Jim Druzik has challenged us all with a conundrum about the Galton
Quincunx being turned upside down. I would submit the following.
     The new distribution of the balls will be a wider and lower bell-shaped
curve than the original. The rational assumes that the second array of chutes
is sufficiently large in number that no ball bumps into the edge of the array.
The result of this second tumbling is the sum of a number of smaller
distribution curves. As each chute in the first array is released to fall into
the new array, each ball will undergo the random left or right deflection that
will result in a small distribution curve. When the next chute is released, the
same will happen except its curve will be slightly displaced from the previous
batch. Since the center chutes of the original distribution have the largest
number of balls per chute, the center of the NEW distribution (i.e. its highest
point) will be in the same position as the original distribution. The new curve
will be wider but lower since the two curves must have the same area (same
number of balls). Would anyone like to join me to continue this drivel over
a bottle of good wine?? It would ease the pain.
     I am struggling with one thing (besides this damn cork). It should be
quite possible to predict how much wider the second curve will be. But at this
hour of night, I'm not coming up with it. Full credit should not be given
without that insight. Ah, the cork is out. A toast to Galton, to Druzik, to
this List and each of you.
               May your distribution curve be NORMAL!!
                  Dave Westneat, Dayton Ohio
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