Silly dicey problem
 

Against my better judgment, I've become involved in a discussion about probability elsewhere (no, it is not the [URL="http://en.wikipedia.org/wiki/Monty_Hall_problem"]Monty Hall[/URL] problem) but it might be too counterintuitive for some to accept.

Scenario:
No trickery - each die is a standard six-sided cube, with each face numbered uniquely 1 to 6, and fair (not altered or loaded).
You need to determine the probability of each die in a pair of dice showing a 2, given certain information.


You throw two fair six-sided dice simultaneously so that each lands by itself in one of two separate boxes, labelled L and R. You cannot see into the boxes, but your partner can. Your partner then truthfully tells you that at least one of the die is showing a 2.

What are the odds that both of the cubes show a 2?

Please explain the process used to arrive at your answer.


I'll post my original response later, if required. I'm more interested in explaining why the incorrect answer is incorrect than anything else.

[SPOILER]I come up with 1/11, others are vehemently supporting the more intuitive 1/6. I've done 3.6MM simulated rolls in Excel; the results support my answer.[/SPOILER]

[spoiler]Of the 36 possible permutations only 11 of those have a 2. And only one of those 11 is 2-2. And since all 11 have the same probability of occurring then the answer is simply 1 in 11.[/spoiler]

Thanks for the response, Retina!

I'll check back later and post my (longish) original reply.

[SPOILER]As far as I can tell, the 1/6 group reasoning is "Well, we know one of the die is a 2, so 1/2 of the possible non-2,2 are impossible, and can be excluded." My response will probably be "No, because you don't KNOW which are impossible, they COULD be possible, and therefore must be included," but I'm not sure such a explanation-free statement will move things forward.[/SPOILER]

Yes, I know this can/will turn into the famous [URL="http://xkcd.com/386/"]XKCD[/URL] comic...

Given that there is atleast one 2, there are 11 possible states: (L=2, R=non-2), (L=non-2, R=2), and (L=2, R=2). Out of which only one is the target state. So 1:11. Or alternatively, we know it is NOT one of the 25 states (L=non-2, R=non-2).

The 1/6 thing is a mistake taken from mis-understanding the problem statement. If your partner had instead said: "The die in the LEFT box is a 2," then there is in fact a 1/6 chance of them both being 2.

Interesting side-question: If knowing which box has the 2 in it gives us a 1/6 chance and not knowing gives us a 1/11 chance, why does knowing which box has the 2 in it [B]increase[/B] the likelihood of having a pair of 2's?


EDIT: Screw it. No spoilers. If someone wants to solve it on their own they haven't read all the way down here. Even if they have, I'm not explaining how 1/11 is correct...

Questions involving the probability of A given B always need to be read more carefully. People either get too focused on the probability of A on its own or misunderstand what exactly B means. The Monty Hall problem is one of the best traps for the second one.

Original post is below the hashes with typo correction and context added to coin flips. I was walking myself through this, which is why it starts with coin flips. I was hoping that it would illustrate how specific information affects the calculations.
########################

I wonder if we should rename the dice problem Schroedinger's Dice.

My rehash:

Toss two fair coins simultaneously so that each one lands by itself in one of two separate boxes, labelled L and R. You cannot see into the boxes, but your partner can. He will truthfully tell you information. [You want to know the likelihood of HH.]

Before your partner says anything, the possible results are:
[code]
___L___|___R___
H | H
H | T
T | H
T | T
[/code]
(or HH HT TH TT for short)

Probability of HH is 1/4 with no other information.

Partner says that one of the coins is H. Now the possible results are:
[code]
___L___|___R___
H | H
H | T
T | H
[/code]
(HH HT TH)

Now the probability of HH is 1/3, since there cannot be two Ts.

New toss, but now your partner says the L coin is H, leaving these possible results:
[code]
___L___|___R___
H | H
H | T
[/code]
(HH HT)
Now the probability of HH is 1/2.


So what if we expand this to the two dice problem. Same setup - two fair dice are thrown simultaneously so that each lands by itself in one of two separate boxes, labelled L and R. You cannot see into the boxes, but your partner can.

