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a nice crunchy probability puzzle
by MaxEntropy_Man Fri Mar 16, 2012 2:46 pm
Three hundred men sit around a circular table. The men are numbered
1–300 and each man has two neighbors. (The neighbors of 1 are 2 and
300, and the neighbors of 300 are 1 and 299.)
There are three hundred waiters, also numbered from 1 to 300. Each
waiter has an urn containing three balls, one lettered L, and C and
one R. Each waiter y draws a ball at random from his urn and if the
ball is lettered L, delivers a dessert to the man to the left of man y. If
the letter is C man y gets the dessert, and if the letter is R the man
to the right of man y gets the dessert. Call a man lucky if he gets
three desserts. Find the greatest possible number of lucky men, and
the probability that this many men are lucky.
this is from the purdue problem of the week series. don't consult the solution there. try to solve it yourself. i will post my solution after 24 hours.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a nice crunchy probability puzzle
by MaxEntropy_Man Fri Mar 16, 2012 8:53 pm
come on you puzzle freaks! take a break from bashing the synthegrator. there is life outside integration and synthesis and heads, butts, and hair partitions!
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a nice crunchy probability puzzle
by Guest Fri Mar 16, 2012 9:09 pm
Just worked it out...
100 and (1/3)^299?
Will type out the reasoning shortly if correct.
100 and (1/3)^299?
Will type out the reasoning shortly if correct.
Guest- Guest
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a nice crunchy probability puzzle
by Guest Fri Mar 16, 2012 10:21 pm
There are 300 desserts in total and a person is lucky if he is served 3. It follows that there can be at most 100 lucky people. Moreover it is clear that there is a configuration that allows for 100 lucky people. For example, it is easy to see that persons 1,4,7,... can all be lucky. In fact, there are a total of 3 such configurations - besides the one mentioned, the other two configurations are 2,5,8,... and 3,6,9.....
The probability of each of these 3 configurations is (1/3)^300 because there are 3 choices available to each waiter, of which only one leads to the desired configuration. As there are 3 such configurations, the probability of 100 lucky people is 3* (1/3)^300 which is equal to (1/3)^299 (almost zero).
The probability of each of these 3 configurations is (1/3)^300 because there are 3 choices available to each waiter, of which only one leads to the desired configuration. As there are 3 such configurations, the probability of 100 lucky people is 3* (1/3)^300 which is equal to (1/3)^299 (almost zero).
Guest- Guest
Re: a nice crunchy probability puzzle
by MaxEntropy_Man Sat Mar 17, 2012 6:54 pm
your answer is of course correct but let me also write down my solution.
without loss of generality i considered person 1 as the central person of interest. let's call the waiter standing behind him Y. so if he gets three desserts, then the persons in front of waiters Y-2, Y-1, Y+1, and Y+2 cannot get three desserts. thus the first person to the left of person 1 who can get three desserts is 298 and to his right the first person who can do so is 4. thus we have the arithmetic progression: 1, 4, 7,...298.
writing the expression for the general term of an arithmetic progression (a = first term=1, k = index, d= common difference = 3) and setting it equal to 298:
a + (k-1).d = 298
this gives, k =100.
the next part i did by analogy. i decided that the situation was like filling 300 spots around a circle with red, green, or yellow balls. the total ways in which this can be accomplished is 3^300, i.e. for every spot we have three color choices. if we treat the maximum men getting three desserts case, by analogy this would correspond to placing red balls around the circle every fourth spot with the middle two spots occupied by green or yellow (but not red). there are exactly three different such arrangements, i.e. starting with a red ball in spot 1 or spot 2 or spot 3. so the required probability is:
3/3^300 = 1/3^299.
as i grow older, i find myself slowing down in algebraic adroitness, but much improved in thinking and problem solving by analogy.
without loss of generality i considered person 1 as the central person of interest. let's call the waiter standing behind him Y. so if he gets three desserts, then the persons in front of waiters Y-2, Y-1, Y+1, and Y+2 cannot get three desserts. thus the first person to the left of person 1 who can get three desserts is 298 and to his right the first person who can do so is 4. thus we have the arithmetic progression: 1, 4, 7,...298.
writing the expression for the general term of an arithmetic progression (a = first term=1, k = index, d= common difference = 3) and setting it equal to 298:
a + (k-1).d = 298
this gives, k =100.
the next part i did by analogy. i decided that the situation was like filling 300 spots around a circle with red, green, or yellow balls. the total ways in which this can be accomplished is 3^300, i.e. for every spot we have three color choices. if we treat the maximum men getting three desserts case, by analogy this would correspond to placing red balls around the circle every fourth spot with the middle two spots occupied by green or yellow (but not red). there are exactly three different such arrangements, i.e. starting with a red ball in spot 1 or spot 2 or spot 3. so the required probability is:
3/3^300 = 1/3^299.
as i grow older, i find myself slowing down in algebraic adroitness, but much improved in thinking and problem solving by analogy.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
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