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

[regnis]

Excellent---finally proof of a completely different question than the one at hand.

[Alan Mendelson]

Regnis is correct. The monetized bet with the Wizard was only meant to monetize the question and was not the original question.

Arc the chances of throwing 2-2 is 1/36.
When one of two dice is known to be a 2 the other die has a 1/6 probability of also being a 2.
You can throw or slam two dice all day and 2-2 will show 1/36 of the time and not 1/11. And not 1/6 either.

[RS__]

Regnis is correct. The monetized bet with the Wizard was only meant to monetize the question and was not the original question.

Arc the chances of throwing 2-2 is 1/36.
When one of two dice is known to be a 2 the other die has a 1/6 probability of also being a 2.
You can throw or slam two dice all day and 2-2 will show 1/36 of the time and not 1/11. And not 1/6 either.

2-2 will show up 1/11 of the time when at least one deuce is rolled.


Singer - there is no purpose to making a video under Alan's conditions. It'd be like me telling you to make a video flipping coins in order to prove your VP strategy works. Flipping coins is not related to your VP strategy working........and rolling one die when another is set/fixed to a deuce is not related to the dice problem.

[Alan Mendelson]

Rolling one die IS how you figure the answer to how the original question was worded. And that's the issue: the wording of the question.

By the way nice video and thanks for doing it. But the video I wanted to see is your justification for the 1/11 answer when one die is a 2. As a matter of fact at the start of the video you said you can't show 11 faces.

One more time: your 1/11 math is good but it's not the answer to the original question. It's all about the wording.

[onyx99]

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?

That's the original question, verbatim, from the original thread. There are two interpretations, one of which yields 1/6 as the correct answer, and one of which yields 1/11. There should be convergence on the correct answer *for each interpretation*. Here they are:

1. Your partner peeked and saw only a single die, which was a 2, and made his statement. (If it hadn't been a 2, he would have said something like, "At least one of the dice is a 5," or whatever.) The probability that both dice are showing a 2 is 1/6.

2. Your partner peeked and saw both dice, at least one of which was a 2, and made his statement. (The number 2 was agreed on beforehand. If neither had been a 2, he would have said something like, "Neither die is a 2.") The probability that both dice are showing a 2 is 1/11.

The first interpretation is unnatural because the question doesn't say he saw only one die; it seems more likely that he'd see both if he peeked. What'd he do, stop looking mid-peek? The second interpretation is unnatural because the question doesn't say that 2 was agreed on beforehand. Why 2? Who asked the guy? It occurs to me that there's actually a third interpretation that is more natural than either:

3. Your partner peeked and saw both dice, at least one of which was a 2, and made his statement. (The number 2 wasn't agreed on beforehand. If he had rolled 3 & 4, for instance, he would've said either "at least one of the dice is a 3" or "at least one of the dice is a 4.") The probability that both dice are showing a 2 is...... 1/6.

Damn it, Alan.

You were right all along.


(For the "math guys": you shouldn't be calculating the probability of a pair of 2s given that there's at least one 2. In the original question, that's not, in fact, the whole situation. You should be calculating the probability of a pair of 2s given that there's at least one 2 AND that your partner stated that there was at least one 2. In the monetized version, in this thread, the fact that your partner is looking for a deuce in particular is written into the bet; the probability in that case is definitely 1/11, which is why no one is taking that bet.)

[regnis]

That's the original question, verbatim, from the original thread. There are two interpretations, one of which yields 1/6 as the correct answer, and one of which yields 1/11. There should be convergence on the correct answer *for each interpretation*. Here they are:

1. Your partner peeked and saw only a single die, which was a 2, and made his statement. (If it hadn't been a 2, he would have said something like, "At least one of the dice is a 5," or whatever.) The probability that both dice are showing a 2 is 1/6.

2. Your partner peeked and saw both dice, at least one of which was a 2, and made his statement. (The number 2 was agreed on beforehand. If neither had been a 2, he would have said something like, "Neither die is a 2.") The probability that both dice are showing a 2 is 1/11.