Before your partner says anything, the possible results are:[code]

___L___|___R___||COUNT||
1 | 1 1
1 | 2 2
1 | 3 3
1 | 4 4
1 | 5 5
1 | 6 6
2 | 1 7
2 | 2 8
2 | 3 9
2 | 4 10
2 | 5 11
2 | 6 12
3 | 1 13
3 | 2 14
3 | 3 15
3 | 4 16
3 | 5 17
3 | 6 18
4 | 1 19
4 | 2 20
4 | 3 21
4 | 4 22
4 | 5 23
4 | 6 24
5 | 1 25
5 | 2 26
5 | 3 27
5 | 4 28
5 | 5 29
5 | 6 30
6 | 1 31
6 | 2 32
6 | 3 33
6 | 4 34
6 | 5 35
6 | 6 36

[/code]Only one combination out of 36 with L2 R2, so 1/36.

Then your partner announces that the L die is a 2.
Cool! Remove all combinations where the L die is not a 2:
[code]
___L___|___R___||COUNT||
2 | 1 1
2 | 2 2
2 | 3 3
2 | 4 4
2 | 5 5
2 | 6 6
[/code]
Now there is one combination out of 6 with L2 R2, so 1/6.

But wait! Your partner screwed up! Due to the arbitrary rules of the game, you cannot be told [i]which[/i] die contains a two, only that [b]A[/b] two has been thrown. So the throws will be repeated, this time with an admonishment to your partner.
A few throws later, your partner announces at least one of the die shows a two. Now what is the probability of BOTH dice showing a two?
Based on your partner's information, you can eliminate all combinations without at least one 2:
[code]
___L___|___R___||COUNT||
1 | 2 1
2 | 1 2
2 | 2 3
2 | 3 4
2 | 4 5
2 | 5 6
2 | 6 7
3 | 2 8
4 | 2 9
5 | 2 10
6 | 2 11
[/code]

Now there is one combination out of 11 with L2 R2 after being told that at least one of the die showed a 2, so 1/11.

Note- Results of 3.6MM (10 runs of 360,000) rolls in Excel:[code]
Any two Both Two both/any ideal 1/11
Totals 1,099,165 100,140 0.091105521 0.09090909[/code]

In my experience, nobody still arguing is likely to be convinced. They will fall into three categories:
1. Those so cocksure of their answer that they don't think about, and probably don't even read, your explanation.
2. Those so mathematically naive they don't really understand the question.
3. Trolls.

If it were me, I would post the clearest explanation I could compose and then withdraw from the discussion. My explanation would be the following diagram. It says the same thing others on this thread have said in words, but a picture has a small chance of reaching some additional people by accessing a different way of thinking.
[CODE]
1 [B]2[/B] 3 4 5 6
+-----------
1 | [B]X[/B]
[B]2[/B] |[B]X X X X X X[/B]
3 | [B]X[/B]
4 | [B]X[/B]
5 | [B]X[/B]
6 | [B]X[/B][/CODE]

I know some wonderful and clever ideas about how to engage the first group. None of these ideas has ever worked for me on the Monty Hall problem, so they must be wonderful and clever in my mind, not in the real world.

William, others on the original thread have posted that exact same figure, as well as colorful variations.

It didn't help.

I've given up trying to convince the non-believers (appellation seems appropriate, since they will not accept mathematics or simulations as proof), and now am more interested in exploring the thought processes that get them to their conclusion. So far, the processes seem to start and stop at "there is one unknown six-sided die, so it must be 1/6", but I may pursue it a bit more.

Thanks to you and everybody else for taking the time to respond to this trivial matter!

P.S. Since my involvement started on a gambling website, now I need to figure out how to monetize their erroneous conclusions. I'm thinking that I should make a game like we roll dice, you pay me $1 for every 2, and I pay you $7 for every pair of 2s (rolls without any 2s are ignored).

A slight modification to the question:

You throw three fair six-sided dice simultaneously so that each lands by itself in one of three separate boxes. You cannot see into the boxes, but your partner can. Your partner then truthfully tells you that at least two of the dice are showing a 2.

What are the odds that all three of the cubes show a 2?

Please explain the process used to arrive at your answer.

Can anyone work out a formula for the generalization to n dice with n-1 known to be showing a 2 or better still with x known to be showing a 2?

[QUOTE=sdbardwick;400220]
P.S. Since my involvement started on a gambling website, now I need to figure out how to monetize their erroneous conclusions. I'm thinking that I should make a game like we roll dice, you pay me $1 for every 2, and I pay you $7 for every pair of 2s (rolls without any 2s are ignored).[/QUOTE]

This. Put your money where your math is. Offering 8 to 1 odds is clearly +EV to you both. :)