An Unlikely Pairing

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 TI-84 plus (http://en.wikipedia.org/wiki/TI-84_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 unstressed-stressed-unstressed-stressed 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 one-hundred-forty alternating unstressed-stressed 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 twenty-nine feet wide by twenty-two 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 twenty-two 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 twenty-one combinations.  Doing the same with twenty-nine feet yields twenty-eight 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. 

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As previously stated, there are twenty-one and twenty-eight different positions in each row/column.  Placing an actor removes three of those, leaving eighteen and twenty-five.  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 thirty-six 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.