The Wizard will bank this bet: 1/6 vs 1/11

[Jihkro]

If you are clearly able to distinguish between the "spinner" and the not spinner, then yes the probability will be 1/6. However, how could you have "chosen" one to be frozen and one to be a "spinner". What if the "frozen" die is not a two and the "spinner" eventually rests on a two? Why would the verifier not have stated "there is at least one two" in that situation. It is as I said in my previous post, you have somehow arbitrarily changed the wording of the question to be "Given that the *red* die lands on a two, what is the probability that both are a two", or "Given that the leftmost die lands on a two, what is the probability that both are a two", or in your wording "Given that only one die is stationary at the time and is a two, what is the probability that after the second comes to rest that both are a two". In each of these alternate wordings, there is some clear distinction being made between the dice, and you are ignoring the possibility that the second die could be a two with the first one not, and still be considered as a scenario where the verifier truthfully states "at least one die is a two."

In the interpretation the 1/11'ers are using, which appears to be the original wording of the question, is that "both dice are tossed" (and both come to rest), and the verifier sees the results of both dice before stating "at least one die is a two."

The two are very different scenarios, and have different probabilities (as explained in my previous post).

[Alan Mendelson]

What if the "frozen" die is not a two and the "spinner" eventually rests on a two?

In which case there is no bet.

[arcimede$]

Is your reading comprehension that bad Arc? It appears so. Go ahead shoot the video with one die fixed as a two and show me your proof that there are eleven other possible faces of which 2-2 appears 1/11 times. Go ahead and do it. Yeah, do it just like the Wizard did it.

There's still nothing in the question about one die "fixed as a 2". All we know is at least one die is a 2. No more, no less. Why do you keep trying to change the question? Why don't you want to test the situation and see what happens? I provided you a simple way to test it. If you want to prove all of us 1/11's wrong, do what I proposed.

I have a feeling Alan now knows he is wrong and simply won't admit it.

[Zedd]

Jihkro thanks for joining and for posting. Unfortunately everything you wrote above has been discussed before.

Our difference is over the wording of the question.

Those of us who say the answer is 1/6 believe that the question simulates the following:

Two dice are thrown on a craps table and one comes to rest on the 2 face. The second die spins like a top. What are the chances that the "spinner" will land on a 2?

We -- the 1/6ers -- see nothing to indicate that given two dice with one frozen to show a 2, could possibly result in 11 faces to be considered.

For the millionth time, the answer to that question is indeed 1/6. No one is arguing against that. Jihkro even explained that in his post (by the way, thanks for jumping in Jihkro).

Alan (and other 1/6ers)…you are failing to realize that the ‘spinner’ question is different from the bet you had described on the very first post of this very thread.

[Zedd]

In which case there is no bet.

Not true.

To quote the first post…

A second person peeks under the cup and truthfully announces if there is at least one deuce showing or not (he will later lift the cup and show to all he wasn't lying).

If "at least one deuce" is announced, the 1/6-ers will wager 1 unit on the premise that there are two deuces under the cup

So if the frozen die is not a deuce and the ‘spinner’ eventually rests on a duece…that DOES meet the condition of “at least one deuce”. In which case, the bet WOULD be on.

[Alan Mendelson]

Whoa Zedd you are mistaken.

If the die that came to a rest is not a 2 and there is still a spinner, the "peeker" would not see a 2.

You see -- this is all about reading comprehension and understanding the question.

The original question IS the spinner situation at a craps table. We know that at least one die is a 2 and the question is the 6 faces on the other die which either is spinning or is yet to be revealed. That's why it's 1/6. The question gives that information.

[RS__]

Ironic how Alan claims others have poor reading comprehension....yet he's discussing something about a die being frozen/set to a 2.

[Alan Mendelson]

Ironic how Alan claims others have poor reading comprehension....yet he's discussing something about a die being frozen/set to a 2.

Because we are told there is a die showing a 2.

[Zedd]

Whoa Zedd you are mistaken.

If the die that came to a rest is not a 2 and there is still a spinner, the "peeker" would not see a 2.

You see -- this is all about reading comprehension and understanding the question.

The original question IS the spinner situation at a craps table. We know that at least one die is a 2 and the question is the 6 faces on the other die which either is spinning or is yet to be revealed. That's why it's 1/6. The question gives that information.

Then the peeker will wait until the spinner has come to a rest. It was agreed that the peeker would be able to see both dice before making the “at least one deuce” announcement.

