If B out of A lots wins, the first person to draw wins with a probability of A/B. The second person divides the case by whether the previous person wins or not, and it can be seen that A/B comes out as a result of (1 - A/B)*(A/(B - 1)+ A/B * (A-1)/(B-1). Is there a way to prove this in a similar way to the above method without such modeling, although we can prove that it is independent of order by corresponding the number of 1 to B to each lots?
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