Two Dice Puzzle: Part Deux

Quote: Ayecarumba

Thanks Dalex64! While I can understand your explanation, I can't shake the feeling that having the information that at least one die is a two, prior to being asked to compute the probability, makes a difference. According to the Wizard, how the dealer peeks makes a difference, but I am not clear why.

Think about the Crazy Casino's Table #2 described above. You approach the table with one die already set aside with the two face up. The dealer assures you that he shook both dice, but you didn't witness it. Theoretically, the die under the cup should be a two once every eleven times, but the dealer looks pretty lazy, and you are pretty sure he doesn't shake them both all the time.

Does it make a difference to you, the bettor? You have the same known starting information in both cases: A die under a cup, and a die that is identified as the one the peeker saw in order to determine that at least one die in the pair was a two".

What is unkown, is the randomness of the shaking beforehand. In the extreme Table #2 case, it is possible that the first die wasn't shaken at all. Since we know that one of the dice is a two before we place our bets, my contention is that it doesn't matter how we got to that point. The unknown now, is what is under the cup. There are only six possible outcomes, each equally likely.

Hi Aya. I can see that you are getting close to understanding this but you are apparently hung up on the 'How the devil can a six sided die have 11 possible outcomes. 11 is a stupid number and we all know the die has six sides. And heck, we can already see one die!!!'

Let's see if I can help by getting rid of that stupid number 11..

Before I do, I'll state again that the 'how we got here' is as important as the 'where we are' and I'll try to demontrate that.

Try this as a thought experiment. Even better, do it in real life. We'll need two dice (identical or different doesn't matter). We'll need a transparent GLASS cup and we'll need a camera and a printer. Real or imagined.

Set the dice out in each of the agreed, possible 11 outcomes with one die to the left and one die to the right.

Each time take two photos, first with no cup and then with the non-deuce covered. There's no 'non-deuce' when you have a pair of deuces, so you will take three photo's that time, one with no cup, one with the right die covered and one with the left die covered.

So.

You'll now have two sets of photo's, a set of 11 photos without cups and a set of 12 photo's with a cup showing.
!2 is a more sensible number to have in our head.

If the 'model' that we photographed is just the pair of dice, there were 11 different photo's. Lets call this set of photo's 'the naked set.'

If the 'model' that we photographed featured a glass cup, then there were 12 different photo's. Let's call this set of photo's the 'decent set'


Print off 'the decent set' of 12.
How many feature a pair of deuces?
2 out of 12. That is great, it's even intuitive.

But now someone cries 'Foul'

'You cheated. there's two pictures here where the dice didn't move at all, didn't change. The only difference is how you chose to place the cup. I'm disqualifying and tearing up one of those pictures because the question was about dice, not about cups.'

They disqualify and tear up the photo with the pair of deuces with the glass covering the right die. They had discretion which of two photo's to disqualify but what the heck.

So, your 2/12 now got whittled down to 1/11

Now, you might start to be a bit convinced, but not totally so. So try this.

You've got 11 photos each showing a revealed deuce and glass covered die. Each was equally likely. NONE has been eliminated.

With this whittled down 'decent set' of 11 photo's, have someone choose one at random, by any means you choose. EVERY PHOTO SHOWS AN UNCOVERED DEUCE.

What are the chances of them choosing a picture with a pair of deuces, with one deuce covered by a glass cup.

Yup. 1/11.
Because 2/12 was just cheating with the cups.

Still not sure. Do the set aside test.

Set the die on the left as a deuce. Do the whole ptotography malarky again.

You will only take exactly 6 photo's featuring a glass cup
You will only take exactly 6 photo's featuring no cup.

No-one cries foul and disqualifies any photo. It will always be 1/6, no doubt, no messing.

Does that help?

Quote: Ayecarumba

Where I am coming from is that as soon as the peeker identifies the die with the two on it, they are no longer both "random" The peeker has established that one die is a two, by observing that at least one die is a two; it no longer matters if it is the first or second die in the combinations 1-2, 2-2, 3-2, 4-2, 5-2, 5-2, 2-1, 2-3, 2-4, 2-5, 2-6, as five of those combinations (whether they be the first or second) have been eliminated. Only six combinations remain:
2 (the die the peeker noted) + 1,2,3,4,5 or 6.

I was about to repeat the same thing.