[Jihkro]

shake and slam the cup on the table.

I'm at a loss as to physically how one die could be "spinning" in the scenario described. Shaken and Slammed. Do dice spin after being slammed? Are both dice at rest normally after a cup is slammed opening down? In the realm of reasonable physics, even if in the split second between slamming the cup and lifting the cup to peek, one (or both) is spinning, if one comes to a stop only milliseconds sooner than the other, could the verifier's mind recognize which came to rest first? How would he decide which was the "spinner" in that case?

The sentence "The one that came to rest first is a two" is the same as "The red die is a two" is the same as "The die furthest to the left is a two" is the same as "The die we rolled (versus the die we spun) is a two" is the same as "The die in vancouver (versus the die in Bangkok) is a two" is the same as...

All of these are different than the following equivalent list:

"There is at least one two (be it the red or the green)" is the same as "There is at least one two (be it the stationary or the spinner's eventual result)" is the same as "There is at least one two (with both dice stationary at the time of announcing the fact)" is the same as "I slammed dice on the table and looked and saw at least one two."

(he will later lift the cup and show to all he wasn't lying).

What happens if the spinner has stopped spinning after he has made his announcement but before lifting the cup and the spinner has come to rest on a two? He will have lifted the cup and the audience/betters see that there is a two and a something else. To the audience's eyes, they won't know that the spinner was the two, and they will then conclude that the person was lying.

[Alan Mendelson]

Then the peeker will wait until the spinner has come to a rest. It was agreed that the peeker would be able to see both dice before making the “at least one deuce” announcement.

You're missing the point. The spinner situation is an analogy for the question. It is not the actual question.

In fact, if this happened at an actual craps table the peeker would wait for the spinner to come to a rest. And then if the spinner came to a rest showing a 2 what then? Well, the question would then center on where the first die came to a rest -- and that would be 1/6, wouldn't it?

Of course it would be -- you said that yourself. When one die is known as a 2 the situation rests with the other die's six faces.

This is why those of us who say the answer is 1/6 just can't figure out how you guys continue to say the answer is 1/11 ??

[Alan Mendelson]

I'm at a loss as to physically how one die could be "spinning" in the scenario described.

Read above. The spinner is an analogy.

[Jihkro]

Read above yourself. You chose to focus on a single part of my statement to which you can respond (rather poorly I might add), and have not responded to any of the rest of the questions or comments I made in my latest post. You continue to work in the situation that the dice are somehow distinguishable (either by color, by spinning status, by location, or by time at which they stop spinning) and declare that the peeker will choose to completely ignore the result of one die (even if it is a two) and only announce the result of the other die.

Again, "What happens if the spinner has stopped spinning after he has made his announcement but before lifting the cup and the spinner has come to rest on a two? He will have lifted the cup and the audience/betters see that there is a two and a something else. To the audience's eyes, they won't know that the spinner was the two, and they will then conclude that the person was lying." There *is* a 1/6 chance that if the peeker was cheating by ignoring the result of the die that stopped second (or equivalent wording) that he will be caught in the act. If I were officiating the proceedings, I would conclude that the peeker was either paid off by the 1/6'er to increase his/her odds by giving an unfair advantage via additional information, or that he was paid off by the 1/11'er in order to make it appear that way. In either case, I would render all previous and future bets cancelled and all money returned until such a time as we could find a fair judge.

"When one (specific) die is known as a two" then yes, the probability is 1/6.

"When *at least one die* is known as a two (nonspecific as to *which*)" then the probability is 1/11.

Toasting in epic bread and feeding the trolls. I don't really care if I'm feeding the trolls. I'm just curious how you refute those statements. Oh, right, by vague responses "its an analogy" and by not directly responding to any specific logical statements.

At the very least respond to the question "Do you understand that if both dice are tossed and both dice are viewed and if the total number of twos shown is either 1 or 2 (regardless of which die shows the two if only one) that the probability is 1/11?" completely disregarding the scenario described in the analogy as having a "spinner."

Whether or not you choose not to respond, I recommend you read http://en.wikipedia.org/wiki/Boy_or_Girl_paradox as this problem is exactly the same as that one. In the boy/girl paradox the full statement reads:

"Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?" The answer to this question is 1/2

"Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" The answer to this question is 1/3

The question you think is being asked is equivalent to the first, "the older child is a girl" is like saying "the stationary die is a two"

The question that is actually being asked is equivalent to the second, "at least one of (disregarding whether it is the older or younger or both) the children is a boy"

Included in the wiki article (here is the link again: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox please read it) is a full explanation of both the ambiguity and the solution, as well as references to textbooks, research studies, and several more similar examples.

