One never really knows what might be under the wallpaper. Redecorating reveals secrets, a veritable archeology of habitation. When my parents sold up and bought a house about 50 years ago, they left few traces. I often look down the street from the main road as I go past, to see if the place is still there, but only once have I walked to the house to see it at close quarters. One day when my nostalgia got the better of me I visited the building to take a closer look at our former dwelling, a ground-floor maisonette in Landport, Portsmouth.
Three bright orange murals
Quite fortuitously, the owners were redecorating. I took the opportunity to knock on the door, but nobody was home. Curtains were down, ladders were up, paint cans, brushes all around, and – surprise, surprise – peeled back wallpaper. I couldn’t resist taking a quick peak through the window at the bedroom where my brother Jon and I slept and dreamed over a ten-year period (see red arrow).
On Jon’s side of the room, as bold as brass, I could see three bright orange paintings – a trumpet, a bugle and a banjo. Immediately I knew these paintings were Jon’s. I knew how Jon would doodle and sketch things. Jon loved jazz. If Jon painted anything, it would have to be these. I was seeing them for the very first time. Jon must have done them after I had moved north in September 1966, shortly before Jon had moved to London as a professional jazz musician.
What are the odds of seeing Jon’s musical murals on my only visit to the building in 50 years?
A) We need to estimate how likely it was that Jon had ‘received approval’ to do these paintings in the first place. It is difficult to put a hard-and-fast figure on this, but it is unlikely to have been more than one in 10.
B) Next, we need to make an assumption about how often the walls were repapered. I expect on average around once every five years, which would be 10 times in 50 years. The probability that the wallpaper was peeled off on any particular day over 50 years (18,250 days) = 10/18250 = .000548 = 5.48 x 10 to the minus 4
C) Next, we need to calculate the probability that I would pay a visit on any particular day over the 29.5 years (10,767 days) that I was living in Britain over this 50-year period, which is 1/10767 = .0000928 = 9.28 x 10 to the minus 5
The odds for P (A +B+C) = 1/10 x (5.48 x 10 to the minus 4) x 9.28 x (10 to the minus 5) = 50.85 x 10 to the minus 10 = 5.08 x 10 to the minus 9
5 chances in one billion
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