Dear Jean,
If N attempts are made at drawing from a lottery
where the probability where the chances of success
from one draw are 1/N, then the probability of
succeeding will be 1-(1-(1/N))**N
That is 1-prob(lose on every draw).
Now (and here i may find myself corrected),
I believe that as N-> infinity this
value converges to 1 - (1/e) or ~0.63
which may be a good approximation in the
Camelot example.
The convergence appears to be be from above,
so the higher N, the lower your chances of
succeeding.
Remember me if you win,
Philip
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