Mathematics and Shakespeare a
combination liable to make members of both academic factions run in fear. They are, however, surprisingly
compatible. Statistics, probability, and
logic pair nicely with verse analysis, staging, and doubling. For the past two weeks this bizarre
intersection has been the focus of my work in the education department. Using
Internet Shakespeare Editions, (http://internetshakespeare.uvic.ca/Foyer/plays.html).
Minitab 16 (http://en.wikipedia.org/wiki/Minitab),
and a TI84 plus (http://en.wikipedia.org/wiki/TI84_Plus_series)
I have determined the probability of randomly generating verse lines, and I
have also discovered why twelve is the ideal maximum number of actors on the
Blackfriars stage. For those with the
guts to face irrational numbers and irregular lines, here are my findings.
Is Iambic Pentameter
a Coincidence?
Coin flips are a favorite
probability simulator of math teachers everywhere. Each flip has two outcomes—heads or
tails—both of 0.5 probability. From
there the textbook, teacher, or worksheet can set up various scenarios to test:
what is the probability of ten flips
yielding eight heads and two tails, six heads in a row, etc? If we replace the coin with a syllable, we
can calculate the likelihood of accidentally writing a regular line of iambic
pentameter. Each of the ten syllables
represents a coin flip and stressed and unstressed syllables represent heads
and tails. This problem assumes there is a 50/50 chance of stressing a
syllable. In order to determine the
probability of randomly generating an unstressedstressedunstressedstressed
pattern for ten syllables in a row, we must multiply the probabilities of each
outcome in the desired order. Since both
unstressed and stressed have the same probability of occurrence—0.5—the problem
looks something like this:
0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5
= 0.000977 or 0.5^10 = 0.000977
In other words, under these
conditions, there is a 0.0977% chance of randomly generating a single line of
iambic pentameter. What is the
probability of generating a full sonnet’s worth of pentameter? In that case we want onehundredforty
alternating unstressedstressed syllables, fourteen lines of ten syllables. The equation is 0.5^140 = 7.175*10^43. For readers unfamiliar with scientific
notation that is 0.0000000000000000000000000000000000000000007175, essentially
zero. Iambic pentameter is not
coincidental.
How Many Staging
Permutations are there for the Blackfriars Playhouse?
The answer to that question
changes according to how many actors are involved. Fortunately, after establishing the number of
positions available to one actor we can create a formula applicable to any
number of people. The stage is
twentynine feet wide by twentytwo feet deep.
For the purposes of this problem we ignore the presence of the gallant
stools. The groups of four X’s represents
the space occupied by an actor, for the purposes of this equation four square
feet.
X

X


X

X


Under these assumptions, there are
588 different positions for a single actor.
21*28 = 588. We multiply by one
less than the actual dimensions because that is how many different positions
two feet (one side of a four foot square) can occupy within 22 or 29 feet. The numbers one through twentytwo represent
to depth of the stage. We are trying to
figure out how many ways there are to place two consecutive feet within that
range, feet one and two, feet two and three, etc. We end up with twentyone combinations. Doing the same with twentynine feet yields
twentyeight combinations.
1.
1 2
2.
2 3
3.
3 4
4.
4 5
5.
5 6
6.
6 7
7.
7 8
8.
8 9
9.
9 10
10. 10 11
11. 11 12
12. 12 13
13. 13 14
14. 14 15
15. 15 16
16. 16 17
17. 17 18
18. 18 19
19. 19 20
20. 20 21
21. 21 22
Things get more complicated for the
second actor. Actor One occupies a four
square foot space somewhere on the stage which Actor Two cannot also
occupy. The 588 possibly positions
assumed the actor could be anywhere on stage, which means several of those
options overlap by one or more units.
This means that we must take out the spot occupied by Actor One and all
overlapping positions when calculating the number of possibilities for Actor Two.
The diagram below illustrates the
need to account for overlap. Each cell is
a square foot. The center Xs represent
the occupied space. The surrounding Ys
indicate parts of other four foot units that overlap with the occupied square.
Y

Y

Y

Y


Y

X

X

Y


Y

X

X

Y


Y

Y

Y

Y


Including the central square there
are nine positions in and around the Xs which Actor Two cannot occupy. 588 – 9 = 579 spaces available to Actor Two. We find the total number of possible stage
pictures by adding Actor One’s possibilities (588) and Actor Two’s possibilities
(579). 588+579 = 11,667. Please note that this equation assumes the
first actor’s position was not on the edge of the stage, as such a position
would reduce the number of positions eliminated after placing Actor One.
If the number of positions
available goes down by nine with each actor added to a scene, we can generate a
formula to calculate the number of staging possibilities. 588+Σ(588(n*9)), where n = the number of
actors on stage previous to the actor being placed. The Σ symbol means “sum of.”
Stage configuration has a huge
affect on ideal blocking. The
Blackfriars Playhouse features a thrust stage, meaning it has audience on three
sides (four if you count the balcony).
Angles are the rule for thrust stages.
As long as actors form diagonal lines, triangles, etc., then every
audience member is able to see at least one of them.
In order to
figure out how many angled combinations there are we follow a similar formula
as the one above but with more positions removed. After removing the occupied position and the
eight surrounding squares from the pool we also remove the column and row
containing the selected position.
X

X


X

X


X

X


X

X

X

X


X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

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X

X

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X

X


X

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X

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X

X


X

X


X

X


X

X


X

X

As previously stated, there are twentyone and twentyeight
different positions in each row/column. Placing
an actor removes three of those, leaving eighteen and twentyfive. 18+25+9 = 52.
Under these conditions the formula becomes 588+Σ(588(n*52)).
Up to a
certain number of actors, the number of staging possibilities grows with each
additional actor on stage as we add some number less than 588 to 588. At and beyond twelve actors, however, the
number of permutations shrinks with each addition. The eleventh actor adds sixteen possible
configurations because 588(52*11) = 16.
The twelfth actor removes thirtysix configurations, 588(52*12) = 36. Scholars debate the exact size of early
modern acting companies; however, the general consensus is that the troupes
were between eleven and fifteen actors.
These calculations support the lower end of that range, at least for
indoor stages like the Blackfriars Playhouse.
The larger outdoor stages such as the Globe may have allowed for larger
casts.
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