Basically as I said before. If you bet prior to seeing either die then it is 11, since you are shaking UNTIL you see at least 1 2, then you only care about the 2nd die.

Quote: OnceDear

Does that help?

that did not help me at all.

Quote: OnceDear

Before I do, I'll state again that the 'how we got here' is as important as the 'where we are'

I think this is where the sides are disagreeing. I don't think it is important at all. Since you are told the outcome of one, we only are figuring the math on the outcome of the 2nd.

At this point I don't think we will ever have an agreement as to what the right answer is.

Do the thought experiment

Quote: GWAE

that did not help me at all.

.....you are told the outcome of one...

NOOOOOOOOO. You are not 'told the outcome of one'. You are told something about the pair of dice.
DO the thought experiment. There IS only one right answer.
Don't trust me. DO THE EXPERIMENT. Prove it to yourself.

Quote: AlanMendelson

To be sure I understand:

If at least one of two dice shows a 2 the bet is on.

If the other die shows a 2, Alan is awarded 8 points.
If the other die does not show a 2, the Wizard is awarded 1 point.

Whoever is ahead by 25 points first wins.

If that is correct, then yes, I am happy to participate. But I want to do it in person and then have lunch.

This could actually be accomplished with the two of us observing a craps game at any casino. We don't even have to bet in the casino game or throw the dice. Personally, I think that would be best.

If you are only using 1 number to qualify the bet and saying the other die is going to pair or not. All other rolls without the prespecified qualifier don't count, then put me down as someone that would like to get 8 to 1 for the pair and I will gladly pay 1to 1 if it is not a pair of 2's or any other number for that matter.

What am I missing here? Never mind, got it now. Wow it matters how you get there.

I deleted because I couldn't see it at first. It does matter how you get there.

Quote: MaxPen

If you are only using 1 number to qualify the bet and saying the other die is going to pair or not.

. You are not.

Quote:

What am I missing here?

About 400 posts of explanation.

Quote: Ayecarumba

Where I am coming from is that as soon as the peeker identifies the die with the two on it, they are no longer both "random" The peeker has established that one die is a two, by observing that at least one die is a two; it no longer matters if it is the first or second die in the combinations 1-2, 2-2, 3-2, 4-2, 5-2, 5-2, 2-1, 2-3, 2-4, 2-5, 2-6, as five of those combinations (whether they be the first or second) have been eliminated. Only six combinations remain:
2 (the die the peeker noted) + 1,2,3,4,5 or 6.

The peeker didn't tell you WHICH die had the 2. That's where you guys get confused.

Quote: Ibeatyouraces

The peeker didn't tell you WHICH die had the 2. That's where you guys get confused.

Exactly. If we cannot identify which 5 options have been eliminated, we cannot call them eliminated.

Again, try this at home and you'll see.

Quote: Ibeatyouraces

Again, try this at home and you'll see.

Why do these doubters not do so???? That beats the hell out of me.
Even doing it for them as a spreadsheet, or a simple list doesn't cut it.

There's none so blind as those who will NOT 6l00dy LOOK

$:o)

Quote: MaxPen

I am willing to put up 200 and the person paying the 8to1 for the pair can quit whenever they have had enough.

I'll take that action!
Unfortunately, I won't be in Vegas until September for G2E...

Quote: OnceDear

Exactly. If we cannot identify which 5 options have been eliminated, we cannot call them eliminated.

Thank you for the detailed explanation. I agree that, as you describe, the odds are 1-11 of the pair. I also agree that how you get there is important. However, I disagree that the five options have not been eliminated under the conditions when the probability question is posed. My position is that it doesn't matter which 5 of the non-paired double options have been eliminated; but when the peeker announces the unchanging "two", one set is eliminated.

Does it matter if the dealer plucks out the "two" he peeked, then shakes the hidden die again? If yes, why? It was unknown before, and it is still just as unknown after the second shake. Maybe the answer to that question is what I need.

Quote: Ayecarumba

Thank you for the detailed explanation. I agree that, as you describe, the odds are 1-11 of the pair. I also agree that how you get there is important. However, I disagree that the five options have not been eliminated under the conditions when the probability question is posed. My position is that it doesn't matter which 5 of the non-paired double options have been eliminated; but when the peeker announces the unchanging "two", one set is eliminated.

Does it matter if the dealer plucks out the "two" he peeked, then shakes the hidden die again? If yes, why? It was unknown before, and it is still just as unknown after the second shake. Maybe the answer to that question is what I need.

