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

[kewl]

If you want me to say "I'm wrong" you are going to have to demonstrate the 1/11 answer using two dice, with one die showing a 2. That is what I asked in the video. Neither miplet or the Wizard showed the 1/11 with one die showing a 2. Both showed a 2 and then proceeded to alter the faces on the die that showed a 2 in order to reach their 1/11 result.

I am not questioning the 1/11 result when you are asking how many combinations of two dice include at least one 2, and of those how many would result in 2-2. That would be 1/11.

But given that at least one of the dice has settled on 2 -- SHOW ME -- using two dice that the chance of 2-2 is 1/11 and not 1/6.

Please, you have to use two physical dice.

I don't WANT you to do anything. It'd be cool though if you and the rest of 1/6 supporters could see the question and the real probabilities for what they are.

My demonstration is as follows:


Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?


Or, you could roll a pair of dice a statistically significant amount of times,disregard all the non 2 rolls and count all the 2x outcomes in relation to all 2-2 outcomes.
What that does is - demonstrate and prove that whenever you have at least one 2 showing the odds for a pair of 2's is 1/11.