The first interpretation is unnatural because the question doesn't say he saw only one die; it seems more likely that he'd see both if he peeked. What'd he do, stop looking mid-peek? The second interpretation is unnatural because the question doesn't say that 2 was agreed on beforehand. Why 2? Who asked the guy? It occurs to me that there's actually a third interpretation that is more natural than either:

3. Your partner peeked and saw both dice, at least one of which was a 2, and made his statement. (The number 2 wasn't agreed on beforehand. If he had rolled 3 & 4, for instance, he would've said either "at least one of the dice is a 3" or "at least one of the dice is a 4.") The probability that both dice are showing a 2 is...... 1/6.

Damn it, Alan.

You were right all along.


(For the "math guys": you shouldn't be calculating the probability of a pair of 2s given that there's at least one 2. In the original question, that's not, in fact, the whole situation. You should be calculating the probability of a pair of 2s given that there's at least one 2 AND that your partner stated that there was at least one 2. In the monetized version, in this thread, the fact that your partner is looking for a deuce in particular is written into the bet; the probability in that case is definitely 1/11, which is why no one is taking that bet.)

Onyx for president

[redietz]

That was very good, onyx. Thanks.

I just wanted to add, parsing this to semi-ridiculous but not incorrect extremes, that technically human beings see things point by point and do not grasp everything in their visual field all at once. The seeing is piecemeal; the interpretation is in the cortex, and that could be piecemeal also. So technically, one would really not "see" a pair of dice together all at once. One die would almost always be seen/interpreted before the other.

[Jihkro]

I'll try once more, and then I'm out.

Again, imagine the scenario with colored dice, one red, and one green. (the color of the dice of course in no way affects the probabilities)

The original question says "at least one of the dice shows a two."

This is equivalent to saying "EITHER the red is a two and the green is not OR the green is a two and the red is not OR the red and green are both a two"

Let us denote each event separately as R, G, B respectively (for red only, green only, both). I.e. "at least one of the dice shows a two" is equivalent to saying "the outcome is either an element of R or G or B"

Notice: given that you know that you are in either B or R (i.e. you know that the red die is a two but you know nothing about the green die) then the odds of being in B compared to R are 1:5, i.e. Pr(B|B or R) = 1/6, because with the red die as a 2, there only five ways to lose and one way to win.

Notice: given that you know that you are in either B or G (i.e. you know that the green die is a two but you know nothing of the red die) then the odds of being in B compared to G are again 1:5, because with the green die as a 2, there are 5 ways to lose and one way to win.

However, we are not given that we are in the event (B or R) (i.e. we are not told the red die is a two) nor are we given that we are in the even (B or G) (i.e. we are not told that the green die is a two), we are instead given that we are in the event (B or R or G) (i.e. we are told that at least one of the dice is a two). In being told "at least one die is a two" this tells us nothing of the color of the die. There are a total of 10 ways to have lost (five ways with a red two, five ways with a green two) and one way in which you could have won. Remember that each of these outcomes are equally likely.

You are time and again insisting that the problem that reads "at least one die is a two" is somehow telling you "the first die is a two" when again, there is no way to distinguish which is the "first" die. Using colors you can have a clear distinction. You are thinking in terms of "the red die is two" and only looking at the ways you can lose in the case that the red die is a two versus the ways you win if the red die is a two. You have completely ignored all of the ways in which you can lose if the green die is a two while the red isn't.

If you toss one die at a time, and the first die is not a two, yes you already know that you haven't won, however you must still toss the second die to see if you have lost. You are in effect refusing to throw the second die and turning all of the not-wins into breakevens when only 5/6 of those not-wins will be breakevens.

The colored dice argument makes it perfectly clear. The 36-sided die argument makes it perfectly clear. The deck of playing cards argument makes it perfectly clear. The ordered tosses argument makes it perfectly clear. The simulations argument makes it perfectly clear.

