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Re: Math/Logic Test for Musicians

Author: Jack Kissinger
Date: 2009-04-10 18:49

A few comments to this point: (Be warned, Brenda, this will probably add cramps to your headache.

)

1. I can't take credit for this quiz. One of the members of the St. Louis Symphony shared it with the president of a community orchestra I play with. He sent it to us.

2. Re: (1):

While the (verifiable) presence in Wikipedia of the term "Mutual Bonds" might constitute conclusive evidence that such bonds do exist, any mathematician/logician worth his/her salt could tell you that the absence of the term from Wikipedia is no guarantee that it does not exist unless Wikipedia is collectively exhaustive, which it is not. Furthermore even if there is no such type of bond as a Mutual Bond, as a quick check with Google will verify, there are a number of companies with the word "Mutual" in their title that could have issued bonds. These might also be identified as "Mutual" bonds where Mutual refers to the name of the issuer rather than the type of bond. This does not necessarily change the answer.

Assuming he goes through 50 boxes of reeds a year and his reeds cost $25 per box it would take about 52 years for him to amass the $1,250,000 Yahoo recently said he would need to retire (1,250,000 = (1250)(FV,.087,52). Unless he's Stanley Drucker or has money from another source, it ain't gonna happen.

3. Re: (4)

It seems to me that (4) might have an answer but we would need a lot of additional information. First of all, does the manager select concerts at random or deliberately. If the latter, if he consciously tries to avoid these three composers, then he probably can do so for most if not all major orchestras. On the other hand, if he consciously tries to attend concerts that include at least one work by one of the three, then he probably can hear them in every concert he attends in any 10-year period.

Assuming the manager's choice of concert is totally random, if we are willing to assume that programming is also totally random and we know the individual probabilities that each time the music director selects a piece, it will be by one of the three composers (MBB), then we can compute the probability that any 4 x 3 x 12 = 120 combinations of works (the number of works our manager will hear over 10 years) will include at least one work by MBB. Alex states that (at the low end) the odds any concert by a major orchestra will include a work by one of our three are 1/4. If the 2009-10 season of the SLSO is representative (and I use them because they provided this quiz), he's right. The odds for SLSO next year are actually somewhere between 36% and 38% depending on whether you calculate on a concert-by-concert or program-by-program basis (since 4 programs are only performed once, 13 are performed twice and 11 are performed 3 times and, BTW, the probablility that one of our icons will included on a program is a direct function of the number of times the program will performed).

Assume then that SLSO's programming is totally random and that there is a 36% probability that a work by MMB will be performed. The probability that at least one work by MBB will be performed is equal to 1 - the probability that no works by MBB will be performed or .64. If we assume that all concerts include three works (which, in reality, they do not), the probability that a work by someone other than MBB will be chosen at random (P[MBB']) is given by the formula:

.64 = P[MBB']^3

So the probability that a work by someone other than MBB will be chosen at random is about .90 and the probability that a work by MBB will be chosen is .10. If the concerts are randomly programmed, the probability that the 120 works our manager hears will include at least one work by MBB is: (1 - (.90)^120) or about .9999968. In other words, he's doomed.
(Assuming that pieces are selected, at least for a given year, without replacement, the theoretically correct approach, IMO makes the problem intractable.)

Of course orchestra program directors do not program randomly. Another way of attacking this problem would be to assume that the manager chooses his concerts each year randomly or, alternatively chooses the weeks he will attend randomly and the SLSO's programming is representative. Using SLSO's upcoming season, even trying to work the problem on a concert by concert basis becomes a computational nightmare, however, because, as I mentioned above, the orchestra does not perform each concert program the same number of times. If, on the other hand we are willing to assume that our manager is with the SLSO and chooses his weeks randomly (with equal probability), we can calculate the probability that he will not encounter MBB next year. To the extent that 2009-10 is representative of our mythical 10-year period (which, as Alex points out is a stretch), we can then extend next year's results to the 10-year period.

According to it's schedule for next year, the SLSO will have a 25-week subscription season. In 16 of those weeks, no work by MBB is performed. In 8 of the weeks, every concert performed includes at least one MBB work. Just to complicate matters, in the remaining week, one of the three concerts performed includes a work by MBB but the other two do not.

The probability that our manager will encounter MBB at least once next year = 1 minus the probability that he will not encounter MBB at all. The probability that he will not encounter MBB equals the sum of the probabilities that:

(1) he does not encounter MBB given that he does not choose the odd week x probability that he does not choose the odd week, and

(2) he does not encounter MBB given that he does choose the odd week x probability that he does choose the odd week.

The former probability is given by:

(17/25)(16/24)(15/23)(14/22)(21/25) = (about) 15.8%

The latter probability is given by:

(16/25)(15/24)(14/23)(13/22)(4/25) = (about) 1.15%

Thus the probability that he does not encounter MBB during the year is about 17%.

Assuming the schedule is about the same each year, and that each year's schedule is independent of the other years (perhaps not true but necessary to make the problem reasonably tractable) the probability that he will not encounter MBB in a 10-year period would be .17^10 = 2 E08. So the probability that he will encounter MBB is .999999998. And he's still doomed.

Chris' solution, while perhaps somewhat cynical, is far more elegant.

Best regards,
jnk

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