If we do the cup scenario and one die is green and the other is red and as I peeked I state "one of them it's a 2". Now you have to guess which one it is. Red or green? Half the time you'll be wrong, so now what?

Quote: Ayecarumba

Does it matter if the dealer plucks out the "two" he peeked, then shakes the hidden die again? If yes, why? It was unknown before, and it is still just as unknown after the second shake. Maybe the answer to that question is what I need.

Yes, it matters.

The odds of rolling double twos, before you have rolled any dice, is 1/36.

The odds of rolling double twos, after you have rolled one two, is 1/6.

I'm trying to use that to demonstrate why re-rolling one of the dice after the fact changes the odds.

Quote: Ayecarumba

However, I disagree that the five options have not been eliminated under the conditions when the probability question is posed.

How, in the way the question is posed, does it eliminate any of the possibilities?

You've agreed that there are 11 possible ways to roll at least one two. (I think you've agreed to that)

I just said the question right there in a way - "I have peeked under the cup and have identified that the dice have landed in one of the 11 possible configurations in which there is at least one two"

does that eliminate 5 of the possibilities?

Does that say something different than "the dealer peeks under the cup and says that at least one of the dice is a two"

I'm trying to get back to: there are 11 possible ways to roll at least one two
and reconcile that with: I see at least one two
in a way that does, or does not, demonstrate that any of the 11 possibilities can be eliminated.

Quote: Ayecarumba

Does it matter if the dealer plucks out the "two" he peeked, then shakes the hidden die again? If yes, why? It was unknown before, and it is still just as unknown after the second shake. Maybe the answer to that question is what I need.

Yes it matters. Because now you are only talking about the probability of getting a two on a single die. Before, it was the probability of the second die being a two when you know that one of a pair of dice is a two. They are two different questions. The important piece of information to understand is that there are 11 ways to roll two dice such that at least one of them is a two. Identifying a two and pulling it out doesn't change that fact.

I am not sure how many different ways it can be explained, with pictures, charts, etc. Posters on these two threads have tried them all. I guess some people just can't get beyond "six sides on a die" no matter what. Please, just get two dice and do it for yourself. It should only take a few minutes to produce enough rolls containing a two. Or look up the Wikipedia article on Boy / Girl paradox that was quoted a few posts back and is essentially the same problem.

I'm currently using camstudio to record random dice throws, paying 8 taking one etc.... It's mind crushingly boring. Didn't help that first throw was 22
$:o)

Quote: OnceDear

I'm currently using camstudio to record random dice throws, paying 8 taking one etc.... It's mind crushingly boring. Didn't help that first throw was 22
$:o)

Guys like Alan and Singer would quit after that first roll :-) I wouldn't blame them lol.

Quote: OnceDear

I'm currently using camstudio to record random dice throws, paying 8 taking one etc.... It's mind crushingly boring. Didn't help that first throw was 22
$:o)

Tee Hee. it's a 168Meg AVI of mindcrushing tedium. All recorded in real time, warts and all.
Available as a download from my server upon request.

For the record,
I set https://www.random.org/dice/ in one half of my screen. That gives random throws of 2 dice, with real time timestamp.
In the other half of my screen I set an excel workbook where I painstakingly recorded the progress of banker starting with 200 and player starting with 200. Banker starts with 200
if 22 then banker paid 8
if at least one 2 showing, but not a pair, player paid 1
Started life with a 22 off the bat.
Player had a few lucky streaks and was occasionally ahead.
It dragged on and on.
and on and on
I kept recording till one of the pots was down to 175
Can you guess who won?

It's so real time, I haven't even played the video yet.



TEE HEE. Like I say. Real time, warts and all. If anyone ever does ask for the link, they'll see that I cock up the payouts a few times. BUT NEVER short change the player. at about roll 33, yod see me go back and fix up some cocked up banker pot changes. It's all live and real. no edit facilities. Even records me posting here in the middle of doing it.

Try it for yourself, ye doubters. OR PM me for a link.

Quote: Ayecarumba

Does it matter if the dealer plucks out the "two" he peeked, then shakes the hidden die again? If yes, why? It was unknown before, and it is still just as unknown after the second shake. Maybe the answer to that question is what I need.

