CE217 ClasS 5 Handout 5
OK, so we’re done with Besikovitch’s Game now.
We’re not done with the class, though…
This is from an exam question a few years ago, in the days when they were multi-choice:
A particular version of Rock/Paper/Scissors uses a points system: gesturing rock costs 1 point,
gesturing paper costs 2 points, gesturing scissors costs 3 points; the winner gains 5 points. To
obtain a high score over a series of games, which gesture should be performed the most often?
(a)
(b)
(c)
(d)
Rock
Paper
Scissors
Rock and scissors equally.
Reminder: paper beats rock, scissors beats paper, rock beats scissors.
So, to figure this out, start by filling in this handy table:
Good
Evil
Rock
Paper
Scissors
Rock
Paper
Scissors
“Easy” way to do this:
0)
Fill the diagonals in with zeroes. If you both do the same thing, the difference
between your scores doesn’t change.
1)
Write next to each of rock/paper/scissors the cost in points to do it.
2)
Make the numbers for Good (which is to say, you) be negative, as they reduce your
score by this amount. The numbers for Evil (your opponent) are positive, because they add to
your score by this amount.
3)
In the six remaining empty spaces, pencil in the sums of the costs for the two
gestures that cross-index each one. This is how much you’d be up if winning the round scored
nothing.
4)
For every match that you win (eg. if Good does scissors and Evil does paper then Good
wins) add in the prize you get for a win. For the other three matches, subtract it.
What does this table tell us? Well, it tells us what the overall gain or loss is for each of the
9 possible combinations of rock, paper and scissors. It does not, however, tell us how many
times we should throw a rock, paper or scissors. Call the number of times we want to throw
rock R, the number of times we want to throw paper P and the number of times we want to
throw scissors S.
We can use the table we’ve just constructed to find the values of R, P and S. The values in
the table act as weightings on the number of times we need to throw each gesture. Although
it looks as if we should throw the highest-value gesture most often, actually that’s not the
case (as we shall see shortly).
I’ve mentioned in the lectures that rock/paper/scissors is zero sum, as what one player wins
the other loses. What this means for our table is that the numbers in each row (and column,
and indeed the whole table) add up to zero. We can use this fact to get a set of simultaneous
equations.
The conventional way to do this is to read the table a row at a time and use the columns as
variables. This means you put R after everything in the rock column, P after everything in the
paper column and S after everything in the scissors column. Go ahead and do that to your
table now.
As I’ve just said, being zero-sum means each row adds up to 0. You should now be in a
position to write down three simultaneous equations (although actually you’ll only need two of
them), one for each row of your table.
So, using R for rock, P for paper, S for scissors, the equations are:
0 R + ____ P + ____S = 0
____ R + 0 P + ____S = 0
____ R + ____ P + 0 S = 0
Yes, you get to fill in the blank bits…
Reminder: the way to read that first line is “the number of times you throw rock times 0,
plus the number of times you throw paper times (what you wrote), plus the number of times
you throw scissors times (what you wrote) is zero”.
You can rewrite each of the above as an equality (so rather than “this-that=0”, you’d put
“this=that”). Go ahead and rewrite the equations in this form. The result will be equations
that look something like this, except you’ll have numbers instead of p1, p2, s1, s2, r1 and r2:
(p1)P = (s1)S
(r1)R = (s2)S
(r2)R = (p2)P
p1 = ____
r1 = ____
r2 = ____
s1 = ____
s2 = ____
p2 = ____
If you’ve done it right, one of those lines will reduce to an equality. However, in general:
P = S(s1)/(p1) = R(r2)/(p2)
This means that the number of times you do P is (s1)/(p1) the number of times you do S, and
(r2)/(p2) the number of times you do R. In other words, the ratio of times you do each one
(in rock, paper, scissor order) is:
(p2)/(r2) : 1 : (p1)/(s1)
For you, this is: ____ : 1 : ____
If you want an answer in integers rather than fractions, do:
(s1)*(p2) : (r2)*(s1): (r2)*(p1)
For you, this is: ____ : ____ : ____
In your case, you can divide by a common factor (if you haven’t already) to get 4:7:4.
Therefore the answer to the question is (b) Paper. Voila!
Quick check: the highest-returning gesture to make looks to be rock, because at best you
can win 7 and at worst you can lose 4. However, if you were to throw rock every time and I
were to throw 7 papers for every 4 scissors (as above), we’d end up drawing. Correct!