[Alan Mendelson]

Jihkro I really don't care to read your stuff. We've heard it all before. Just show me a video with two dice: one showing a 2 and tell me how you arrived at 1/11 without touching the die showing a 2.

thanks for stopping by.

[indignant99]

Actually, it's not wrong. It's the same thing we've been arguing about. One die is known to be a two, and the question is what is showing on the six sides of the second die.

Here is the assisting graphic that you're needing.
OneHitWonder can provide you an excellent explanation.

[onyx99]

Alan, you seem dug in on the idea that if the probability is 1/11, then someone should be able to show eleven equally likely outcomes using a single die. And no one can... but this is a red herring. Once you're told that one of the dice is a 2, the six possible outcomes of a single die are no longer equally likely.

I can understand Alan's intuition and common sense here. After all, the information you're being given is just that there's at least one 2. That's the exact *same* information you'd have if the 2 were set aside in the first place and only one die were rolled under the cup. So how can the probabilities be different in the two situations?? But this (like the Monty Hall problem) is one of those situations where what plain common sense tells you is wrong, and you need to do some work to understand how.

Here are a sequence of simple statements that are all true, and where each one (to me) follows pretty directly from the previous one.

1. You roll two dice in plain view. Getting an 11 is twice as likely as getting a 12.
2. Your friend rolls two dice under a cup. It is twice as likely that there's an 11 under the cup as a 12.
3. Your always-honest friend rolls two dice under a cup and peeks. You ask, "Did you get at least 11?" He says, "yes". Only 11 and 12 are now possible, and it's still twice as likely that there's an 11 under the cup as a 12.
4. He rolls and peeks. You ask, "Is it true that one of the dice is a 6 and the other is at least a 5?" He says, "yes". Only 11 (5&6) and 12 (6&6) are now possible; 11 is still twice as likely as 12. (So the probability it's a 12 is 1/3.)
5. Same as the previous story, except you then say, "Show me a 6." He moves the cup to reveal a 6. The remaining die must be either a 5 (if the roll was 11) or a 6 (if the roll was 12), but these aren't equally likely: 5 is twice as likely as 6, and the probability there's a 6 under the cup is 1/3.
6. He rolls and peeks. You say, "If it's true that one of the dice is a 2, show me a 2." He moves the cup to reveal a 2. The remaining die can be 1-6, but these aren't equally likely: 1 and 3-6 are each twice as likely as 2, so the probability there's a 2 under the cup is 1/11.

If you accept the first statement and reject the last, then there's some transition in there that you don't accept. Which one, and why? To me, the fourth statement is okay, and the fifth is counterintuitive... but put side by side, they are clearly equivalent statements. That particular transition helps me get past the (erroneous) common-sense feeling that the *information* should be enough to determine the *probabilities*. Maybe you get tripped up somewhere else.

[OneHitWonder]

Kewl you wrote "it's done before the roll." That alters reality.

The original problem specifically states that two dice were rolled and at least one is showing a 2. Whatever you thought prior to that no longer matters. Now you have to deal with at least one die showing a two. The question now centers on the second die in the problem which may or may not be a two and the chances for that second die to be a 2 is 1/6.

By figuring probability before the roll alters the reality of the question. You are changing the facts and the conditions of the problem.

As I have said a hundred times, your 1/11 is "good math" but for a different question -- perhaps for a question that asks: "how many different combinations of two dice have at least one 2 and of those combinations how many would show 2-2?" That would be your 1/11 answer.

This is pretty good Alan.

The only change I would make is to substitute "how many different combinations of two dice involve the number 2 and of those..." for "how many different combinations of two dice have at least one 2 and of those...". Exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets.

Alan, you did a good thing by pointing out that this problem isn't written for the 1/11 chance answer in theory. The original problem supplies only a specific roll, or rolls of a specific nature; and nothing has been restricted about the number 2 beyond the "peeker's" specific observation.