It doesn't matter if the dealer ONLY plucks out the two that he sees. But if he rolls the other dice again, we are in the 'set one dice' scenario and the odds become 1/6

Quote: OnceDear

Tee Hee. it's a 168Meg AVI of mindcrushing tedium. All recorded in real time, warts and all.
Available as a download from my server upon request.

For the record,
I set https://www.random.org/dice/ in one half of my screen. That gives random throws of 2 dice, with real time timestamp.
In the other half of my screen I set an excel workbook where I painstakingly recorded the progress of banker starting with 200 and player starting with 200.
iff 22 then banker paid 8
if at least one 2 showing, but not a pair, player paid 1
Started life with a 22 off the bat.
Player had a few lucky streaks and was occasionally ahead.
It dragged on and on.
and on and on
I kept recording till one of the pots was down to 175
Can you guess who won?

It's so real time, I haven't even played the video yet.

Hehe, thanks OnceDear. Did it come out to about 1 pair for every 11 non-pairs?

Quote: Ayecarumba

Hehe, thanks OnceDear. Did it come out to about 1 pair for every 11 non-pairs?

You know,
I havent checked... Will do so now

Quote: DJTeddyBear

I'll take that action!
Unfortunately, I won't be in Vegas until September for G2E...

I see it now. Took awhile for me to grasp that it matters how one gets there. I will gladly buy you a lunch though, when you are in town.

Quote: MaxPen

Quote: DJTeddyBear

I'll take that action!
Unfortunately, I won't be in Vegas until September for G2E...

I see it now. Took awhile for me to grasp that it matters how one gets there. I will gladly buy you a lunch though, when you are in town.

Sigh, it was a good (pipe?) dream when it lasted.

Don't worry about the lunch. Just join the gang for the G2E meet-up / dinner followed by gambling. (Details should be announced sometime in September.)

Quote: Ibeatyouraces

The peeker didn't tell you WHICH die had the 2. That's where you guys get confused.

I am not confused all. If there are two dice, and at least one die shows a 2 then the question is what does the other die show?

2 -1 = 1

2(dice) - 1(die) = 1(die)

It doesn't matter which of the two dice that one die is. One die has six faces, numbered 1-6.

By the way, my lunch bet is with the Wizard.

I will gladly take a bet that pays 8 points for a 1/6 chance. That is a good bet.

Quote: AlanMendelson

I am not confused all. If there are two dice, and at least one die shows a 2 then the question is what does the other die show?

2 -1 = 1

2(dice) - 1(die) = 1(die)

It doesn't matter which of the two dice that one die is. One die has six faces, numbered 1-6.

By the way, my lunch bet is with the Wizard.

I will gladly take a bet that pays 8 points for a 1/6 chance. That is a good bet.

It makes a huge difference. The wizard has an 18.18% edge against you. You're pretty much playing a slot machine. If I thought different, I'd be betting on your side for huge dollars and I'm not.

Quote: Dalex64

The odds of rolling double twos, before you have rolled any dice, is 1/36.

The odds of rolling double twos, after you have rolled one two, is 1/6.

Amazing, you gave the answer to the question. Yet you don't understand you gave the answer to the question???

Look here's the problem: You have to answer the question as it is written. The question is written in a particular way. But you guys who say the answer is 1/11 are not answering the question that is written. You are adding your own interpretation that is not there in plain, simple English. Sorry. You fail in reading comprehension.

You guys are playing to 25 points. Alan needs 2-2 to show up 4 times. Wizard needs either 2-1, 1-2, 2-3, 3-2, 2-4, 4-2, 2-5, 5-2, 2-6 or 6-2 to show up 25 times in no particular order. I'll bet on the wiz!

Ahhhh damn. I am going to the other side I think.
The example about making a red and blue die did it for me. So if I am understanding it correctly it would work like this.

If you have a red and blue, roll until the red die is a 2, then odds on blue would be 1-6. So even though you know 1 die is a 2, it would matter that you have 1-11 ways to get to to 2-2.

I think the 2nd question said a 2 was on the felt. So now it is not possible to change the outcome of that die so it is 1-6.

My wording did not work out right but you know what I mean. It's hard to post on forum while driving.

Quote: GWAE

... It's hard to post on forum while driving.

I hope you're joking!

I think it best to restate the problems (as I have done in the original two-dice puzzle). I have modified the explanation of Table 1, and have split Table 2 into Tables 2A and 2B with more explanatory matter. I have left Table 3 as is. Assume the dealer always tells the truth.

