I think Barry nailed why some folks thing this is a 1 in 6 probability.
Taking Barry's example a little further, let's say you have 2 dice and roll one of the die until you get a 2. Then, you roll the second die. What is the probability that the second die will be a 2? Clearly, 1 in 6 times.
The challenge is to understand that the above situation is way, way different than the rules for the Dice Challenge. In the dice challenge, you roll both dice together and there are 3 different possible outcomes. Outcome 1 is neither die is a 2. Probability is 25/36. Outcome 2 is that both dice are a 2. Probability is 1/36. Outcome 3 is that exactly one die is a 2. Probability is 10/36. For this challenge, Outcome 1 is a don't care. Outcome 2 is the winning roll and Outcome 3 is the losing roll. From the subset of Outcome 2 and Outcome 3, you win 1 time in 11 rolls.
Interesting problem. You could further break down Outcome 3 into the individual cases when die A is the 2 vs when die B is the 2, but that might just confuse the issue.
In probability, when in doubt, enumerate. If Alan lists all the possible combinations ( expanding the Outcome1, 2 and 3 above) and counting winners and losers , he should see what the correct answer will be. If he doesn't want to do that fairly simple exercise, I don't think he will ever get it.
More importantly, how do I get in on this wager :).
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Posted by: greeklandjohnny@aol.com
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