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During Act V, Scene 2 of Hamlet, Hamlet’s uncle Cladius wagers that Hamlet will win in a duel with Laertes. In order to make sure that Hamlet participates, Cladius “places a valuable stake, at odds, on Hamlet’s skill” and sends the courtier Osric to tell Hamlet:
Hamlet: …But, on: six Barbary horses against six French swords, their assigns, and three liberal-conceited carriages; that’s the French bet against the Danish?
Osric: The King, Sir, hath laid, that in a dozen passes between yourself and him, he shall not exceed you by three hits: he hath laid on twelve for nine; and it would come to immediate trial… .
Hamlet: Let the foils be brought… and the King hold his purpose, I will win. For him if I can; if not I will gain nothing but my shame and the odd hits.
In other plays, Shakespeare shows a vast knowledge of the rules and fencing, and he is famous for his attention to detail. However, the terms of the wager were confusing and critics did not want to “bestir themselves in the cold unpoetick dabble of calculation”. For over 400 years this passage and in particular the conditions under which Claudius wins the wager remained a mystery.
There are three believable interpretations. In a match of 12 passes, Claudius loses and Laertes wins if:
Version I: Laertes’ total hits exceed Hamlet’s by at least 3 (Hamlet wins if he makes 5 hits before Laertes makes 8 hits).
Version II: Laertes is the first to make 3 successive hits (whoever first makes three successive hits wins; if no one does in 12 passes, then Hamlet is the winner).
Version III: Laertes is the first to achieve a total of 3 more hits than his opponent (the match is won by the player who first obtains an excess of 3 hits more than his opponent; if this “excess of 3” does not occur in 12 passes, then Hamlet wins by default).
Sprinchorn, a drama professor at Vassar, considers the first two versions, and Saunders, a math professor at Washington State Univeristy, considers the third version and explains why it is the version that Shakespeare used.
Assume that both participants are equally matched at fencing. Then consider flipping a fair coin labelled H and L a total of 12 times. The probability that L will occur at least 8 times is
Then the odds for Laertes are only 1 for 4 instead of the 9 for 12 given.
Assume that both participants are equally matched at fencing. Then consider flipping a fair coin labelled H and L a total of 12 times. What is probability that neither L nor H occurs 3 times in a row?
The number of possible trials remaining is 4096 – 466 = 3630. By symmetry 1815 have a triplet of L first. The probability of Laertes winning is
and probability of Hamlet winning is 0.557.
This is closer, but then the odds for Laertes are only 3 for 7 instead of the 9 for 12 given.
Let Pr[D] be the probability that a default occurs; in other words that in 12 passes
neither is ever ahead by 3. Let Pr[H] and Pr[L] be the probabilities that Hamlet or Laertes respectively win by scoring 3 more hits than the other. Then assume that
Pr[H] + Pr[L] + Pr[D]=1 and Pr[H] = Pr[L].
giving the equitable probability of 4 for 7 that Claudius wins.
Saunders, Sam C., “Could Shakespeare Have Calculated the Odds in Hamlet’s Wager?” The Oxfordian. Volume X (2007): 20-34 http://shakespeare-oxford.com/wp-content/oxfordian/Saunders-Wager.pdf
Shakespeare, William. “Hamlet.” Arden Shakespeare. Ed. Harold Jenkins. Walton-on-Thames, Surrey:
Sprinchorn, Evert. “The Odds on Hamlet.” Columbia U. Forum 8 (1964): 41-5; reprinted in The Practice of
Modern Literary Scholarship. Ed. Sheldon P. Zinter. Glenview, IL: Scott, 1966.