Store-bought raspberries. Watery, tasteless and without the element of danger. Image: Flickr/LaGrandeFarmersMarket cc license
The berry bushes down the block are heavy with red, ripe raspberries. And so I believe it is my family’s duty to relieve the bushes of some of their burden, lest the weighty berries do long term damage to the plants themselves. Unfortunately, the plants’ best interests are hampered by their owner who bogarts these berries, likely for some lame use like fruit cake. And the plants are tall enough to obscure my kids from the house, but not tall enough to obscure me, I send them to poach berries. You know, for the greater good.
Here was this week’s puzzle:
If Leif and Kestrel are willing to give me a 20% cut, I’m willing to turn a blind eye. Imagine that Leif picks five berries per poach and Kestrel picks three berries per poach and that they attempt to poach once every day, with the exception of any day just after they’ve been caught. Now imagine that each time they poach berries, they have a 15% chance of getting caught. How many berries can I expect to eat each each week, averaged over time?
There were many answers and much debate, but correctness depended on one of the classic tricks of probability: splitting the situation into its possible states and running a probability for each. Andy wrote it this way:
1. They go poaching and don’t get caught (probability = 0.85x)
2. They go poaching and get caught (probability = 0.15x)
3. They don’t go poaching at all, because they got caught yesterday (probability = 1-x)
Blaine and Felicia (who happen to be this week’s winners of the $50
ThinkGeek gift certificate), wrote the following splendid solution:
Let p be the chance that you ARE sitting out.
Let q be the chance you are NOT sitting out.
Together these are mutually exclusive and therefore add up to 100% (or mathematically we say 1)
p + q = 1
p = 1 – q
The chance you are NOT sitting out, but CAUGHT is 0.15q
The chance you are NOT sitting out, and SUCCESSFULLY POACHED is 0.85q
We know that eventually the two values p and 0.15q end up being the same, so equate them
p = 0.15q
Substitute in 1-q:
1 – q = 0.15q
Rearrange:
1 = 1.15q
q = 1/1.15
The chance we are caught that day is 0.15q
0.15q = 0.15(1/1.15) = 0.15/1.15 = 15/115 = 3/23
And the chance we poach some berries is 0.85q
0.85q = 0.85(1/1.15) = 0.85/1.15 = 85/115 = 17/23
For the sake of completeness that means you have:
3/23 = chance child is sitting out
3/23 = chance child was caught today
17/23 = chance child was able to poach successfully
Now multiply this last number by 8 berries attempted times 7 days and then times 1/5 (20% commission) to get the expected number of berries poached each week.
Berries per week = 17/23 x 8 x 7 x 1/5
If you reduce that to a fraction you end up with:
952/115 = 8 32/115 berries each week
Answer:
8 32/115 berries each week (decimal value = 8.27826086956521739130434… underlined portion repeats indefinitely)
But my favorite answer by far was Breana’s, who wrote that, “My mom gave me this problem to exercise my summer-numbing, soon-to-be 9th grade brain. So I dug out my calculator and got to solving.” She also promises that, “our berries are a tad sour yet, but if you lived nearby, we’d share w/ L&K happily and your take would improve drastically.” Thanks, Breana and please keep playing the puzzle!
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