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

[onyx99]

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.

I don't mean to be rude, but is English your native language? The definition of the set is, in fact, worded to work for both the general and the specific. The question "did you get at least 11?" (assuming 'get' is understood to mean 'generate by rolling the dice', and 'at least 11' is understood to mean 'a configuration of the dice with at least 11 total pips on the top faces') is a straightforward yes-or-no question... no tricks. The set defined by 'at least 11', intersected with the set of possible rolls, is { 11, 12 }. And saying "I got at least 11" is the same as saying "I got an element of the set defined by 'at least 11'," with the advantage that it's how people actually talk.

[Zedd]

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.

I believe you’re one of the first 1/6ers to attempt actual math. Too bad you’re wrong.

If it is given that one or both of the dice shows a deuce, then 6/11 times the left die will be a deuce. Likewise, 6/11 times the right die will be a deuce.

Then (to kind of follow your approach), when the left die is a deuce the other die will not show a 2 with 5/6 chance; and when it is the right die showing a deuce, the other die will not show a 2 with 5/6 chance. Then the calculation for NOT getting a pair of deuces:

(6/11 * 5/6) + (6/11 * 5/6) = 10/11

Therefore, the probability to get a pair of deuces:

1 – (10/11) = 1/11

[OneHitWonder]

I don't mean to be rude, but is English your native language? The definition of the set is, in fact, worded to work for both the general and the specific. The question "did you get at least 11?" (assuming 'get' is understood to mean 'generate by rolling the dice', and 'at least 11' is understood to mean 'a configuration of the dice with at least 11 total pips on the top faces') is a straightforward yes-or-no question... no tricks. The set defined by 'at least 11', intersected with the set of possible rolls, is { 11, 12 }.

I certainly don't dispute that "at least 11" is { 11, 12 }.

And saying "I got at least 11" is the same as saying "I got an element of the set defined by 'at least 11'," with the advantage that it's how people actually talk.

Then, people talk deceptively, or in ignorance. Claiming "one or more prizes" in a lottery of a single known prize is wrong. One might attach EV's to the "or more" part, and discover a constant deficiency in winnings.

[OneHitWonder]

Too bad you’re wrong.

Not so fast.

If it is given that one or both of the dice shows a deuce, then 6/11 times the left die will be a deuce. Likewise, 6/11 times the right die will be a deuce.

Yes, so I must have meant when the "peeker" goes from the left and the right sides at the same time. 6/11 = 6/11, so half the time from the left, and half the time from the right.

Then (to kind of follow your approach), when the left die is a deuce the other die will not show a 2 with 5/6 chance; and when it is the right die showing a deuce, the other die will not show a 2 with 5/6 chance. Then the calculation for NOT getting a pair of deuces:

(6/11 * 5/6) + (6/11 * 5/6) = 10/11

Therefore, the probability to get a pair of deuces:

1 – (10/11) = 1/11

Depends how you look at it. In any event, the 1/11 is is made up of twice the basic 1/6 answer.

Five-to-one money odds paid twice on a roll of 2-2 is the same as ten-to-one money odds paid once on a 2-2. And, ten losses of 1 unit is twice five losses of 1 unit on the other rolls.

[OneHitWonder]

Okay. This is about specific rolls, whether summed by probability.

The double 1/6 chance game of what to do when 2-2 has been rolled is like the 1/11 chance answer chart. The single game or roll can do also the double game of the 2-2's roll; but the double game can not do the singles' games (by coming apart while remaining the double game).

Specific rolls happen consecutively. You have to work with that. And, in practice, it's as meaningless to do row and column of chart together as it is to speak of all the rolls of one 2 and two 2's.

[onyx99]

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.

This doesn't take a few days to figure out. As many have pointed out, it's an elementary exercise in a first-year probability course to show that the answer is 1/11. It might take a few days to convince you, or Alan, that you're wrong... but then again, it might take forever. I'm not planning to spend forever. I wrote down what I think is a logical progression that, if followed, may help you find the spot where your intuition is leading you astray. But you have to be *trying* to find that spot. I can't understand it for you.

That being said, if you're still convinced you've got the right end of this particular stick, all I can say is... wanna bet?

