Hmm, looks like there's no takers on this one, so here's the answer, in case anyone is interested... it's really not that hard ;-)
Okay, if there is to be exactly one tie in 5 throws, we can say two things:
1.) That tie must occur on the 1st, 2nd, 3rd, 4th or 5th throw.
2.) If two players threw 5 times, and they tied exactly once, the tie would be equally likely to have been in the 1st, 2nd, 3rd, 4th, or 5th rounds.
Since it's equally likely that the tie could occur in any particular round, all we have to do is calculate the probably that the two players will tie exactly once AND that tie will occur in the first round, multiply that by five, and we have our answer. We already know that the odds of a tie on a single throw are 1 in 3, and the odds of a non-tie are 2 in 3, and we know that we can calculate the probability of a particular set of outcomes in a series of events by multiplying the probabilities together.
So the answer is:
1 2 2 2 2 80
--- x --- x --- x --- x --- x 5 = ----- = 32.9%
3 3 3 3 3 243
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