David Jenkins
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 posted on 5/10/11 at 07:07 PM |
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Another puzzle to think about...
I really annoyed a lot of people last time I put a puzzle up... so here's another one!
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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
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RichardK
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| posted on 5/10/11 at 07:17 PM |
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Put half of the white beads in each first and then half of the black bead in each on top of those. So all the black beads are on the top??
Gallery updated 11/01/2011
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Confused but excited.
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| posted on 5/10/11 at 07:19 PM |
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As the assistant has his eyes shut, drop the wizard first, then his assistant and leg it. Sorted.
I'm no good at maths.
Tell them about the bent treacle edges!
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Ninehigh
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| posted on 5/10/11 at 07:27 PM |
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quote:
Originally posted by RichardK
Put half of the white beads in each first and then half of the black bead in each on top of those. So all the black beads are on the top??
+1
I don't see any way, persuming it's a random bead from a random vase, to make it any better than 50-50
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SteveWallace
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| posted on 5/10/11 at 07:28 PM |
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Put one black bead in one of the jars and all of the rest in the other. That way, if he picks the jar with the black bead (50% proability based on
number of jars) you go free, if he picks a bead from the other jar you have a 99/199 (nearly 50%) chance of going free, meaning that you have nearly a
75% chance of going free.
By the way, if by a random chance I've got it right, no I have not seen it before - honest
Alternatively, put an equal number of beads in each jar (so that they have equal weight) and then hit each of them over the head with one jar and leg
it.
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RichardK
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| posted on 5/10/11 at 07:28 PM |
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And no jiggy jiggy shakey, shakey is going on!!
Gallery updated 11/01/2011
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ken555
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| posted on 5/10/11 at 07:30 PM |
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Place one black in a vase the rest in the other.
50% chance of picking first vase.
How ever if he chooses the 2nd vase.
49.74.....% of picking black (99/199)
Over all odds 74.87% of black
[Edited on 5-10-11 by ken555]
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ken555
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| posted on 5/10/11 at 07:32 PM |
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beaten while typing and working it out 
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SteveWallace
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| posted on 5/10/11 at 07:34 PM |
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you were more accurate though as I didn't do mine to two decimal places
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David Jenkins
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| posted on 5/10/11 at 07:37 PM |
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Steve got there first! Well done...
(the actual value is 74.87%)
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RichardK
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| posted on 5/10/11 at 07:41 PM |
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stupid quiz anyway
Gallery updated 11/01/2011
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steve m
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| posted on 5/10/11 at 07:59 PM |
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You could paint the white beads black ?
or put half equal white beads in each vase and fill the vase up with pva,
wait to dry and then fill the remainder up with black beads, ?
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Ninehigh
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| posted on 5/10/11 at 08:07 PM |
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How have you got a 75% chance of getting a black from a 50% chance of a 50% chance?
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JoelP
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| posted on 5/10/11 at 08:34 PM |
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quote:
Originally posted by Ninehigh
How have you got a 75% chance of getting a black from a 50% chance of a 50% chance?
Thats probability for you, not always what it appears to be. By adding two jars and seperating the beads, it doesnt have to be 50/50.
You have a 50% chance of 100%, and 50% chance of nearly 50%., and obviously 50 and 100 average at 75%
[Edited on 5/10/11 by JoelP]
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