[Alan Mendelson]

Please... all of you who say the answer is 1/11:

Show me your video with at least one die with the 2 face up and how you can count 11 possible faces for consideration. You can use your cell phone and post on YouTube. It is free to do.

Have two dice on a table and with either die showing a 2 (remember the original problem says at least one die is a 2) count for me the 11 faces that fit your 1/11 answer. Remember, in the real world when a die settles on a 2 you cannot change the value or face on that die.

Show me the 11 faces. Now go to work.

[onyx99]

Have two dice on a table and with either die showing a 2 (remember the original problem says at least one die is a 2) count for me the 11 faces that fit your 1/11 answer. Remember, in the real world when a die settles on a 2 you cannot change the value or face on that die.

Show me the 11 faces. Now go to work.

There aren't 11 different faces for the hidden die to be showing, of course. There are six. That tells you nothing AT ALL about the probability of each face unless you assume they're all equally likely. (Spoiler: they're not.) Here's the million-dollar question: WHY do you think you can assume that?

[Alan Mendelson]

There aren't 11 different faces for the hidden die to be showing, of course. There are six.

Exactly. And that's the answer to the original question. The original question is not asking for your "probability gymnastics." Sure the word is used -- but it's not used in a strict context.

I get it. Regnis gets it. Redeitz gets it. Singer gets it. A dozen of my close and intimate friends on Facebook get it. Only your "math guys" don't get it. You see the word probability and you lunge for your textbooks. You can't see the forest for the trees.

As I said early on over on the WOV forum you guys "overthought" the question. In reality it is a simple math question in which you look at one die and count the faces on one die. No graphs, charts, spreadsheets, advanced theory is needed.

And the only way to justify your graphs, charts, spreadsheets, probability math, and so forth is by altering the main condition of the problem which is that you have at least one die already showing a 2 and that die should remain showing a 2.

You guys dug the Panama Canal when all you needed was a drainage ditch.

[Jihkro]

How about a thirty-six sided die? Rolling one thirty-six sided die is the exact same probabilistically as rolling two six-sided dice. We can show you all eleven faces of the thirty-six sided die without needing to touch.

If the thirty-six sided die has faces labeled (1,1), (1,2), (1,3), ... (1,6),(2,1), (2,2), ..., (3,1),...(6,1),...(6,6), then the eleven faces are those with either a 2 in the first entry or a 2 in the second entry. The probability of this thirty-six sided die landing on a specific face (say for example 2,2) is exactly the same as the probability of two distinct dice (with a clearly defined order) (say for example one red die and one green die) with the first distinct die (say for example the red die) corresponding to the first number and the second die (the green) corresponding to the second number.

If you complain about the fact that the thirty-six sided die doesn't have faces in a more usual labeling, I can remind you that there are several dice used in several common games which use nontraditional face labels, such as backgammon doubling dice, poker dice, etc. If you absolutely insist that the dice faces must be 1,2,...,36 that is fine too. Let the number on the d36 be x. Then x can correspond to a red d6 value of floor((x-1)/6)+1, and a green d6 value of ((x-1)mod6) +1. I.e. exactly the labeling described earlier (1,1), (1,2), (1,3),...

Roll the d36. If it lands on any of the faces (1,2),(2,1), (2,2), (2,3),...(3,2),(4,2),...,(6,2), then the bet is on as "there is at least one two." As each and every one of the 11 faces was equally likely, being told that the d36 landed on one of those 11 faces, the probability that it is in fact the face labeled (2,2) is... you guessed what I'd say... 1/11.

In using a d36, there is absolutely no way to have one "stationary" and one "spinner."



How do you refute this argument? In what way is the throwing of a d36 as opposed to two d6's somehow different? (its not for the record)

---
also... "and you lunge for your textbooks"... implying that we don't have such elementary math memorized already...

[regnis]

Ok---my final post on this subject. I roll a 2. What are the odds it becomes a 6 so that we can satisfy the 1-11ers?

I set my die on 2 one thousand times and it hasn't changed even once. Is my sample size too small??

I set one die on 2 another 1000 times. Still no change.

Let me know when my sample size is enough.

