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Evil wizard, 2 identical vases, beads

You have been captured by an evil wizard. However the wizard also likes maths, and sets you a challenge. You have two identical vases, and 100 white beads and 100 black beads. You must arrange the beads however you like between the two vases. The only condition is that no vase can be empty. And all the beads must be in the vases.

The wizard will then get his assistant to choose a single bead from one of the vases. The assistant will pick purely at random, and will not peek! If the assistant picks a black bead, you will go free. If the assistant picks a white bead... well, let's not go there. Obviously you would like the assistant to pick a black bead.

The question is this. How do you arrange the beads so as to give yourself the best chance of freedom?

There are no tricks. Just mathematics.

The answer will be posted next week

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