Originally Posted by Jihkro View Post
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.