Engineers and scientists

I have experienced such a "condition" recently, so I documented its stages - for fear of relapse :‑)

Last Saturday, my older brother asked me about the probability of "something" related to a lottery that is played here in Greece. The lottery is called "Joker" (and no, it has nothing to do with the Bat mobile). He described it to me as a set of balls with numbers from 1 to 40; I was told that they are thoroughly mixed in each lottery, and that 5 of them are randomly picked. Then another number (the "Joker") is randomly picked from a separate set of balls numbered from 1 to 20. If you have predicted all 6 of them, you instantly become a millionaire.

"What is the chance that the joker is equal to one of the 5 others?", I was asked.

It's been 15 years since I passed the probability and statistics exam in the University, but... like I said: "relapse" :‑)

In theory...

When we calculate the odds for a logical AND of events, the overall probability of the combined event is the product of the probabilities of the events. At least, that's how I remember it...

P(A and B and C and D and E) = P(A) x P(B) x P(C) x P(D) x P(E)

• A is the event of the 1st of the five not being equal to the joker
• B is the event of the 2nd of the five not being equal to the joker
• C is the event of the 3rd of the five not being equal to the joker
• D is the event of the 4th of the five not being equal to the joker
• E is the event of the 5th of the five not being equal to the joker

    p(A) = 39/40
    p(B) = ...

The second one (B) is a tad more difficult, though: Event B says that the second ball is not equal to the joker - this means that

either event (A) took place and now, event (B) takes place
    or
event (A) didn't take place and now, event (B) takes place

either
(B1) the 1st ball was not the joker and neither is the 2nd one
or
(B2) the 1st ball was equal to the joker and the 2nd one is not

P(B) = P(B1 or B2) = P(B1) + P(B2) - P(B1 and B2)

P(B1) = P(1stEscapesJoker) x 
  P(2ndEscapesJoker given that 1stAlsoEscaped) = 39/40 x 38/39

Similarly,...

P(B2) = P(1stIsJoker) x 
  P(2ndEscapesJoker given that 1stIsJoker) = 1/40 x 39/39

So, what is P(B)?

    P(B) = P(B1) + P(B2) - P(B1 and B2)  =

      39   38      1   39
    = -- x --  +  -- x --  -  0     =
      40   39     40   39

      38    1     39
    = -- + --  =  --
      40   40     40

Let's do it once more, for P(C):

    P(C) = P(third ball not equal to the joker) =
         = P(C1 or C2 or C3) =
         = P(C1) + P(C2) + P(C3) - 
	    - P(C1 and C2) - P(C2 and C3) - P(C1 and C3) +
            + P(C1 and C2 and C3)

Where:
    C1 = first escapes, second escapes, third escapes
    C2 = first escapes, second matches, third escapes
    C3 = first matches, second escapes, third escapes

and therefore,...

            39   38   37     37
    P(C1) = -- x -- x --  =  --
            40   39   38     40

            39    1   38      1
    P(C2) = -- x -- x --  =  --
            40   39   38     40

             1   39   38      1
    P(C3) = -- x -- x --  =  --
            40   39   38     40

which, finally, leads us to...

    P(C) = P(C1) + P(C2) + P(C3) - 
	    - P(C1 and C2) - P(C2 and C3) - P(C1 and C3) +
            + P(C1 and C2 and C3) =
         
             37       1       1
         =   --  +   --  +   --  - 0 - 0 - 0 + 0  = 
             40      40      40

             39
         =   --
             40

So, what is the chance of ALL 5 balls escaping the joker (the "logical AND")? The product of the 5 probabilities, so...

                          39  5
    P(all 5 escaping) = ( -- )
                          40

P(one or more of them equal to joker) = 
    1 - P(all 5 escaping) = 1 - (39/40)^5 = .1189043

Now let's put the theory to the test - Python to the rescue!

In practice...

#!/usr/bin/env python

success = 0
tries = 0
for joker in xrange(1, 21):            # From 1 to 20
    for first in xrange(1, 41):        # From 1 to 40
        for second in xrange(1, 41):   # From 1 to 40
            if first == second:
                continue         # Ignore invalid lotteries
            if first != joker and second != joker:
                success += 1
            tries += 1

print float(success)/tries

P(A and B) = P(A) x P(B) = (39/40)^2 = 0.950625

But...

bash$ python experiment1.py
0.95

Don't read further down immediately - think this through, reread... treat this as a puzzle!

No, it is not a Python "floating point precision" issue...

Let's try to see what Python gives us for each of P(A) and P(B) (i.e. the chance of the first one escaping the joker, and - separately - the chance of the second one escaping the joker):

#!/usr/bin/env python

success1 = 0
success2 = 0
tries = 0
for joker in xrange(1, 21):
    for first in xrange(1, 41):
        for second in xrange(1, 41):
            if first == second:
                continue
            if first != joker:
                success1 += 1
            if second != joker:
                success2 += 1
            tries += 1

print float(success1)/tries
print float(success2)/tries

bash$ python experiment2.py 
0.975
0.975

Hmm...

Science and engineering make up...

    P(A and B) = P(A) x P(B) = (39/40)^2 = 0.950625

Time to dig up the textbooks... (boxes, dust, cough)

There, I found it...

Independent?

They are most definitely NOT independent! Whether the second one escapes the joker VERY MUCH DEPENDS on whether the first one did! If the first one was equal to the joker, the second one SURELY will NOT (since it cannot be equal to the first one)! Don't get confused by the fact that we calculated the probabilities of each event and found them to be the same; the important point is that the outcome of the first event has implications on the second one.

Now what?

Wait a minute... By knowing P(A) and P(B), we also know P(not A) and P(not B)...

    P(first  is equal to joker)  = P(not A) = 1-P(A) = 1/40
    P(second is equal to joker)  = P(not B) = 1-P(B) = 1/40

P(1stIsJoker or 2ndIsJoker) = 
    P(1stIsJoker) + P(2ndIsJoker) - P(1stIsJoker and 2ndIsJoker)

Finally, I got it!

    P(1stIsJoker or 2ndIsJoker) = 
	P(1stIsJoker) + P(2ndIsJoker) = 1/40 + 1/40 = 2/40

    P(both first and second escape) = 
	1-P(1stIsJoker or 2ndIsJoker) = 1-(2/40) = 0.95

Engineering: Bulldozing around till you get your answer...

    P(firstEscapes and secondEscapes) = P(B1) = 
    
    (see above in Theory section)

                  39   38     38
                = -- x --  =  --  =  0.95
                  40   39     40 

p(1st and 2nd and 3rd and 4th and 5th escape the joker) =

   p(1stEscapes) x 
    p(2ndEscapes given 1stEscaped) x 
     p(3rdEscapes given 1stAnd2ndEscaped) x
      p(4thEscapes given 1stAnd2ndAnd3rdEscaped) x 
       p(5thEscapes given 1stAnd2ndAnd3rdAnd4thEscaped) =
  
                  39   38   37   36   35     35  
                = -- x -- x -- x -- x --  =  -- 
                  40   39   38   37   36     40

    and thus, the answer to our problem is... 
                                                    35    5
        P(one of the 5 is equal to the joker) = 1 - -- = --
                                                    40   40

• If you do play the Joker lottery, don't select as a joker one of the 5 other numbers you play - or better yet, don't pay the "stupidity tax" that is all lotteries.
• Don't ask engineers scientific questions! They are VERY susceptible to "me-almost-scientist" relapses... And they will feel illusions of grandeur in the process - have mercy :‑)

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