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This is a variation on the coupon collector's problem. In the case when all the probabilities are equal ($a=b=c=1/3$), the expected number of spins is $3\bigl(1 +\frac12 + \frac13) = \frac{11}2.$A spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.
What is the expected number of spins required until all three outcomes are seen?
There's an easy way and a harder way to do this. Guess which I did first.
Stupid stupid mistake! My method was correct but I left out a $+$ sign, converting a sum into a product. The expression $$1 + \frac1{b+c}\Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr)$$Your formula does give the right answer for a=b=c but not for other possibilities.
For comparison purposes:
a=1/2, b=1/3, c=1/6 should give 73/10
and
a=9/20, b=9/20, c=1/10 should give 353/33.