[DOWRY]

I modelled the problem by numbering the girls 1 to 100, then generated a random dowry between $500 and $1,000. I then calculated a progressive mean which, as expected, approached closer and closer to $750. I then calculated a progressive Standard Deviation which very quickly dropped below $150 and then slowly approached $133. The simple formula might then be to accept the dowry which is greater than AVG + 2 x STDDEV. However, I found that this hurdle was more often over $1,000 and therefore no dowry passed the hurdle. Trial and error found AVG + 1.7 x STDDEV usually produced acceptance of the greatest dowry. However, my stats capability does not allow me to calculate the best value for the multiple of STDDEV, nor the resultant probability of success. Graph from XLS

Dowry model.pdfattached.

[Find the missing combination]

Merry Christmas, wolfgang!

TGNUH

Merry Xmas to you both.

RDNXJ

[Tale of two clocks]

The clocks will be lined up again in 90 days. The only way to get from January to May in 90 days is from 1/30 to 5/1 on a leap year. Thus, Robert's birthday was in 1933 (turning 47 in 1980) and Arthur is one year older.

Almost, ndd. 90 days is only possible from 31 January to 1 May when February has 28 days. Thus Robert is turning 47 in a non-leap year which is 1979 and therefore Robert is older than Arthur.

[The elusive chord]

I was just pointing out that Bertrand left out that method for choosing a chord. However,

Pick a random point inside the circle and spin a random angle relative to the radius and pivoting on the point creating a chord.

I agree with CaptainEd. Jaynes' solution to the Bertrand paradox shows that random radius (method 2) is the only robust theoretical means of choosing random chord.

[easy one]

double the smaller number, add to the larger

7*8 = (2x7) + 8 = 22

That won't work, 'cause 4*9 would be 17 not 21.

A variation to that theme which would work is: if second number is smaller, then [first number] + twice [second number], else, [first number] + twice second number] - 1. That is quite inelegant, and does not solve for instance of both numbers the same. Back to thinking further outside cube.