If your logic was correct, then the bet should have been a smart one to make. Try what I did and take 100 chips and see how many times you end with 50+ in the green compared to 50 in the red. Use a cup if you want. Use a friend to shout out "at least one is a two" before ante'ing a chip if you want. Neither of those will make a difference, afterall if you can't trust yourself, who can you trust? Write a computer program to run the simulation for you if you don't want to spend the time throwing dice yourself. Tell one of your employees to write the program for you if you don't have the ability to code such a program. Yes, you will win occasionally, however it pales in comparison to the number of times you lose. If it was a smart bet, you should have been winning well over half the time, not less than one percent of the time. Repeat the process where instead of going 50+ in the green versus 50 in the red, try it with 200+ in the green versus 200 in the red. The numbers get even worse. Again, if it was a smart bet it should have been well over half of the time that you come out on top, not just a fraction of a fraction of a percent.

"Alan has been asked to perform a test, and we know that will result in a 1in11 conclusion." -- If we all know that it will result in a 1in11 conclusion, then why is this still being debated?

---

@onyx, the probability for interpretation 3 is still 1/11, for exactly the same reasons already mentioned.

[onyx99]

One die would almost always be seen/interpreted before the other.

Definitely true; but I don't think that changes anything in my post. In interpretation 2, your partner has to see/interpret both dice before reporting in the negative, so it doesn't matter if he sees them at the same time or not. In interpretation 1, he only sees one die at all, so this doesn't come up. And in interpretation 3, it's fine if he reports immediately on the first die he sees/interprets; or it's fine if he sees them both and then chooses one at random on which to report.

[regnis]

I'll try once more, and then I'm out.

Again, imagine the scenario with colored dice, one red, and one green. (the color of the dice of course in no way affects the probabilities)

The original question says "at least one of the dice shows a two."

This is equivalent to saying "EITHER the red is a two and the green is not OR the green is a two and the red is not OR the red and green are both a two"

Let us denote each event separately as R, G, B respectively (for red only, green only, both). I.e. "at least one of the dice shows a two" is equivalent to saying "the outcome is either an element of R or G or B"

Notice: given that you know that you are in either B or R (i.e. you know that the red die is a two but you know nothing about the green die) then the odds of being in B compared to R are 1:5, i.e. Pr(B|B or R) = 1/6, because with the red die as a 2, there only five ways to lose and one way to win.

Notice: given that you know that you are in either B or G (i.e. you know that the green die is a two but you know nothing of the red die) then the odds of being in B compared to G are again 1:5, because with the green die as a 2, there are 5 ways to lose and one way to win.

However, we are not given that we are in the event (B or R) (i.e. we are not told the red die is a two) nor are we given that we are in the even (B or G) (i.e. we are not told that the green die is a two), we are instead given that we are in the event (B or R or G) (i.e. we are told that at least one of the dice is a two). In being told "at least one die is a two" this tells us nothing of the color of the die. There are a total of 10 ways to have lost (five ways with a red two, five ways with a green two) and one way in which you could have won. Remember that each of these outcomes are equally likely.

You are time and again insisting that the problem that reads "at least one die is a two" is somehow telling you "the first die is a two" when again, there is no way to distinguish which is the "first" die. Using colors you can have a clear distinction. You are thinking in terms of "the red die is two" and only looking at the ways you can lose in the case that the red die is a two versus the ways you win if the red die is a two. You have completely ignored all of the ways in which you can lose if the green die is a two while the red isn't.

If you toss one die at a time, and the first die is not a two, yes you already know that you haven't won, however you must still toss the second die to see if you have lost. You are in effect refusing to throw the second die and turning all of the not-wins into breakevens when only 5/6 of those not-wins will be breakevens.

The colored dice argument makes it perfectly clear. The 36-sided die argument makes it perfectly clear. The deck of playing cards argument makes it perfectly clear. The ordered tosses argument makes it perfectly clear. The simulations argument makes it perfectly clear.

