Episode #709

[daniel1948]

Where I am, it’s now February 12.  I want to be the first, hopefully, to wish everyone a Happy Darwin Day ( or a Happy Lincoln Day if you’ll prefer).

Damn! It’s only 7:40 a.m. on February 11 here. And you’re already at the 12th. You’re in too much of a hurry. We’re more laid back about things where I am.

[Zerowantuthri]

When the rogues were talking about Shakespeare Cara mentioned we know who Shakespeare was, that he was a dude, that we know who he was married to and so on.

It is surprising to most people that we know very little about who Shakespeare was (yes he existed and was a man and we know who he married but for the most celebrated playwright of all time you'd think we'd know more.

All we know for certain is that Shaxpere, Shaxberd, or Shakespear, was born in Stratford in 1564, that he was an actor whose name is printed, with the names of his fellow actors, in the collected edition of his plays in 1623. We know that he married Anne Hathaway, and died in 1616, according to legend, on his birthday, St George's Day. The so-called "Stratfordian" case for Shakespeare rests on these, and a few other facts, but basically, that's it.

SOURCE: https://www.theguardian.com/culture/2010/mar/14/who-wrote-shakespeare-james-shapiro

[DevoutCatalyst]

The trolley would run over the fat man and continue on and kill the others, too. Congratulations, hero. 

This is basically what I said when I first heard this version of the trolley problem. Except I was a bit less certain. I said: “How do you know the fat man will stop the trolley? Maybe he misses the tracks, and is killed to no purpose. Or is crippled for life, also to no purpose.” But basically all versions of the trolley problem suffer from the same uncertainty: How do you ever know that your action will have the expected effect?

The trolley seen here is in Ostrava, Czech Republic. It's made by Stadler, the Tango NF2 model:



The vehicle weighs 36,835 kg empty. Were there a man heavy enough to derail that I wouldn't have the strength to make him budge. Plus he might judo me onto the tracks leaving me not half the person I used to be just a few moments ago.

[bachfiend]

When the rogues were talking about Shakespeare Cara mentioned we know who Shakespeare was, that he was a dude, that we know who he was married to and so on.

It is surprising to most people that we know very little about who Shakespeare was (yes he existed and was a man and we know who he married but for the most celebrated playwright of all time you'd think we'd know more.

All we know for certain is that Shaxpere, Shaxberd, or Shakespear, was born in Stratford in 1564, that he was an actor whose name is printed, with the names of his fellow actors, in the collected edition of his plays in 1623. We know that he married Anne Hathaway, and died in 1616, according to legend, on his birthday, St George's Day. The so-called "Stratfordian" case for Shakespeare rests on these, and a few other facts, but basically, that's it.

SOURCE: https://www.theguardian.com/culture/2010/mar/14/who-wrote-shakespeare-james-shapiro

I’m pretty certain I saw Shakespeare’s grave a few years ago when I visited Stratford-upon-Avon.  I could have visited Anne Hathaway’s cottage but I didn’t bother.

I went to a staging of ‘Hamlet’ by the Royal Shakespeare Company, and I’m very ashamed to admit that it was the first Shakespeare play I’d ever seen.  It was very long (I was extremely grateful for the interval) and ‘interesting.’  It was set in a modern African country with the actors armed with AK-47s, but talking in Elizabethan English.

I’m sometimes bemused by directors feeling the need to update the classics.  I went to a Wagner Ring Cycle in Leipzig last year.  In ‘Siegfried,’. Mime was conspiring to get Siegfried to kill Fafner with his sword, despite having his own AK-47, and could have done it himself.

DevoutCatalyst,

I’ve always imagined that the trolley in the runaway trolley problem was a very small one, perhaps one of those hand operated ones, small enough to make it plausible to be derailed by running into a large person.

If you’re going to imagine a very implausible scenario, then you can imagine anything.  Giving as an answer ‘it couldn’t happen’ wouldn’t be accepted by the investigators.

[DevoutCatalyst]

I’ve always imagined that the trolley in the runaway trolley problem was a very small one, perhaps one of those hand operated ones, small enough to make it plausible to be derailed by running into a large person.

But big enough to kill 5 people?

[2397]

With how many different ways Shaxpur spelled his name, maybe he was multiple people. All married to Anne Hathaway.

Anne's Bards.

[bachfiend]

I’ve always imagined that the trolley in the runaway trolley problem was a very small one, perhaps one of those hand operated ones, small enough to make it plausible to be derailed by running into a large person.

But big enough to kill 5 people?

Well, then you have to ‘imagine’ that it’s also large enough to kill 5 people too.  As I said, it’s not a very plausible scenario. 

People generally judge acts of commission more harshly than acts of omission.  Deliberately causing the death of one person is generally judged more harshly that allowing the death of 5.

