seanie

You've changed the scenario.

If I don't get punched there is no 6.

If Alice punches me and Bob doesn't, I know that the blue dice is a 6 and the red one isn't. I don't need to calculate the probabilities of the other dice being a 6. It's zero.

If they both punch me I know that the red dice is certainly a 6, and I that there is a possibility that both are 6 (you didn't specify exactly what Alice did if they were both 6). I still only know that one dice has rolled 6.

If Bob punches me and Alice doesn't I punch Bob.

But this scenario isn't analogous to A, B or C.

seanie

An attempt to explain my reasoning.

Possible combinations of 2 dice A and B.

A B

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

As highlighted there are indeed 11 possible combinations involving a 6 in the total of 36 equally likely outcomes. If you ask me the chances of rolling a combination involving a 6 I'd say 11 in 36. Fair enough.

Going back to the original post at least one of the two dice rolled is a 6. Without knowing which dice is definitely a 6 what can I infer about the other dice?

Well I know that either A or B have a 6 on it.

If it's on A I look at my list to see the number of possible combinations in which A is 6.

I see that out of 36 equally likely combinations there are 6 in total, only one of which is double 6. So if A is 6 the chances of B being 6 are 1 in 6.

If however it's on B I look at my list to see the number of possible combinations in which B is 6.

I see that out of 36 equally likely combinations there are 6 in total, only one of which is double 6. So if B is 6 the chances of A being 6 are 1 in 6.

Going back to the 11 out of 36 combinations involving a 6.

A B

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

That's where we started. But now the identity of one of the dice has been determined as a 6.

I don't know which one.

But if it's A I discount the first 5 combinations. They're not logically possible.

If it's B I discount the last 5 combinations. They're not logically possible.

I don't need to know if A or B is in fact the dice with a 6 I've been told about. I'm left with a 1 in 6 chance of the other dice being a 6 in either case.

Why am I wrong?

beausoleil

Because you don't know whether it is A or B, and not knowing makes a difference - all possibilities are still in play.

If you knew it was A you could discard 5, if you knew it was B you could discard 5, but you don't know whether it is A or B, so you can't discard any of the 11.

Try the experiment - with the coins it is easy! In fact, suppose it goes perfectly 12 throws yielding:

HH
HH
HH
HT
HT
HT
TH
TH
TH
TT
TT
TT

When at least one head is thrown it is accompanied by a tail 6 times and by a head 3 times. Yet each coin is random 50-50 chance of coming down heads or tails.

I am

Because you are counting the 6-6 combination twice. Once for die A and once for die B. Thus you calculae the odds to be 2/12.

seanie

I'm not trying to be stubborn. I have never studied probabilities, I know that people's intuition for probabilities is generally very poor, and that I have a minority opinion on this thread. My intuition says that I'm probably wrong.

But I want to know why. I don't see the flaw in my reasoning.

I struggle a bit with some of the number crunching so try this for a scenario.

A week from now I'm in a room with Bob and Alice. They're absentmindedly tossing one dice each, again and again. They're not even paying attention to the outcome.

Each time they throw a dice there is a 1 in 6 chance of it being a 6 correct?

I'm watch them toss the dice simultaneously when something clicks! This thread pops back into my brain! As the dice bounce I realise I have an opportunity. When the dice come to rest I shout...

"STOP!"

Bob and Alice turn and stare at me.

"Keep looking at me!" I say. "we're going to conduct an experiment".

"DAVE!" I cry.

Dave comes in from the room next door.

"Dave, could you do us a favour? Both Bob and Alice have a dice in front of them. None of us have looked or are going to look at either of them. Could you look at both of them for me? Then having done that could you whisper to me what only one of them is? Either one it doesn't matter. And don't tell me whether it's Bob's or Alice's".

"Sure thing Seanie." replies Dave.

As Dave goes about his business I muse on the probabilities. I know that every time Alice threw the dice the odds of a 6 were 1 in 6. Every time Bob threw the dice the odds of a 6 were 1 in 6. I know the odds of a double 6 are 1 in 36. No controversy there. I wonder what Dave will tell me.

Dave ambles back, leans over and whispers...

"One of them is a 6 Seanie."

Wow......the odds of a double 6 have shortened dramatically. But to what?

Dave's a trustworthy guy so I know one of the dice is a 6. What are the odds that the other dice could be a 6? Well up until Dave whispered in my ear I was certain that each dice had a 1 in 6 chance of being a 6. Every single time up till now that Bob or Alice rolled their dice a 6 was a 1 in 6 chance.

Now you're telling me that wasn't the case. Actually one of the dice only ever had a 1 in 11 chance of being a 6. Which dice? The dice Dave didn't tell me about, whoever it belongs to.

Mmmmmmmmm....... I'm not convinced.

seanie

Not at all.

If I knew it was A i can discard 5. We agree.

If I knew it was B i can discard 5. We agree.

But I don't know whether it's A or B so I can't discard any? Nonsense.

I know it's A or B. One or the other. Either way I can discard 5.

Not knowing whether it's A or B merely means that strictly speaking I don't know which particular 5 I should discard.

Should 6-1 or 1-6 go? 6-2 or 2-6?

I don't know.

I don't care.

Because I know for a logical certainty that one set of 5 must go.

Which leaves me 1 in 6.

Principia

Really, Seanie, the best way is for you to get out two dice and start experimenting. Short of having a person next to you and giving you a mini-course on probabilities, this course of action is perhaps the easiest. Discussing something you have no intuition over, in a message board under such a disjointed fashion is only going to confuse you.

Good luck.

Bill Snedden

Wow. A relatively simple scenario gone to two pages. Just more proof of how un-intuitive statistics can be.

Dare we start talking about Monty Hall?

Bill

Grizzly

Try this. First start by defining your sample space. What are the mutually exclusive conditions by which our sample space is defined.

1) the first die is a 6 but the second one is not.
2) the second die is a 6 but the first one is not.
3) they are both 6's.

Only the last condition will yield a hit. The first and the second are misses. These are mutually exclusive events that cover the entire sample space. This means you can figure out the probability each event, add them up, and you will get the total probablity. What are the exhaustive events of the first scenario?

61
62
63
64
65


What are the exhaustive events of the second scenario?

16
26
36
46
56

What is the the only event that occurs in the third scenario?

66

We have now defined our sample space

61
62
63
64
65
16
26
36
46
56
66

It has 11 elements. Now, out of our sample space, how many are “hits”? Just one – the last 1. Or an odds of 1 out of 11 that if one of the die is a 6, then the other die will be a six.

Grizzly

Bill, I teach graduate level statistics (in a psychology department). The Monty Hall problem has brought many of my students to their knees.