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

[RS__]

If I roll the dice 66 times and in each roll, at least one of the dice is a 2....how many times can I expect both dice to be a 2? PS: I'm not setting one die to a 2 or anything like that -- just giving them a typical roll where both dice are random.


If you answered 11....just try it out for yourself.

[Alan Mendelson]

If I roll the dice 66 times and in each roll, at least one of the dice is a 2....how many times can I expect both dice to be a 2? PS: I'm not setting one die to a 2 or anything like that -- just giving them a typical roll where both dice are random.


If you answered 11....just try it out for yourself.

Yes it should be 11 or 1/6 times but you are rolling real physical dice and you could get zero or 66 of the "hard 4" and we all know that.

More importantly: JUSTIFY how the Wizard can "double count" the die showing a 2 when answering the Original Question.

[RS__]

Yes it should be 11 or 1/6 times but you are rolling real physical dice and you could get zero or 66 of the "hard 4" and we all know that.

I don't care what "could" happen. We all know what could happen. I'm asking what is expected. Ok. So your answer is 1/6 (11/66) for this case, yes?

More importantly: JUSTIFY how the Wizard can "double count" the die showing a 2 when answering the Original Question.

That's where you're wrong (again). The "die showing a 2" is NOT being double-counted. Actually, it is being removed, giving us 1/11 NOT 2/12 (or 1/6).


This is like the many similar examples/scenarios I've posted....none of which you've even bothered to think about. Let's try this again:

Let's say I have 20 cans on my table. Each can has a 50% chance of being sprite (green) or 50% chance of being coke (red). I ask you to pick a can at random. Before you pick a can, I know you have a 50% chance at picking a red can. Still with me?

Now, let's say before you join me, you're in another room. I remove 3 of the red cans from the table. So now there are 17 cans on the table. If you pick a can at random, what's the chance it'll be red?

Remember -- a can has a 50% chance of being red and 50% chance of being green. This is true of cans in general. However, I have now removed 3 red cans from the table. The chance of you picking a red can, at random, is no longer 50%.

If you don't believe me, you can test this out. You don't need cans, you can use something else. Something you have a lot of.


The reason why this is similar to the 2-dice problem is because the can I'm removing is directly related to the can I want you to pick (or rather, the can that I'm figuring the probability of). In the same way, with the dice problem, I am removing a 2, then trying to figure the probability of the other die being a 2.



Now, let's look at another similar but not identical scenario.

Let's say all the same stuff is true about the cans. But this time, I still remove the 3 red cans. You come in the room, and I ask you "If you pick a can at random, and you pick either green or red [at random], what is the chance/probability you will choose a can of the same color you've picked?" In this situation, it is [oddly enough] -- 50% chance. This is because the color I'm "searching for" is not directly related to the color I have removed. 50% of the time (you pick color red) your chances of choosing a red can decrease, and the other 50% of the time (when you pick color green) your chances of choosing a green can increase.


Let's say I roll 2 dice, you can't see them. I only will "count" a roll if at least one of the dice is a 2 [and you don't know this]. I ask you "pick a random number between 1 and 6". You have a 1/6 chance of being correct [the number you pick is the value of the other die]. In the can problem, if you chose red, you'd have less than 50% chance of being right. And if you chose green, you'd have a greater than 50% chance of being correct. But the sum is still 50%. This is the same for the dice problem. If you choose a 2, you have less than 1/6 chance of being correct. If you choose a non-2, you'd have greater than 1/6 in being correct.

[Alan Mendelson]

The "die showing a 2" is NOT being double-counted. Actually, it is being removed, giving us 1/11 NOT 2/12 (or 1/6).

I am sure that you made an error here -- a mistype, or you misspoke. Because when you remove one die from a two dice problem there are only six faces remaining on the die that remains.

I didn't have to get an advanced degree for this deduction. I just looked at a die that I keep in my top desk drawer.

I love all of your little examples of math problems but the question is specific. I don't really care about cans, or shirts, or even your fancy graphs and charts and "expectations." Because this is a specific question and a specific problem.

