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

[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]

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.

[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.

[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]

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.

[OneHitWonder]

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.

You're welcome, but thank you. Many of these sorts of "paradoxes" I have already thought about a lot before now in one form or another.

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.

Sometimes, we have "it" in the (sequential) language of our thoughts, and must struggle to put "it" into math; sometimes, the other way around. Sometimes, "it" isn't about language at all; but our (random-access or simultaneous) spatial senses instead.

I leave you with an anecdote. One of the curious things, discovered upon examination after his death, about Einstein's brain was that the two lobes - lingual left and spatial right - were closer together than normal. This could have facilitated his ability to perform accelerated calculations. Princeton had hired him just for that purpose later on in his life. It was noted as well that he had the brain of a much younger man without the usual plaque of someone of 75.

[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]

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.

[kewl]

No I don't understand.

Yes, I know.

Go ahead and roll the dice. It will give you your proof eventually.

[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.