If your logic was correct, then the bet should have been a smart one to make. Try what I did and take 100 chips and see how many times you end with 50+ in the green compared to 50 in the red. Use a cup if you want. Use a friend to shout out "at least one is a two" before ante'ing a chip if you want. Neither of those will make a difference, afterall if you can't trust yourself, who can you trust? Write a computer program to run the simulation for you if you don't want to spend the time throwing dice yourself. Tell one of your employees to write the program for you if you don't have the ability to code such a program. Yes, you will win occasionally, however it pales in comparison to the number of times you lose. If it was a smart bet, you should have been winning well over half the time, not less than one percent of the time. Repeat the process where instead of going 50+ in the green versus 50 in the red, try it with 200+ in the green versus 200 in the red. The numbers get even worse. Again, if it was a smart bet it should have been well over half of the time that you come out on top, not just a fraction of a fraction of a percent.

"Alan has been asked to perform a test, and we know that will result in a 1in11 conclusion." -- If we all know that it will result in a 1in11 conclusion, then why is this still being debated?

You are still hung up on throws where there is no 2. The facts are that there is a 2----therefore we only need to know what's on the second die. And it does not matter whether the first die, second die, blue die, red die, etc. is the die with a 2. Whichever has a 2, the odds of the other one being a 2 are 1 of 6.

[Alan Mendelson]

Definitely true; but I don't think that changes anything in my post. In interpretation 2, your partner has to see/interpret both dice before reporting in the negative, so it doesn't matter if he sees them at the same time or not. In interpretation 1, he only sees one die at all, so this doesn't come up. And in interpretation 3, it's fine if he reports immediately on the first die he sees/interprets; or it's fine if he sees them both and then chooses one at random on which to report.

Onyx thank you.

Yes, we see one die at a time and then we make the connection (the roll) of the two dice. But it doesn't matter, as Onyx pointed out.

[Zedd]

The problem is not the original ambiguous one. I’ve said this on page 6 in the other thread:

But, the original question is indeed incapable of solution without making assumptions on how the condition of 'at least one deuce' was met.

However, after we had cleared up this ambiguity and established that the number IS agreed upon in advance--Alan still believes the answer is 1/6. This is simply wrong.

The problem is the one explained very clearly on page 1 of this thread.

Again, Alan (and other 1/6ers) believes the answer to this problem is 1/6.

Again, wrong. It is 1/11.

Alan has been wrong all along.

[onyx99]

@onyx, the probability for interpretation 3 is still 1/11, for exactly the same reasons already mentioned.

@Jihkro, you need to think interpretation 3 through more carefully. The peeker is twice as likely to announce "At least one of the dice is a 2" in the case where the roll is a pair of 2s (where he couldn't have said anything else) as he is in the case where it's a 2 and a 5, say (where he could have announced instead that "at least one of the dice is a 5"). Conditioning on what the peeker chose to say changes the posterior probability to 1/6.

[Rob.Singer]

I'll try once more, and then I'm out.

Again, imagine the scenario with colored dice, one red, and one green. (the color of the dice of course in no way affects the probabilities)

The original question says "at least one of the dice shows a two."

This is equivalent to saying "EITHER the red is a two and the green is not OR the green is a two and the red is not OR the red and green are both a two"

Let us denote each event separately as R, G, B respectively (for red only, green only, both). I.e. "at least one of the dice shows a two" is equivalent to saying "the outcome is either an element of R or G or B"

Notice: given that you know that you are in either B or R (i.e. you know that the red die is a two but you know nothing about the green die) then the odds of being in B compared to R are 1:5, i.e. Pr(B|B or R) = 1/6, because with the red die as a 2, there only five ways to lose and one way to win.

Notice: given that you know that you are in either B or G (i.e. you know that the green die is a two but you know nothing of the red die) then the odds of being in B compared to G are again 1:5, because with the green die as a 2, there are 5 ways to lose and one way to win.

However, we are not given that we are in the event (B or R) (i.e. we are not told the red die is a two) nor are we given that we are in the even (B or G) (i.e. we are not told that the green die is a two), we are instead given that we are in the event (B or R or G) (i.e. we are told that at least one of the dice is a two). In being told "at least one die is a two" this tells us nothing of the color of the die. There are a total of 10 ways to have lost (five ways with a red two, five ways with a green two) and one way in which you could have won. Remember that each of these outcomes are equally likely.