Table 1: The dealer places two six-sided dice in a cup, shakes them, and then slams the cup down on the table, hiding the results. He must say either, "At least one of the dice is a two," or,"Neither die is a two." He then peeks under the cup, and states, "At least one of the dice is a two. Please place your bets. The only wagers available are 'Total 4' and 'Total not 4.'"

Table 2A: You approach the table and see that there is a die with the number two up, next to an upside down cup. The dealer says to you, "I shook 'em until at least one was a two, and then I removed a two from the cup. Under the cup is the other die. As you can see, at least one of the dice is a two. Please place your bets. The only wagers available are 'Total 4' and 'Total not 4.'"

Table 2B: You approach the table and see that there is a die with the number two up, next to an upside down cup. The dealer says to you, "I shook 'em until the die closer to me was a two, and then I removed it from the cup. Under the cup is the other die. As you can see, at least one of the dice is a two. Please place your bets. The only wagers available are 'Total 4' and 'Total not 4.'"

Table 3: You approach what appears to be a standard Craps table, and see that it is not Craps, but a game where the only wagers available are, "Total 4 (at least one die a two)" and "Total not 4 (at least one die a two)". The felt also states, "All rolls without at least one die a two 'en prison'". The dealer explains that this means that any roll without at least one die a two will be "locked up" until a roll with at least one die a two. She asks you to "Please place your bets". You make a wager. The stick then dumps a bowl with five dice and pushes them to you so that you can pick two and shoot em.

Quote: Ibeatyouraces

I hope you're joking!

Well kinda. I wasn't actually driving, I was at a red light. I put phone down when it changed.

Quote: AlanMendelson

Amazing, you gave the answer to the question. Yet you don't understand you gave the answer to the question???

Look here's the problem: You have to answer the question as it is written. The question is written in a particular way. But you guys who say the answer is 1/11 are not answering the question that is written. You are adding your own interpretation that is not there in plain, simple English. Sorry. You fail in reading comprehension.

I disagree.

Have you tried it yet?

Quote: GWAE

Ahhhh damn. I am going to the other side I think.
The example about making a red and blue die did it for me. So if I am understanding it correctly it would work like this.

If you have a red and blue, roll until the red die is a 2, then odds on blue would be 1-6. So even though you know 1 die is a 2, it would matter that you have 1-11 ways to get to to 2-2.

I think the 2nd question said a 2 was on the felt. So now it is not possible to change the outcome of that die so it is 1-6.

It matters why the one 2 is on the felt. Both dice are rolled at the same time. I was ready to throw money at this, until that sunk in. If the 2 was placed and then the second die was rolled it would be 1 in 6. The Wizard has a tricky prop with this one. :-)

Edited below.

Quote: MaxPen

It matters why the one 2 is on the felt. Both dice are rolled at the same time. I was ready to throw money at this, until that sunk in. If the 2 was placed and then the second die was rolled it would be 1 in 6. The Wizard has a tricky prop with this one. :-)

We've laid out every possible combination a 2 can show up. Somehow some people just can't see this. The cup, peeker and anything else is irrelevant.

RE: SDSDNSR probability quiz.

Quote: GWAE

If you have a red and blue, roll until the red die is a 2, then odds on blue would be 1-6.

Stop right there. That's it.

1/11 refers to the combinations of dice with at least one 2 showing. That's a different question and answer.

Once one die is known to be a 2, the odds of the second die are always 1/6.

GWAE don't make the mistake of using "Answer B" (1/11) for "Question A" (on one die what is the chance that one face in 6 will show?).

Do you think "the dealer peeks under the cup and says at least one of the dice is a two" and "the dealer peeks under the cup and states that he sees one of the 11 possible combinations of dice with at least one two showing" mean different things?

In both cases, isn't one of the dice known to be a two?

It does not make sense intuitively. But there is a difference between:

1 - Dealer/peaker only says something when a "2" is present, and says nothing when neither die is a 2.

2 - Dealer/peaker says something every time, like "there is at least one ___ present" (could be 1, 2, 3, 4, 5, OR 6).


#1 - there is a 1/11 chance that both dice are a 2 / the same.

#2 - there is a 1/6 chance that both dice are the same.



You can make Venn diagrams, do simulations on the computer or in real life, grab a set of dice and look at each way they can be arranged, you can do a proof or work it out logically or however you want to solve it. But in the end, #1 is 1/11 and #2 is 1/6.