[regnis]

Wait--while typing this I knocked it over. Now it is a 5.

Now I get it.

[regnis]

Go Hawks!!!!!

[arcimede$]

Please... all of you who say the answer is 1/11:

Show me your video with at least one die with the 2 face up and how you can count 11 possible faces for consideration. You can use your cell phone and post on YouTube. It is free to do.

Have two dice on a table and with either die showing a 2 (remember the original problem says at least one die is a 2) count for me the 11 faces that fit your 1/11 answer. Remember, in the real world when a die settles on a 2 you cannot change the value or face on that die.

Show me the 11 faces. Now go to work.

Have you run the simple test yet? At most a half hour of your time and you can prove us all wrong. What are you waiting for?

[Alan Mendelson]

Arc, I ran the test. I have one die showing a 2. I have another die and 1 out of six times when I roll it, it also shows a 2.

[Rob.Singer]

What I enjoy seeing is the math guys getting so frustrated over Alan's repeated very simple common sense request that this collection of geniuses put their swollen heads together with just a little of that AP/+EV cash they so often theorize about, and come up with that video where at least one of the two dies shows a 2. Real world reality guys....stop stalling. Even EvenBob could do it while he's sitting around picking his toes. So enuf with the game of dodge ball. SHOW us what you're made of and stop stop talking probabilities and expectations. It's time you all graduated from that world of make believe.

[RS__]

It's time you all graduated from that world of make believe.

Ironic you say this.

[arcimede$]

Arc, I ran the test. I have one die showing a 2. I have another die and 1 out of six times when I roll it, it also shows a 2.

Alan, that is not what the question asks. There are only two conditions in question

"You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?"

So, once again. Throw the dice, mark one row when the first condition is met, mark a second row when the second condition is met. When you've run 100+ tests then divide the first row by the secon row and tell us what you get. Couldn't be simpler. The only possible reason you are avoiding the test is you already know you are wrong.

[Rob.Singer]

Ironic you say this.

More ironic appears to be how you (and arci, etc. etc) keep dodging the request for a simple video by--this time by deflecting. Alan has been asked to perform a test, and we know that will result in a 1in11 conclusion. However, he's asked you ego-driven, self-proclaimed know-it-all's to produce a simple video that supports the much more mundane & comprehensive interpretation of the question long ago and many times, and you can't even figure out how to do that.

Theory: don't order it from any menu, and don't try to deposit +EV phantom bucks into any bank.

[Jihkro]

Since the analytical proof and our "EV phantom bucks" arguments aren't working, here is 20 minutes of me rolling the dice moving chips into or out of the box with the payouts as described.



I started with 50 blue chips in the box and 50 red chips on the table. If exactly two deuces rolled, I moved eight chips out of the box into the pile. If exactly one deuce rolled, I moved one chip into the box from the pile. If no deuces rolled, I did not move any chips. I intended to and was perfectly willing to continue the video until either all red chips were put away (the player was 50 units "in the red") or until all blue chips were in the pile (the player was 50 units "in the green"). I.e. a 9-1 payout. My phone ran out of storage in the process, but in the effort of keeping this as a single take (to avoid argument that I only wanted to show video that supports my argument), I'm uploading anyways.

If you don't wish to watch the video and hear me talk, the recording stopped at the player being 27 units in the red with that payout rate. I went back and counted how many times double-deuce rolled and used that to figure out how many single-deuce rolled. In this video double-deuce rolled a total of five times compared to 69 single deuce rolls and an estimated 240 total rolls of the dice. Had the probability been 1/6 as claimed, this result is a whopping 2.2 standard deviations below the mean. It is highly unlikely that anyone shooting a video would be patient enough to wait for such a result. Considering the actual probability of 1/11, this result occurred only 0.68 standard deviations below the mean, well within the realm of likelihood.

I admit that this is *not* conclusive proof. Lucky streaks (and unlucky streaks) can and always will eventually occur. Mathematicians are well aware of this fact, which is why we develop analytical tools and methods to approach problems. The only reason I chose to do it is because what *is* conclusive proof has been outright ignored, so a physical demonstration might at least make you rethink it a little bit.