I am

Wrong.

If you showed *BOTH* dice to Bob, and he said "atleast one of these dice is a six" you would then know that the chances of both being a six is 1/11.

If the identity of the first die is predetermined, the chance for the second six is a "fair roll" - 1/6.

[overall the chances for a 6-6 are 1/36 (1/6 * 1/6)]

[ July 03, 2002: Message edited by: I am ]</p>

seanie

Sorry I'm still not following this. I'll stick with 6-6 for now. A double 6 is a 1 in 36 chance correct? The consensus appears to be that having seen one 6 the chances of the other dice being a 6 should be calculated at 1 in 11, because these are the following possibilities to choose from.

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

Now looking at those combinations it seems to me some can be logically dismissed. Let's say the left column represents dice A and the right column dice B OK?

I'm looking at a dice that's rolled a 6 and I'm trying to work out what the odds for the other dice are.

Well if the dice I'm looking at is dice A I can dismiss the first 5 combinations immediately. Any combination in which dice A isn't 6 is clearly wrong. That only leaves me with 6 possibilities. So I'd calculate the odds of the other dice being 6 at 1 in 6.

Maybe I'm not looking at dice A. It's actually dice B. In which case I can dismiss the last 5 combinations. Again they're impossible. Again I'm left with 6 options. A 1 in 6 chance.

It doesn't matter if the dice I'm looking at is A or B. It doesn't matter whether I think it's A or B. It doesn't matter if they're rolled simultaneously or sequentially. The chances of an unloaded 6 sided dice turning up a 6 are always 1 in 6.

seanie

Why is it wrong?

A - Bob sees 1 dice that is a 6. He is in the position of knowing that 'at least one of these dice is a 6'.

B - Bob sees both dice and correctly tells me that that one is a 6. I am in a position of knowing that 'at least one of these dice is a 6'.

Bob in A possessess the same knowledge as I do in B. How would we calculate the odds differently?

beausoleil

The point is that you don't know whether it is A or B - all you know is that there was at least one 6.

That's why only those 11 and all of those 11 options are relevant and the answer is 1 in 11.

What you say about an unloaded die is correct. But we're not talking about the roll of an unloaded die - we're talking about rolling two die and then eliminating some of the possible outcomes through being told some information about the result.

One more go! If you roll two dice, and look at only one, it doesn't matter what it is - the other one is completely independent. What makes this case different from that is that somebody has looked at both dice and given you information about the combination - that it contains one 6. There are 11 such combinations and only 1 has a 6.

[ July 04, 2002: Message edited by: beausoleil ]</p>

beausoleil

In case A, Bob knows that one of the dice selected at random is a 6. He knows nothing about the other dice.

Case B, you know that at least one of the dice is a 6.

Combinations selected in case B are excluded from case A, so the odds are different.

(I misunderstood an earlier post - when you said you showed dice A to Bob I assumed you meant that you chose to show him A because it was a 6)

seanie

I don't need to know whether the dice is A or B.

Whether it's A or B I can logically dismiss 5 of the 11 combonations.

seanie

How does that work? These are separate events.

In case A Bob knows one dice is a 6 and he knows nothing about the other dice.

In case B I know one dice is a 6 and know nothing about the other dice.

How would we calculate the odds differently?

seanie

Just to clarify what I'm saying.

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

Left column dice A. Right column dice B.

If the 6 I'm looking at is on A I can logically dismiss the first 5 combinations.

If the 6 I'm looking at is on B I can logically dismiss the last 5 options.

In either case I'm left with 6 possible combinations. A 1 in 6 chance.

Whether the dice I'm looking is A or B is completely unimportant. I don't need to know.

Because whether it's A or B the result is the same. I can exclude 5 and be left with 6 possible combinations.

1 in 6.

seanie

I don't get this at all.

If I look at one of the dice I get information about the combination. In know one number in the combination.

If someone else looks at one of the dice and tells me what it is I get information about the combination. I know one number in the combination.

If someone else looks at both of the dice and tells me what one of them is I get information about the combination. I know one number in the combination.

In each case I get the same amount of information. 1 number in the combination.

How can I proceed to calculate the odds differently?

For any specified 2 digit outcome (be it 6-6, 5-6, 6-5, 2-3 or whatever) if I know that one of the two dice matches the combination then I can confidently predict that the odds of the other completing the specified combination is 1 in 6.

beausoleil

(A, B, C added by me)

Case A and B are different from C.

In A and B, someone looks at a dice and tells you if it is a 6.

In C, someone looks at two dice and tells you if one of them is a 6.

Suppose they are both happening at the same time - a red and a blue dice are being thrown. Bob looks at the red dice and punches you if it is a 6. Alice looks at both dice and punches you if either red OR blue is a 6.

Saying that you have the same information in A B and C is the same as saying that Alice never punches you if Bob doesn't.

If you still don't believe the answer is 1 in 11, I suggest you do the experiment. Roll two dice, every time either one is a 6 write down the number on the other. For double 6 you write down 6 once. If neither is a 6, don't write anything down. Count up the statistics. You only need to do it a few times to show the answer isn't 1 in 6.

Or do it with coins. Throw two coins. Every time one of them is heads write down whether the other is heads or tails. If neither is heads, write nothing down. Tails will outnumber heads by two to one in what you write down.