The challenge in grasping this is that many see the second die as independent of the first one, so that it appears that it really doesn't matter what the first die is, the probability of the second one "must" be 1 in 6, since the second die doesn't "know" anything about the first die and lands independently of the first one. While I can clearly see the correct solution on this one, my own intuition on the "Monty Hall" version of essentially the same problem made it difficult for me to finally give in, and as someone with (I thought) a good instinct of basic probability, I could only take solace in the fact that some leading mathematicians also failed to "get it." Perhaps the many posts similarly reflect this (and, of course, many of the posts are more about the "real" bet than the hypothetical one originally posted).
I think the idea of labeling the two dice is an excellent way to understand this.
If you have "Die A" and "Die B", it's one situation when you say "If Die A is a 1, what are the chances that Die B is a 1", and another situation when you say "If Die A OR Die B is a 1, what are the chances that the other one is also a 1". The way this is presented, the second situation is clearly the one we're talking about.
--BG
1b. Re: The Dice Challenge
I can't believe that there are so many posts on this. It is simple 5th grade probability. Alan can't be that dumb (can he?). I do have an out (forgive my poker terminology): Alan can say he wanted the 2 dice labeled 1 and 2, the dice to be rolled simultaneously and if die 1 showed a 2 the chance that the other die (die 2) would show a 2 would be 1 in 6.
[Non-text portions of this message have been removed]
Posted by: Barry Glazer <b.glazer@att.net>
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