To view also a specific roll theoretically - in the probability or "how often" sense (given that the combinations of two dice involve the number 2) - ask how often will the left die involve a 2. Half the time; the other half of the time the right die will involve a 2. Hence, half the time the other die will show a 2 given the number 2 somewhere. When it's the left die, the other die will show a 2 with 1/6 chance; and, when it's the right die, the other die will show a 2 with 1/6 chance. Now, we put this into a calculation as did the 1/11 chance answerers. There are two parts to this calculation:

(1/2 X 1/6) + (1/2 X 1/6) = 1/12 + 1/12 = 1/6.

This is the specific roll or rolls theoretical counterpart to the 1/11 chance answer. The way to perform the calculation if going by a specific roll or rolls in theory. All specific rolls considered, in theory, in terms of which side the roll or rolls involve the number 2.

This calculation may be done also within the 1/11 chance calculation:

(5/11 X 1/6) + (5/11 X 1/6) + {(1/22 X 1/6) + (1/22 X 1/6)} = 1/6.
___________________________________________

Note that the only way to arrive at the 1/6 chance answer in and of itself for a specific roll or rolls is to find a 2 on the only die looked at. This is the exact manner in which the original problem is written.

Again, exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets. The reason to require "one die or the other is a 2" - the same as involving the number 2 - instead of "at least one two". "One die or the other is a 2" would allow for the specific roll or rolls to read the same as the general roll or rolls, though waiting on a 2 would still be a requirement in advance. This is not the original problem as written.

[onyx99]

One more thing... one might argue that you *do* have extra information in the case where a 2 is set aside ("frozen") from the start: namely, you know *which* die is a guaranteed 2. Perhaps that extra information changes the probability to 1/6. This is sort of true, and bears on the business with spinners and so on; but it's also vague and a little misleading, in that it reinforces the misconception that "same information = same probabilities". Let there be a red die and a green die, and tell your friend to show you a 2 if there is one (with no color bias), and let your friend reveal a red 2. You now have *exactly* the same definitive information ("the red die is a 2") as you would if the red die were set aside as a 2 and only the green die were rolled. And yet (since the addition of the colors is purely cosmetic), the probability that the green die is a 2, in this situation, remains 1/11.

I find this scenario, if anything, even more counterintuitive than the original one. The assumption of no color bias is key here. If your friend shows you the green whenever he can, the probability that the green die is a 2 given that he showed you the red die is *zero*. If he shows you the red die whenever he can, then the probability that the green die is a 2 given that he showed you the red die is 1/6. This extra complexity is swept under the rug... and the no-bias condition is enforced... when the dice are indistinguishable.

[OneHitWonder]

3. Your always-honest friend rolls two dice under a cup and peeks. You ask, "Did you get at least 11?" He says, "yes". Only 11 and 12 are now possible.

No, one of either the 11 or the 12 is no longer possible.

The "peeker" saw either the 11 or the 12. To be honest, if the "peeker" saw the 11, then he can not say or the 12. To be honest, if the "peeker" saw the 12, then he can not say or the 11.

He may say that what he saw is an element of the set defined by at least the 11. But, he may not honestly say that the specific roll or element of the set defined by at least the the 11 which he saw is that set. Can't roll or get a set of rolls on one roll. Unless the definition of the set is specially worded to work for both the general and the specific.

[OneHitWonder]

One more thing... one might argue that you *do* have extra information in the case where a 2 is set aside ("frozen") from the start: namely, you know *which* die is a guaranteed 2. Perhaps that extra information changes the probability to 1/6. This is sort of true, and bears on the business with spinners and so on; but it's also vague and a little misleading, in that it reinforces the misconception that "same information = same probabilities". Let there be a red die and a green die, and tell your friend to show you a 2 if there is one (with no color bias), and let your friend reveal a red 2. You now have *exactly* the same definitive information ("the red die is a 2") as you would if the red die were set aside as a 2 and only the green die were rolled. And yet (since the addition of the colors is purely cosmetic), the probability that the green die is a 2, in this situation, remains 1/11.

I find this scenario, if anything, even more counterintuitive than the original one. The assumption of no color bias is key here. If your friend shows you the green whenever he can, the probability that the green die is a 2 given that he showed you the red die is *zero*. If he shows you the red die whenever he can, then the probability that the green die is a 2 given that he showed you the red die is 1/6. This extra complexity is swept under the rug... and the no-bias condition is enforced... when the dice are indistinguishable.

You will have to spend a few days thinking about this, and elaborating more carefully.

It's not fair to the rest of us who have already put in the time on this to have to straighten all of that out.

Thanks.