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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Comments and Answers forx Discussion 1}
\centerline{September 16, 2005}
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\noindent
The easiest way to ``go wrong'' in counting is to 
take an approach that ends up counting the same thing
more than once and not realize you are doing that.  For 
example, on question A part 3 of this discussion, several 
groups seemed to be thinking like this in counting the 
number of ``one-pair'' poker hands (possibly with
some ``adjustments''):
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{\it The first card in the hand can be any one of the 
52 cards in the deck, then to make the pair, we take
one of the other three cards of the same denomination
as the first one, then three other cards of different
denominations:}
$$52 \cdot 3 \cdot 48 \cdot 44 \cdot 40.$$

\noindent
Unfortunately, this is {\it not correct} for the following reasons:
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\item{1)}  You are getting each possible pair in $2 = 2!$
different ways in the $52 \cdot 3$.  (Say the pair 
is the 3 of hearts and the 3 of spades.  You're counting
the pair once if the first card is the 3 of hearts, and
also a second time if the first is the 3 of spades).
\item{2)}  Similarly, you're getting every combination of
three other cards in the hand in $3! = 6$ different ways
depending on the order in which they are selected.
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\noindent
The correct answer is easier to see if you think
of first picking the denomination for the pair, 
then the two suits of the pair, then the denominations
and suits of the other three cards, taking into account
that the hand is an unordered set:
$${13\choose 1}\cdot {4\choose 2} \cdot {1\over 3!}\left({12\choose 1}\cdot 4
{12\choose 1}\cdot 4 \cdot {11\choose 1}\cdot 4 \cdot {10\choose 1}\cdot 4\right)$$
The probability of getting a 5-card hand with exactly one pair should
be 
$${{13\choose 1}\cdot {4\choose 2} \cdot {1\over 3!}\left({12\choose 1}\cdot 4
{12\choose 1}\cdot 4 \cdot {11\choose 1}\cdot 4 \cdot {10\choose 1}\cdot 4\right)
\over {52\choose 5}} \doteq .423$$
(You have a $42.3\%$ chance of getting one if 5 cards are drawn randomly.)

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\noindent
{\it Other Answers}
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\noindent
A.
\item{1)} $4\cdot 13 = 52$  
\item{2)}  ${52\choose 5}$ for the unordered hands,
$P^{52}_5 = 52\cdot 51\cdot 50 \cdot 49 \cdot 48$ for the ordered
hands.  Note that 
$${52\choose 5} = {1\over 5!}\cdot 52\cdot 51\cdot 50 \cdot 49 \cdot 48.$$
\item{4)} ${13\choose 5}\cdot 4$ (4 choices of suit, and ${13\choose 5}$
different combinations of 5 denominations in that suit).
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\noindent
B.  
\item{1)} There are $140 + {140\choose 2} = 9870$ different majors and pairs
of two majors.  (Note: Majoring in English and Sociology isn't
different from Sociology and English.)
\item{2)}  There are $26^2 + 26^3 = 18252$ different codes of two
or three letters.
\item{3)}  The current system can continue to be used since there
are more than enough codes for all majors and pairs of majors.
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\noindent
C.
\item{1)}  With 3 equal prizes, the probability of 
2 or 3 team members winning a prize is
$${{25\choose 2}975 + {25\choose 3}\over {1000\choose 3}} \doteq .0018$$
\item{2)}  With 3 different prizes we should keep track
who wins what:
$$\matrix{ 100 & 35 & 15\cr
           team & team & nonteam\cr
           team & nonteam & team\cr
           nonteam & team & team\cr
           team & team & team \cr}$$
In each of the first three cases, we get $25\cdot 24\cdot 975$ possible
sets of winners, in the last $25\cdot 24\cdot 23$.  The probability
is 
$${3\cdot  25\cdot 24\cdot 975 + 25\cdot 24\cdot 23\over 1000\cdot 999\cdot 998}$$
If you look at this carefully, you'll see that it equals the answer
to part 1.  It actually {\it doesn't matter} for the probability, since
the denominators of 6 cancel.  Another way to say this:  For each way
of assigning winners as in part 1, there are exactly 6 ways to
assign the 3 different prizes as in part 2.  So the probabilities 
are the same either way(!)
\bye

