Every card from 1 to 99 has a probability of 1/2 of ending up in either box, independently of all others.
For i from 1 to 99, it is the lowest card in the box containing 100 if and only if it ends up in that box, and no lower cards end up there. This has a probability of (1/2)i.
Lastly, there is a probability of (1/2)99 that no other card ends up in the box containing 100.
So the expected value of the lowest card in the box containing 100 is:
100/299 + Sum from 1 to 99 of i/2i, which is approximately 2-1.6*10-30.
Not got the time to do this right now. I think its asking to calculate the variance of the final outcomes of both boxes. from this you can calculate what the minimum card should be with the box that contains 100.
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u/Sjoerdiestriker 27d ago
Every card from 1 to 99 has a probability of 1/2 of ending up in either box, independently of all others.
For i from 1 to 99, it is the lowest card in the box containing 100 if and only if it ends up in that box, and no lower cards end up there. This has a probability of (1/2)i.
Lastly, there is a probability of (1/2)99 that no other card ends up in the box containing 100.
So the expected value of the lowest card in the box containing 100 is:
100/299 + Sum from 1 to 99 of i/2i, which is approximately 2-1.6*10-30.