seanie

I've looked at the original question again.

Is there something ambiguous about the wording that is leading me astray? Am I inferring something I'm not entitled to?

I understand where 1 in 11 comes from. There are 11 combinations involving 6 and only 1 is double 6. I get that. If someone had rolled a dice a thousand times and kept a note of every outcome involving a 6 and then asked me the odds of picking a 6-6 from those results at random I would say yes, the odds of me getting that result are probably going to be around 1 in 11. I get that.

But I still can't get away from the notion that the answer to the original question is 1 in 6.

Please help.

parkdalian

This little execise sent me way down the vertical thinking path as well. Part of the problem is the wording. It doesn't give you the exact procedure for revealing the six.

Here's another way of looking at it.

Procedure A:
A friend rolls the dice behind a partition and agrees to let you know if there is a six by handing you a die with a six. After a few rolls, he hands you a six. What he has just done is weeded out a six, thus shifting the odds against upward that the unknown die is not a six. 1 in 11 are the odds of a second 6 because 10 out of 11 times in this procedure, your friend will have handed you the only 6 out of the two dice. Also, the odds are greater that he will hand you a 6 doing it this way, occurring 11 out of 36 rolls.

Note that there is no difference whether he hands you the die or not (simply saying "there is at least one 6" suffices), it gives you the same information either way. You never know if that was the first or second die he's referring to and your friend might have rolled a non six die first, knocking the odds down to zero from your friend's point of view.

Procedure B:
A friend rolls the dice behind a partition and agrees to hand you a die randomly chosen from the two. In this case it would probably take more rolls than procedure A for the person to hand you a six, in fact 6 in 36 (or 1 in 6) rolls to show you a six. The odds of a six for the second unknown die are indeed 1 in 6 because the odds aren't tainted by a biased selection process. Procedure B is exactly the same as rolling two dice, one after the other.

In procedure A: 11 in 36 chance of handing you a 6

In procedure B: 6 in 36 chance of handing you a 6

There is your 5 missing combinations again.

To word it another way, imagine you like cherries, and your friend likes chocolates. You go to a dispensing machine that has a 50-50 chance of dispensing either cherries or chocolates. your friend says to you that he has the money for 2 treats from the machine, so he will select first from the two dispensed treats, whatever they may be. If 2 cherries come up, you get your cherry treat. If 1 chocolate and 1 cherry come up you still get a cherry because you friend will always choose a chocolate. You're out of luck only if two chocolates come up. So your odds are better than 50-50 (3 out of 4) of you getting a cherry since your friend is skewing the odds in your favor by plucking out the chocolates first, analogous to procedure A.

Phscs

1/11

The point is the word "assuming".

We don't "know" - we "assume".

If we throw two dice, and we are shown (or told) that one is 6, then the probability of the other being 6, is 1/6.

In this question however, we dont "know". We assume - before, during, or after - that one of them is 6.

In this case we are left with two groups, dice 1 is 6, and dice 2 can be 1-6, the second group is dice 2 being 6, and dice 1 being 1-5 (6-6 already covered). 11 possibilities to consider.

Notice that when we KNOW as opposed to ASSUME, then the groups disappear - we have a 6 in our hand - and there is only one dice left.

When we ASSUME, then there are still two dices left to consider, giving 11 possibilities.

In terms of information given, there is a big difference between the two:

Knowledge: We know one dice is 6, and we know which one. Two pieces of information.

Assume: We know one dice is 6, but we do not know which one. One piece of information.

Does this seem sensible?

regards

-phscs

beausoleil

I give you two boxes, one has £10,000 in it. If you knew which one it was in, you could discard the other. But the odds of choosing the right box are still 1 in 2.

seanie

Thanks for those last posts. In fact thanks to all who posted.

My mind is still foggy but I think I can see something coherent beggining to emerge. I might get it in the end.

Parkdalian thanks.

I'll have to think about your post but it seems sensible (and it gives me hope I'm not completely stupid.)

Thank you Phscs.

But I don't quite follow this;

Firstly I can know one dice is a 6 without knowing which one. It can still constitute 1 piece of information. Dave told me one was a 6. He didn't tell me which one.

