Originally Posted by Alan Mendelson View Post
No. I am not saying it has a probability of 1/11. But I am saying he will probably win.

Are you aware of the specifics of the bet that I have with the Wizard? I tried to meet up with him in Vegas a couple of weeks ago when I was in town to meet with a client but I could not make it in time.

First of all none of you in the 1/11 camp has yet to read the "original problem" in the same way that those of us who say it's 1/6 read the problem.

You guys can't even tell me which die is showing the 2. Those of us who say it's 1/6 say it doesn't matter which die is the 2.

So, when you guys tell me it doesn't matter which die is the 2, OR, if you can tell me which die it is, then we can discuss a different bet.

Zedd you came to this forum late in the discussion. Early in the discussion I said the original problem resembled this event at a craps table:

Two dice are thrown and one die immediately comes to rest showing a 2. The second die becomes a spinner and spins like a top. What are the chances that the "spinner" will also land on a 2?

If you don't view this as the same as the original question, we are going in different directions on the same road.
I’m aware of the specifics of the bet…which, as far as I know, is basically what you had described in your original post. These are the same specifics that OnceDear used in his spreadsheet.

I’m also aware of the original question. And you’re right; it can be read in different ways. The original question is ambiguous because it’s unclear how the condition of ‘at least one deuce’ was satisfied. Therefore, we must make assumptions on what sample space to use in our calculations. With this particular problem—one assumption produces an answer of 1/6 while another assumption produces 1/11.

BUT…this ambiguity is eradicated in your original post. We now know that the condition is satisfied only when EITHER one of the two dice is a deuce. This is different from when a specific die is a two—which is a more difficult condition to meet. Therefore, we must look at all two-dice combinations with either dice being a deuce (there are 11 of them).

1/11 is the absolute, no doubt about it, correct answer to the question in your OP.