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

[kewl]

Now we have the "geniuses" arguing with each other. Theories flying all over the place.

Who's letting you here, Troll? Oh, yes, Alan is. Sad.

[OneHitWonder]

Now we have the "geniuses" arguing with each other. Theories flying all over the place.

Lol, from the guy with 3,374 or so posts. And not a valid argument or theory among them.

What's your iq? Which test? Just curious.

[OneHitWonder]

Who's letting you here, Troll? Oh, yes, Alan is. Sad.

R. Singer (like a bird) is essentially harmless. Not the same thing as AP's who think that they ARE mathematicians/physicists.

[Alan Mendelson]

Read the question one more time.
It is 1/6 on one die, yes. But what this has to do with the question?
You throw two dice.

Yes two dice were thrown and we are told at least one is a 2. The answer then depends on one die which has six faces.

[Alan Mendelson]

You throw two dice. There are eleven ways they can land as 2-x.

Why would you even consider one of the dice to determine your probability?
Don't you see there is so many ways you can arrive to 2-x, that your 1 face out of six is no longer relevant? Don't you see that when the blue die is 2, there are 6 ways the red die can land and vice versa? Don't you see that these possibilities are intrinsic to every single throw?

All of this doesn't matter in answering the original question. Sorry. It's always been simple.

[kewl]

Yes two dice were thrown and we are told at least one is a 2. The answer then depends on one die which has six faces.

No!
You get confused by the statement at least one is 2. Its not like saying the red/blue is 2. It is one OR the other. That makes possible either one is. Twice the more combos. On every single throw.

[Alan Mendelson]

There are only two dice. It doesn't make a difference which die is the 2 because the answer is always 1/6 for any die in a two dice problem when you only consider one die.

Your 1/11 is a Rube Goldberg answer.

[Alan Mendelson]

No!
You get confused by the statement at least one is 2. Its not like saying the red/blue is 2. It is one OR the other. That makes possible either one is. Twice the more combos. On every single throw.

Sorry there is no reason to consider 11 combinations. You have a 2. To get 2-2 you need 1/6 faces on another die.

[regnis]

No!
You get confused by the statement at least one is 2. Its not like saying the red/blue is 2. It is one OR the other. That makes possible either one is. Twice the more combos. On every single throw.

OK--"it's one or the other". So let's say "one" is a 2. How many faces are on the "other"? Envelope please----6. Chance of a 2--1 of 6.

[Alan Mendelson]

No!
You get confused by the statement at least one is 2. Its not like saying the red/blue is 2. It is one OR the other. That makes possible either one is. Twice the more combos. On every single throw.

Let me go over it again, to emphasize what regnis told you:

Once a 2 has been identified on either die you only have to consider six faces on one die. And if a 2 is on both dice it is still the same: you only have to consider the six sides on one die.

No one said anything about counting up the faces on two dice and adding together the possible combinations -- this is needless. Not only is it needless but it's an exercise that is not required to answer the original question.

[OneHitWonder]

Twice the more combos. On every single throw.

Not on every single throw. Single throws are the one-2 rolls by column or row but not both, or 2-2.

[OneHitWonder]

No one said anything about counting up the faces on two dice and adding together the possible combinations -- this is needless. Not only is it needless but it's an exercise that is not required to answer the original question.

Absolutely correct.

Either way, we agree on this.

[redietz]

OneHit, you've definitely grown on me. I think pretty much anything I had to say on this is in better hands. Thanks, and good luck.

I'll limit my future comments to (1) the personal value this topic has had for me and (2) the irony that some math folks are blind to their competency limitations with language, but that they get all snarky pointing out the math limitations of others. Math has its own language, but language has its own math, in a sense. Language is not the first language of many of these folks.

[kewl]

I don't understand why is this so hard to comprehend.

You have a bag with 36 balls - 25 red and 11 blue. One of the blue balls has a smiley face on it.
You draw a random ball out of the bag. If its red you disregard it and put it back in the bag, cause it is the blue ones you're after.
And every time you draw a blue one, the chances you have drawn the one with the smiley is 1/11.

Or,

You find a magical pair of dice which rolls perfect distribution of all 36 possible throws one after the other.
You roll them 36 times and you get 11 2-x and exactly one 2-2.

Do it on your own. Roll dice many times. Count. Report the results here. And be shocked.

[kewl]

OK--"it's one or the other". So let's say "one" is a 2. How many faces are on the "other"? Envelope please----6. Chance of a 2--1 of 6.

Do you understand what "OR" means? You can't say "its one that is 2" without accounting there are twice as much combos for that "one". Subtract one combo (2-2) so you don't double count it and there you go.

[kewl]

Let me go over it again, to emphasize what regnis told you:

Once a 2 has been identified on either die you only have to consider six faces on one die. And if a 2 is on both dice it is still the same: you only have to consider the six sides on one die.

No one said anything about counting up the faces on two dice and adding together the possible combinations -- this is needless. Not only is it needless but it's an exercise that is not required to answer the original question.

In order for you to "identify" a die you need to include the chances it (the die) has to become a 2. You throw two dice. Either die has equal chance to become a 2. And it will fulfill it's chances - half the time one will become 2 with 1/6 frequency and half the time the other. During each half the other die will be a non deuce 5/6 of the time.

[Alan Mendelson]

kewl you just don't understand? This is not about a bag with 36 balls or any combination of 11 dice with a 2.

This is about TWO DICE and we know there is a 2 on at least one of them.

And the question is what would it take for us to have 2-2 showing on both dice? Really -- in plain, simple English that's the question.

And if we have 2 on one die, the other die in the problem has a 1/6 chance of also being a two.

That's all this question asked. You guys got all wrapped up in your math and statistics and probabilities and forgot to look at the simple question that has a simple answer.

And as proof I asked the WOV forum to show videos of your thought process with two dice. And there it was for everyone to see how you compounded the question and came up with the wrong answer.

Yes, tell the Wizard he's wrong too.

[Alan Mendelson]

In order for you to "identify" a die you need to include the chances it (the die) has to become a 2.

this is the most ridiculous comment I've read yet.

[kewl]

this is the most ridiculous comment I've read yet.

Don't you understand that since either die can be your "2", you now have twice as much chances to see a 2-x, compared to what you would have if you were only looking for a specific die (only the red or only the blue die) ?

[Alan Mendelson]

Don't you understand that since either die can be your "2", you now have twice as much chances to see a 2-x, compared to what you would have if you were only looking for a specific die (only the red or only the blue die) ?

No I don't understand. When one die is a 2, there are 6 chances on the other die. It doesn't matter which die is the 2. And if both dice are showing a 2 it was still a 1/6 chance that either die is showing a 2.