Dalex:

- What matters is what happens when the dealer/peeker does not see a 2 on neither dice (or he sees one of the 25 non-deuce combinations). THAT is what matters. Weird.

- No, you do not know that "ONE" of the dice is a 2. You know that "ONE OR THE OTHER" dice is a 2.

I think I'm starting to understand Alan's and others thinking. Let me ask this to see if I'm right?

A) You roll the dice and at least one ends up a 2. What is the probability that the "other" is a 2?
B) You roll the dice and at least one ends up a 2. What is the probability that "both" are a 2?


,

Quote: Ibeatyouraces

I think I'm starting to understand Alan's and others thinking. Let me ask this to see if I'm right?

A) You roll the dice and at least one ends up a 2. What is the probability that the "other" is a 2?
B) You roll the dice and at least one ends up a 2. What is the probability that "both" are a 2?


,

Both are the same thing.

The (dis)ambiguity comes from what would happen if neither die (dice?) were a deuce. Or rather, the full context of the dice roll.

You need it in context to figure out what it's asking (or what the true answer is).

Quote: RS

Both are the same thing.

The (dis)ambiguity comes from what would happen if neither die (dice?) were a deuce. Or rather, the full context of the dice roll.

You need it in context to figure out what it's asking (or what the true answer is).

I know they are the same, but some people do interpret them differently.

Quote: RS

...
2 - Dealer/peaker says something every time, like "there is at least one ___ present" (could be 1, 2, 3, 4, 5, OR 6).
...
#2 - there is a 1/6 chance that both dice are the same.
...

Nope. That is 1/11 also. The announced pip-count doesn't matter.
If one of the dice is an "X" (that's either one), then it's 1/11 that the partner die is also an "X." We've discussed deuces until we're blue in the face, but it's the exact same scenario for any pip-count that could be called.

I wonder how many times Alan retook this shot???

http://youtu.be/3jDKXRhgWyA

Quote: indignant99

Nope. That is 1/11 also. The announced pip-count doesn't matter.
If one of the dice is an "X" (that's either one), then it's 1/11 that the partner die is also an "X." We've discussed deuces until we're blue in the face, but it's the exact same scenario for any pip-count that could be called.

That's a tricky one indeed. And I hate to say it Indi, but you are wrong this time.

If the announcer announces every time and always makes an assertion 'At least one of the dice is a [x]' where x is a value 1-6 chosen to be just enough for the assertion to be true, then we have a new problem: That assertion did not NEED both dice to be peeked at.

The Banker had considerable discretion and may have based his assertion on the first die he saw. We cannot discount that possibility.

The problem then effectively becomes 'what are the chances that the dice formed a pair', and that is 6/36 = 1/6.

What's happened is that a number of throws which would have been 'push' have now come into play every time.

Quote: indignant99

Nope. That is 1/11 also. The announced pip-count doesn't matter.
If one of the dice is an "X" (that's either one), then it's 1/11 that the partner die is also an "X." We've discussed deuces until we're blue in the face, but it's the exact same scenario for any pip-count that could be called.

I disagree. Perhaps I'm wrong (I don't think I am).

There are 36 ways to roll a pair of dice, 6 of which are doubles. 1/6 of the rolls will be doubles. If I roll a pair of dice 36 times (and get a perfectly even distribution), 30 times the dice will not match. 6 times the dice will match.

It doesn't matter which die is picked -- the other will have a 1 in 6 chance of being the same as the picked/called die.


EDIT: This is different than the "only count a roll with 'X' (specifically 1, 2, 3, 4, 5, or 6) in it, disregard everything else). In the "only count rolls with an 'X'" scenario, you're limiting yourself to only 11 possibilities. If you count all rolls and pick either die at random to be the "at least one X is showing", then the chance of both dice (or 'the other die') being the same would be 1/6.

Double Edit: Anyone who disagrees: I'd be willing to make a wager (or a whole bunch of wagers for this scenario) if you find yourself in Las Vegas at an 8:1 payoff.

I still disagree with you, OnceDear.
Regardless whether ONE die were viewed, or BOTH, an announcement is made. The announcement is "X." And now, FIVE possible pairs are ruled out. What has not been ruled out...

Left=X X=Right
Left=X non-X=Right
Left=non-X X=Right

I will surrender, IF you say that it's always and only the LEFT die viewed.