Levy-Gritter E-mail

	 Akash Levy <akashlevy@gmail.com>
Question about tonk probabilities
3 messages
Akash Levy <akashlevy@gmail.com>	 Sun, May 19, 2013 at 4:35 PM
To: mgritter@cs.stanford.edu
Dear Dr. Gritter,

My name is Akash Levy and I saw your post about tonk probabilities on the twoplustwo forum (http://forumserver.twoplustwo.com/21/draw-other-poker/profitably-dropping-tonk-315444/) from about 5 years ago. As a beginner statistics student at Allderdice high school, I was curious about how you produced the probabilities in your post. Can you please give me a quick explanation of how your probability calculator worked? How did you calculate the number of hands that were lower than any specified hand value? Thanks a lot.

--
Sincerely,
Akash
Mark Gritter <mgritter@cs.stanford.edu>	 Sun, May 19, 2013 at 5:35 PM
To: Akash Levy <akashlevy@gmail.com>
Hi, Akash.

I have the code from that experiment around, but unfortunately it is
not a particularly simple calculation.  I used a generating function
to calculate the number of possibilities.

You can think of the problem as the ways to choose a fixed value, from
a multiset { 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, ...., 10, 10, 10, 10,
10, 10, 10, 10, 10 } representing the value of every card.  (We could
take some cards out to represent what we hold, if we are interested in
the distribution of the opponent's hand.)  It is thus similar to
problems like "given N nickels and P pennies, how many ways can I make
change for a dollar?" except that we have more options (1 through 10)
and a fixed number of each value (4 for everything but 10, 16 for
10's) and a limitation on the number of total coins.

The second answer here:
http://math.stackexchange.com/questions/15521/making-change-for-a-dollar-and-other-number-partitioning-problems
might be helpful.  I don't have a good reference on generating
functions handy--- most examples consider infinite sequences, which
aren't relevant here.

The generating function G(x,y) is the product of many terms
1+y*x^(a_k) where a_k is the successive values of the multiset, that
is,

G(x,y) = (1+yx)(1+yx)(1+yx)(1+yx)(1+yx)(1+yx^2)(1+yx^2)....(1+yx^10)

The term of this very big polynomial with the exponent y^5 represent
the ways to pick 5 cards.  Of those, the coefficients on items with
x^N represent the number of ways to make a point value of N.  The
"intuition" here is that for each card, we can either not include it
(the "1" part of each term) or we can include it and cause the number
of cards and the value of go up ("y" by one and "x" by the value of
the card).  Multiplication of the terms gives us all possible
combinations.

You might be able to translate that reasoning into some more direct
counting approach, but it would probably be a quite complicated
recurrence.

For an easier example, suppose you just had {1, 1, 1, 2, 2, 2} in the
deck and a 2-card hand, then the generating function is

G(x,y) = (1+yx)(1+yx)(1+yx)(1+yx^2)(1+yx^2)(1+yx^2)

Using Wolfram Alpha to do the algebra, that expands to:
x^9 y^6+3 x^8 y^5+3 x^7 y^5+3 x^7 y^4+9 x^6 y^4+x^6 y^3+3 x^5 y^4+9
x^5 y^3+9 x^4 y^3+3 x^4 y^2+x^3 y^3+9 x^3 y^2+3 x^2 y^2+3 x^2 y+3 x
y+1
if we look at the terms for y^2 they are
3 x^4 y^2
9 x^3 y^2
3 x^2 y^2

So there are 3 ways to make 4 points, 9 ways to make 3 points, and 3
ways to make 2 points.  We can check that since there are thee ways to
pick two from (2, 2, 2), three ways to pick two from (1, 1, 1), and to
make 3 points we need one card from each value, so there are 9 ways.

I have attached the python code I used, which comes with no guarantee
that it is actually correct.  :)

Mark Gritter
(not Dr., unfortunately)
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		tonk.py
3K
Akash Levy <akashlevy@gmail.com>	 Sun, May 19, 2013 at 6:15 PM
To: Mark Gritter <mgritter@cs.stanford.edu>
Thanks a lot. For my statistics class's final project, I modeled this situation with a normal model but I don't think the answer was accurate. The problem is complicated by the fact that there are four cards that count as 10 points which skews the curve to the left and makes any predictions about when you should drop invalid. Another complicating factor, is the lack of independence in the deck. Your example with the two card hand really made the explanation much easier to understand. I don't know much about combinatorics but I think I get the gist of what a generating function is. I'm going to try to modify your Python code to calculate the probabilities of getting caught with a seven-card hand instead of a five-card hand. It shouldn't be too hard... I'm going to change the parameter for Y to 7 and regenerate the values. Again, thanks for the help; I really appreciate it.
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--
Sincerely,
Akash