You are time and again insisting that the problem that reads "at least one die is a two" is somehow telling you "the first die is a two" when again, there is no way to distinguish which is the "first" die. Using colors you can have a clear distinction. You are thinking in terms of "the red die is two" and only looking at the ways you can lose in the case that the red die is a two versus the ways you win if the red die is a two. You have completely ignored all of the ways in which you can lose if the green die is a two while the red isn't.

If you toss one die at a time, and the first die is not a two, yes you already know that you haven't won, however you must still toss the second die to see if you have lost. You are in effect refusing to throw the second die and turning all of the not-wins into breakevens when only 5/6 of those not-wins will be breakevens.

The colored dice argument makes it perfectly clear. The 36-sided die argument makes it perfectly clear. The deck of playing cards argument makes it perfectly clear. The ordered tosses argument makes it perfectly clear. The simulations argument makes it perfectly clear.

If your logic was correct, then the bet should have been a smart one to make. Try what I did and take 100 chips and see how many times you end with 50+ in the green compared to 50 in the red. Use a cup if you want. Use a friend to shout out "at least one is a two" before ante'ing a chip if you want. Neither of those will make a difference, afterall if you can't trust yourself, who can you trust? Write a computer program to run the simulation for you if you don't want to spend the time throwing dice yourself. Tell one of your employees to write the program for you if you don't have the ability to code such a program. Yes, you will win occasionally, however it pales in comparison to the number of times you lose. If it was a smart bet, you should have been winning well over half the time, not less than one percent of the time. Repeat the process where instead of going 50+ in the green versus 50 in the red, try it with 200+ in the green versus 200 in the red. The numbers get even worse. Again, if it was a smart bet it should have been well over half of the time that you come out on top, not just a fraction of a fraction of a percent.

"Alan has been asked to perform a test, and we know that will result in a 1in11 conclusion." -- If we all know that it will result in a 1in11 conclusion, then why is this still being debated?

---

@onyx, the probability for interpretation 3 is still 1/11, for exactly the same reasons already mentioned.

Now there's a post taken straight out of "I just can't comprehend the original question's wording" heaven. Again. Right in-line with RS__s' interpretation used as an excuse not to provide the requested video.

[OneHitWonder]

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.

You haven't written anything to further enlighten us. Nor even tried to answer my question to you about the lottery? Would you rather enter a lottery with one prize, or a lottery with one or more prizes? Not the same options.

Now there's a post taken straight out of "I just can't comprehend the original question's wording" heaven. Again. Right in-line with RS__s' interpretation used as an excuse not to provide the requested video.

My thought exactly.

The first interpretation is unnatural because the question doesn't say he saw only one die; it seems more likely that he'd see both if he peeked. What'd he do, stop looking mid-peek?

Who cares how many of the dice were read? Makes no difference to the 1/6 chance answer w/o the announcement of the number 2 in advance.

I think that you better use a few days to think about this, and try to redo your assumptions.

No point in digging into your responses - and non-replies - further.

Sorry.

[OneHitWonder]

The problem is not the original ambiguous one. I’ve said this on page 6 in the other thread:


The problem is the one explained very clearly on page 1 of this thread.

Again, Alan (and other 1/6ers) believes the answer to this problem is 1/6.

Again, wrong. It is 1/11.

Alan has been wrong all along.

Okay! If you say so.

... You being the veridical paradox guy with a math background in Wikipedia. And the backdoor calculation to the specific problem. Lol.

[Zedd]

Okay! If you say so.

... You being the veridical paradox guy with a math background in Wikipedia. And the backdoor calculation to the specific problem. Lol.

I did not use Wikipedia. Your reading comprehension is quite poor—which is ironic since this is what you are shouting at others for.

I have the correct calculations for the problem in post 1 of this thread.

