Jar π has 800 red candies
Jar π has 600 green candies
π) Move one cup of red candies from jar π to jar π and mix.
π) Move one cup of mixed candies from jar π to jar π and mix.
π) Move one cup of candies from each of jar π and jar π to a bowl and mix.
π) Randomly select one candy from the bowl.
Is the selected candy from the bowl more likely to be red or green?
For clarity, "one cup" effectively means the same number of candies are transferred at each step.
3 edits
@joe-shmo saidWell, red should always be more likely, but the proportion of red in the final bowl should increase with the "cup size".
Jar π has 800 red candies
Jar π has 600 green candies
π) Move one cup of red candies from jar π to jar π and mix.
π) Move one cup of mixed candies from jar π to jar π and mix.
π) Move one cup of candies from each of jar π and jar π to a bowl and mix.
π) Randomly select one candy from the bowl.
Is the selected candy from the bowl more likely to be red or g ...[text shortened]...
For clarity, "one cup" effectively means the same number of candies are transferred at each step.
We can use the following expected value function:
nM/N
Where:
n is the cup size
M is the number of red or green candies in the jar
N is the total number of candies in the jar
This finds the mean number of the candies of your choice that will be in each transfer. Using the means eliminates the issue of random variation for the purposes of this question. Obviously you don't need the expected value function for the first transfer (when all the candies will be red).
Using that function it can be shown that when a cup is set to the minimum size (1), the expected value of red candies in the final bowl is 2405/2404 and the expected value of green candies is 2403/2404, which gives probability of 2405/4808 for red.
When the cup size is set to the maximum (600), the expected value of red candies in the final bowl is 675 versus 525 for green, so a probability of 9/16.
Edit: Fixed some typos (wrote this on my phone in bed - sorry!)
1 edit
@ashiitaka saidCorrect!
Well, red should always be more likely, but the proportion of red in the final bowl should increase with the "cup size".
We can use the following expected value function:
nM/N
Where:
n is the cup size
M is the number of red or green candies in the jar
N is the total number of candies in the jar
This finds the mean number of the candies o ...[text shortened]... green, so a probability of 9/16.
Edit: Fixed some typos (wrote this on my phone in bed - sorry!)
I chose to follow the the expected value of green candy in each cup.
let:
π₯ = cup "size"
π( πΊ_a ) = probability of drawing a green candy in jar π after each step
π( πΊ_b ) = probability of drawing a green candy in jar π after each step
π) Move one cup of red candies from jar π to jar π and mix.
π( πΊ_b ) = 600 / ( 600 + π₯ )
π) Move one cup of mixed candies from jar π to jar πand mix.
We are again moving back π₯ candies into jar π, so there will be 800 candies in the jar.
The expected value for the number of greens being mixed in from the transfer:
πΈ( πΊ_b ) = π₯ β§ 600 / ( 600 + π₯ )
π( πΊ_a ) = πΈ( πΊ_b ) / 800 = 3/4 β§ π₯ / ( 600 + π₯ )
π) Move one cup of candies from each of jar π and jar π to a bowl and mix.
πΈ( πΊ_a ) = π₯ β§ ( 3/4 β§ π₯ / ( 600 + π₯ ) )
πΈ( πΊ_b ) = π₯ β§ 600 / ( 600 + π₯ )
π) Randomly select one candy from the bowl.
Finally the probability of drawing a Green out of the bowl is then:
π( πΊ_bowl ) = ( πΈ( πΊ_a ) + πΈ( πΊ_b ) ) / ( 2π₯ )
= 1/2 ( 3/4 π₯ + 600 ) / ( π₯ + 600 )
We can see that for all π₯ > 0, 3/4 π₯ + 600 < π₯ + 600 implies that ;
π( πΊ_bowl ) < 1/2
Thus its is more likely we draw red from the bowl .