Secondly if I assume one dice is 6 I don't actually know one dice is a 6. It's speculation. There is no actual information content about whether there is a 6.
But assuming just means I'm going to proceed as if I knew there was a 6. Assuming it's a 6 merely means I'm going to approach the calculations as if I knew.

Does that make sense?

And Thank you Beausoleil.

That last one you gave me was a doozy. The logic I've been using up till now doesn't seem to work at all with that one. I'm stumped. I'll have to go away and think some more.

Thanks again.

Phscs

Hi Seanie!

Consider this then:

"IF we through two dices, and one of them is 6, what is the probability of the second being 6 as well?"

Notice the lack of knowledge - we havent actually thrown the dices yet - either dice may be 6.

Compare now to this:

"We have thrown to dices, one of them IS 6, what is the chance of the second being 6 as well?"

The difference is not in the number on the dice, but in the number of dices itself.

In the first case we are forced to consider possible combinations of TWO dices, while in the second, we only consider the remaining dice.

It comes down to specific versus general - one dice is 6, but is it a specific dice, or can it be either? If it can be either, then there are 11 combinations, one of which has 2 sixes.

Do the statistical experiment like this: Throw 2 dices 10,000 times, and write down all values which contain at least one 6.

In the group you have written down, 1/11 will contain 2 sixes.

Do another experiment, where you ADD information. Another person throws the dices, and every time a 6 turns up, he asks you: "what is the value of the other dice".

You will have 1/6 chance to be right every time!

regards

-phscs

Grizzly

Seanie,

When trying to work out probability problems, I find it helpful to always keep in the back of my mind the idea of sample space and hits. Whenever I run into a probability problem, my first reaction is "what are all the possible outcomes" and "of those possible outcomes, how many are hits". This sample space is typically the denominator and the number of hits is the numerator.

As Parkdalian so excellently described, the problem is one of defining the sample space (how many times your friend is going to hand you the die). The hit in this problem is always 1 (two sixes). When your friend hands you a die whenever a 6 appears, there are 11 instances when this can happen. When your friend hands you the die whenever the first die is a 6, there are only 6 instances when this can happen. In the second case, 5 misses have been removed from your sample space.

Probabilities are not always intuitive - many people struggle with them. If they were easy, we wouldn't have courses that focused on probability!

Cheers

[ July 05, 2002: Message edited by: Grizzly ]</p>

seanie

So is this really down to the interpretation of the question?

What are the most parsimonius rephrasings of the original post in order to make it clear which of the two procedures to follow? (the procedures as outlined by Parkladian).

Can anyone help?

seanie

On second thoughts I'll try and work that out myself.

I've imposed on you all enough.

Cheers.

Principia

You're a gambler, you said?

Try thinking of it this way. You place a bet that the next dice roll is double 6s. And to boot you have a friend that is helping you cheat, by taking a peek at the dice rolls and giving you hints.

Would you rather:

A) the friend tell you nothing about the dice roll (i.e. odds are still 1 in 36)

B) the friend tell you that he sees at least one 6 (i.e. odds are now 1 in 11, a bit less)

C) the friend hand you one die that is a 6
(i.e. odds are now 1 in 6 since you only have to guess the remaining hidden die, even less!)


The whole point is that in B the friend gives incomplete information, which is represented by having a whole lot more choices (i.e. 11 total) to choose from. Part of the missing formation in B which is present in C is which die is the 6. So in B, not only do you have to guess which die is a 6 that your friend was talking about, you also have to figure out the number on the other die. The 'guessing which die your friend is talking about' part lowers the probability by about a half from C.

To see this, have a friend cover up a dice that he knows is a 6 in one hand (doesn't tell you which), and tosses another dice and covers that one too. Now, think about what goes through your mind as you consider your bet. You are thinking, 'OK, I have a 6 in there, and the other is a complete unknown. So, I am going to take a guess that one hand has the original 6 my friend told me about, and take another guess that the other hand has a 6, too.' In the first guess you'd be right half the time. In the the second guess you'd be right a sixth of the time. But you won't be right a twelth of the time altogether. You get a freebie when you screw up which hand, and there are two sixes anyway (so you can't tell which hand had the original six). Try this at home with a friend.

Good luck.

[ July 05, 2002: Message edited by: Scientiae[retired] ]</p>