[Jihkro]

It occurs to me that there's actually a third interpretation that is more natural than either:

3. Your partner peeked and saw both dice, at least one of which was a 2, and made his statement. (The number 2 wasn't agreed on beforehand. If he had rolled 3 & 4, for instance, he would've said either "at least one of the dice is a 3" or "at least one of the dice is a 4.") The probability that both dice are showing a 2 is...... 1/6.

Damn it, Alan.

You were right all along.

The edit button seems to disappear after a time, so one final post. I will admit to being wrong about the third interpretation's probability. If indeed rolling a 2-3, the peeker has a chance of saying "there is at least one three" despite there being a two on the table, yes the probability will become 1/6. The difference is between the questions: "find the probability Pr(A|B)" and "find the probability Pr(A|B and C)", where A - there are two deuces, B - there is at least one deuce, C - the peeker announces "there is at least one deuce", where the peeker in this scenario is only going to reveal information about one of the visible numbers.

I agree then that the probability in that interpretation is 1/6, and if that is indeed the interpretation that Alan has been working off of, then I apologize to him and other 1/6'ers. To other math guys like me, to explain why this interpretation has a probability 1/6, it is again easiest to consider an equiprobable sample space, so consider two differently colored dice (red and green) and a coin. The dice are cast and the coin is flipped. If the coin comes up heads, then the peeker announces "there is at least one --whatever number appears on the red die--", else he announces "...--whatever number appears on the green die--." Instead of our samplespace consisting of 36 events, now it consists of 72 events. Those events that have the announcer saying "there is at least one deuce," will be (in order: coin-red-green) T-1-2, T-2-2, T-3-2, T-4-2, T-5-2, T-6-2, H-2-1, H-2-2, H-2-3, H-2-4, H-2-5, H-2-6, in which case there are 2 winning scenarios out of 12 possible, for a probability of 2/12 = 1/6.

As an amusing aside, if the peeker isn't inclined to announce about one of the numbers appearing specifically, but instead chooses a number 1-6 uniformly at random, peeks at the two dice and announces either "There are no -randomly chosen number-" or "There is at least one -randomly chosen number-", given that his announcement was about there being at least one number 2, what is the probability that there are in fact two deuces then? what is the probability that there are in fact two deuces given that his announcement was that "there are no 5's"?

I will admit and agree that the original wording of the question does not address the peeker's reasoning for making the announcement "there is at least one two" and is therefore ambiguous, but giving the peeker a strict protocol to follow (such as always gives information *only* about the number of twos present and never about any other number) will rectify the problem. It was under this interpretation that I had been working since arriving to discuss the problem as that was how it was presented and discussed previously.

---
(answers to my aside questions with the peeker choosing a number 1-6 uniformly at random to announce information about: "at least one 2": 1/11, "there are no 5's": 1/25)

[OneHitWonder]

I did not use Wikipedia. Your reading comprehension is quite poor—which is ironic since this is what you are shouting at others for.

I have the correct calculations for the problem in post 1 of this thread.

You did reference Wikipedia, and your simplistic calculations may as well be found there. I didn't "shout" at anyone, except to re-exclaim your own written self-proclaimed correctness from the start.

As for my own abilities, you have yet to match those in reply. You seem to go quite silent after I reply to point out your own mistakes, omissions, and further lack of comprehension in realizing those. Eg, you wrote veridical; I asked how so; and you wrote that Wikipedia thinks it's veridical because a lot of unnamed persons say counter intuitive. Lol.

In any event, I don't recall where I used the words nonsense, fart, dunce cap, or fretted over others' levels of comprehension within a topic of argument. The stuff of losers.

Fine, YOU have all the CORRECT answers for the "problem in post 1 of this thread". Go fly a kite; and have your kids write it on your tombstone after you're gone. So does everyone else on the gambling forums.

Most of us have better things to do. It's nice to learn something, even from the losers; and wonderful when from the winners. And either way, move onward and forward.

[Alan Mendelson]

I just want to say thank you to everyone. I did learn some things here including some probability math. And welcome again to our new forum members.

Maybe now we can all get back to something familiar like dissecting Total Rewards? LOL