Again, I am going to ask you to tell me why you have to "double count" the die with a 2?? Answer.

[RS__]

I already responded, Alan.

It is clear that you are either trolling or a real big fart.

Teaching someone smart is easy. Teaching someone dumb is hard. Teaching someone ignorant is impossible.

[Rob.Singer]

And it continues....The math people who cannot impose their will of probability & theory on those who choose to view things and events in all their glorious reality, always resort to vulgarity & name calling because they can't get their way.

That's why wizard's forum constantly uses lengthy threads on subjects heavily laced with theory: it's the SAFE way to bring something up--anonymously and theoretically. If most of them actually went to casinos and played a hundred bucks, they'd pee their pants.

[Alan Mendelson]

RS___ will be taking a week off for his contribution above.

[redietz]

I have my doubts that RS is in a teaching profession.

[indignant99]

I can't make the graphic clearer, without showing duplicate pictures of those same (one red and one blue) dice. I wouldn't like to do that. You wouldn't, either.

I had to commit the crime of showing the BLUE DEUCE twice.
I had to commit the crime of showing the RED DEUCE twice, too.

[indignant99]

My last post had a graphic. After editing my graphic, the original link (had worked), is now pointing into "the void."

[Alan Mendelson]

I like the "or" graphics. 1/6.

[indignant99]

I have tried to improve the upper half of my graphic.
Retain the bottom half - Alan prefers the bottom half.

[OneHitWonder]

I like the "or" graphics. 1/6.

Looks like television circuit schematics?

[OneHitWonder]

Teaching someone smart is easy. Teaching someone dumb is hard. Teaching someone ignorant is impossible.

Geniuses teach themselves. They have a mind of their own.

[Alan Mendelson]

Looks like television circuit schematics?

When I was in high school, I had to go to a school to get a FCC First Class RadioTelephone Operators License so I could work at radio stations and do double-duty baby-sitting the transmitter and being the DJ or news reporter or sports reporter. I worked my way through high school and college doing that. At school we had to study schematics and memorize charts. I had no idea what I was looking at. But I passed the tests at the FCC on the first try. First the third class, then the second class, and then the first class license. Then one day I am working at a radio station and the transmitter failed and I had to change the "main plate" said the Chief Engineer in a phone call. And I said "what's that?" He talked me through it over the phone. I still have no idea what a main plate is. By the way, that was in 1970.

[indignant99]

Looks like television circuit schematics?

When I was in high school, I had to go to a school to get a FCC First Class RadioTelephone Operators License so I could work at radio stations and do double-duty baby-sitting the transmitter and being the DJ or news reporter or sports reporter. I worked my way through high school and college doing that. At school we had to study schematics and memorize charts. I had no idea what I was looking at. But I passed the tests at the FCC on the first try. First the third class, then the second class, and then the first class license. Then one day I am working at a radio station and the transmitter failed and I had to change the "main plate" said the Chief Engineer in a phone call. And I said "what's that?" He talked me through it over the phone. I still have no idea what a main plate is. By the way, that was in 1970.

Gentlemen, thank you. Your comments have helped.
I understand where I am prone displaying unclear graphics.
I am trying my best to be clear. (I'm failing.)
Welcome your critique, on a fresh graphic.

[regnis]

Now eliminate the other options on the die that already showed a 2, as unlike a leopard, it does not change its spots.

1 of 6

[indignant99]

Now eliminate the other options on the die that already showed a 2,
as unlike a leopard, it does not change its spots. 1 of 6

I don't think I showed extra faces on the die that showed a 2.
Please put a big X or circle slash on "the options" to eliminate.

[Rob.Singer]

Alan, now your friend mickeycrimm is bringing up the two dice problem on vpFree, and badmouthing you in the process. For some reason he chose to only give the 1in11 explanation while ignoring why you concluded it was 1in6. He was his usual drunken self of course, and the first responder spotted that right off the bat. But don't forget, this is an "AP" who regularly pulls in $596.97 every day from those well-known loosely goosey Montana machines! (And of course, he also lives with his mommy.....)

[Alan Mendelson]

I don't read VP Free nor do I care what he writes there.