[1/8] In the 1978 rock-opera of Faust, by Richard O’Brien and Tom Stoppard, there’s a wonderful duet between Faust and Mephistopheles where they’re negotiating the contract for Faust’s soul.
The Devil wants Faust to toss a 6-sided die every hour, and if the last 6 numbers seen
The Devil wants Faust to toss a 6-sided die every hour, and if the last 6 numbers seen
[2/8] are 6 5 4 3 2 1, his soul must descend to Hell.
Faust says, sure, but then if the numbers:
1 2 3 4 5 6
ever come up, his soul should ascend to Heaven.
Mephisto is having none of that, so they haggle. Faust argues that the infernal sequence should be:
6 6 6 6 6 6
Faust says, sure, but then if the numbers:
1 2 3 4 5 6
ever come up, his soul should ascend to Heaven.
Mephisto is having none of that, so they haggle. Faust argues that the infernal sequence should be:
6 6 6 6 6 6
[3/8] for symbolic reasons, and says, hey, this is no more nor less likely than the original choice.
So the Devil agrees.
Then Faust sings a monologue rejoicing, because he just gave himself a nearly 20% increase in his time on Earth enjoying his superpowers!
The drama critic
So the Devil agrees.
Then Faust sings a monologue rejoicing, because he just gave himself a nearly 20% increase in his time on Earth enjoying his superpowers!
The drama critic
[4/8] for the New York Times wrote a scathing attack on the play, claiming that Faust clearly gained no advantage! Despite Martin Gardner and others coming to Stoppard’s defence, controversy raged for months.
But it’s easy to see why Stoppard was right.
When the die has been
But it’s easy to see why Stoppard was right.
When the die has been
[5/8] tossed just 6 times, clearly either 666666 and 654321 are equally likely. But suppose neither comes up that soon.
Then there are 6^6–1 ways the first 6 throws might have gone. But if the fatal sequence is 666666, then the only ones that could damn on the 7th toss are
Then there are 6^6–1 ways the first 6 throws might have gone. But if the fatal sequence is 666666, then the only ones that could damn on the 7th toss are
[6/8] the *five* possibilities where the first toss was 1,...,5 and the rest were all 6s.
However, if the fatal sequence is 654321, then there are *six* possibilities to damn on the 7th: the first toss can be anything from 1,...,6, and the rest must be 65432.
This discrepancy
However, if the fatal sequence is 654321, then there are *six* possibilities to damn on the 7th: the first toss can be anything from 1,...,6, and the rest must be 65432.
This discrepancy
[7/8] between the two cases persists, and leads to the expected number of hours Faust will survive for the sequence 654321 being:
6^6
while for 666666 the expected number of hours is:
6^6 + 6^5 + 6^4 + 6^3 + 6^2 + 6
which is a fraction below 20% longer.
6^6
while for 666666 the expected number of hours is:
6^6 + 6^5 + 6^4 + 6^3 + 6^2 + 6
which is a fraction below 20% longer.
[8/8] Stoppard rigorously proved these formulas in a letter to the Times, but sadly they declined to publish it.
He proved that when k numbers of the target had shown up, the expectation values E_k for Faust’s remaining hours of survival obeyed the linear equations below:
He proved that when k numbers of the target had shown up, the expectation values E_k for Faust’s remaining hours of survival obeyed the linear equations below:
