**6 6	second**

[/QUOTE]

are not equally likely options to the other options. There are two steps in that selection, the die roll with it’s inherent probabilities, and then the selection of which to reveal. Note that for the other options, all iterations with that outcome provide the same choice of which to reveal, but for (6,6), half the time you pick one and half the time you pick the other.

@BigT

This is adding information not stated in the problem. At no point does it say that they picked one die at random and read the number. The number 6 is given in the problem without any means for it to be different, and thus is fixed. Any simulation must keep the 6, or else it is modeling a different problem. And once you do require there to be a 6, the 1/11 interpretation is the only one that works.

I do not understand why such an explanation is still being promoted, when one of the basic rules of word problems is that all required information is given.

@misterdls

We can put each die face of the 11 listed combinations into one of two “modes”. Either
the die face shows 6 or it doesn’t. If it shows 6 we’ll say it’s a “mode6” face.
If it isn’t a 6 then we will call it a “modex” face. The actual number shown on a
modex die in the box isn’t important in our deliberations here.

Now consider the initial dice throw. After the throw and BEFORE seeing if either die
is a 6 (and therefore mode6), the content of each of the 36 combinations will ALWAYS
consist of one of the following mode combinations:

Green Red
mode6 mode6
mode6 modex
modex mode6
modex modex.

4 combinations, each equally likely. If one of the 4 combinations are removed
from this list, the remaining 3 combinations remain equally likely.

False. That’s just like saying “Either my lottery ticket is a winner or a loser, so I have a 50% chance of winning.”