If you turn over a well shuffled deck of cards one card at a time, what is the probability that you'll see at least one of each of the denominations of the numeric cards, ace through ten (1 - 10) before seeing any face card (J, Q, K)?
Here are the elements of the calculation.
Call the non-face cards “pip cards”.
For the event to happen, P(happens) = 1- P(doesn’t happen)
Need the sum of the probabilities of elimination on draws
1-9 added to the probability it doesn’t happen on
draws 10 through 37 (37 is the longest possible successful run. E.g.,
we drew all of the pip cards except any 10s, and then drew a 10.)
For each draw i, there is a the probably of termination by a
face card of P_T = [12 / (53-i) ] ^ i
Alternatively, assuming we drew all pips, P_I is the probability
of the insufficiency of the drawn pips.
The probability of failure on pip draw i>9 is equal to the probability of
the absence of any one type of numbered pip.
We compute the sum of the probabilities we have not drawn a
1 or a 2… … or a 10. This is 10 x the probability we have not drawn
a 1 as since all numbers are interchangeable.
After draw i, there were a = (40 choose i) equally likely ways to
choose i pips and there were b = (36 choose i) ways to
not draw a 1. The probability of having not drawn a
1 is (b / a) and the probability of having not drawn
one out of numbers 1-10 is 10 (b/a)
Above are the elements of a computation. To be continued…
Edited on May 3, 2019, 12:50 pm
A start....