More plausible scenarios can be devised.  Imagine you’re First Officer William Murdoch in charge on the bridge of the Titanic and you’re informed that there’s an iceberg very close directly ahead.  You’ve got only seconds to decide - should you attempt to steer away from the iceberg, probably not succeeding, or do nothing and run into it headon?  It seems to be a reversal of the act of commission/act of omission, but it’s not.  Attempting to steer away is the natural reflexive act.  Deliberately running headon into the iceberg is the act of commission, requiring strong will.

Most people would probably think that attempting to steer away from the iceberg was the correct one, despite knowing that it caused the death of 1500+ people.  And that deliberately running headon into the iceberg was wrong, despite ships surviving such collisions with no problems save a dented bow.  You could surmise that some people on the ship could have been killed in a headon collision if they happened to be descending stairways at the time of the collision.  Or leaning over the railing wondering what that rapidly approaching iceberg shaped object was...

[daniel1948]

I think there’s an assumption that trolleys are extremely easy to derail, that a pebble on the track would do it, and a fat man would be more than enough, but that anyone hit head on is a goner. I attribute this to the trolley-problem’s inventor’s being more concerned with creating moral dilemmas than with allowing for the realities of mass and momentum.

For a different choice in a similar situation, The Cowboy Fireman:

[bachfiend]

With how many different ways Shaxpur spelled his name, maybe he was multiple people. All married to Anne Hathaway.

Anne's Bards.

People, before Spellcheck, didn’t bother much with their spellings of names.  Adolf Hitler’s grandfather was a Hiedler.  His father changed his name to Hitler from Schicklgruber, which is unfortunate.  I don’t think Heil Schicklgruber would have caught on.  (Have I broken Godwin’s Law?)

[fuzzyMarmot]

To see this more easily, consider a planet where there are only three days per year, and try to find the probability that, in a group of three people, at least two people will share a birthday.

bachfiend's reasoning would lead to a result of 1-(2/3)^3=19/27.

The correct answer is 1-(2/3)*(1/3)=7/9.

The key error is the following: Just because there are three pairs, and a pair has probability 2/3 of not sharing a birthday, it doesn't mean you get to say that the probability of all three pairs not sharing birthdays is (2/3)^3. That multiplicative property of probabilities is only valid for independent events.

No, it wouldn’t.  My reasoning would go along the line that if there’s three people One, Two and Three, and that if there’s only three days in the year, then One could have a birthday on any of the three days, and the chance is therefore 3/3, Two could have a birthday on either of the other two days, so as to not share a birthday, so the chance is therefore 2/3, and Three must have a birthday on the remaining, again not to share a birthday with One and Two, with a chance of 1/3.  So the odds are 3/3 x 2/3 x 1/3 or 6/27 or 2/9.  And hence the chance of at least two sharing a birthday
is 1 - 2/9 ie 7/9.  Which is the answer I’m supposed to have got.

Oh, wait.  I now see it.  My original explanation was apparently the correct one - it’s the product of 364/365, 363/365, ...  I shouldn’t have taken any notice of the simplified method of calculation.

I'm glad that you were able to identify the error in your explanation. Everyone makes mistakes; the key is admitting when one is wrong.

[daniel1948]

...
I went to a staging of ‘Hamlet’ by the Royal Shakespeare Company, and I’m very ashamed to admit that it was the first Shakespeare play I’d ever seen.  It was very long (I was extremely grateful for the interval) and ‘interesting.’  It was set in a modern African country with the actors armed with AK-47s, but talking in Elizabethan English.

I’m sometimes bemused by directors feeling the need to update the classics.  I went to a Wagner Ring Cycle in Leipzig last year.  In ‘Siegfried,’. Mime was conspiring to get Siegfried to kill Fafner with his sword, despite having his own AK-47, and could have done it himself.

Interesting. I saw a performance of Romeo and Juliette by the Royal Shakespeare Company, in which, at the beginning of the play, the guards have machine guns and shoot them off to break up the fight. But the rest of the play was as the Bard wrote it. Juliette was white, and played by an actress much older than the character’s 13 years, and Romeo was a Rastafarian. But none of the language was changed. It was an amazing performance. I could have done without the machine guns, but after that first scene they didn’t get in the way of the play.

There are semi-mythological figures whose historicity may be questioned, and Steve would do well to get a historian on the show. But I think that we know enough about Shakespeare to regard him as a real person. BTW, at the time he was writing, English spelling had not yet been standardized. There’s nothing odd, in the context of the time, in the variability of the spelling of his name. We’ve standardized it now, but it’s a mistake to make a big deal about the different spellings used then, even by the man himself.

[arthwollipot]

I haven't listened to this episode yet (it's at the top of my queue), but I'd like to address a pet peeve of mine that I'm absolutely sure will annoy the crap out of me when I get to the segment.

I, and every Australian and British person I know, understand that Robin Hood has two names. First name Robin, surname Hood. Robin Hood. Americans, however, in almost all cases, pronounce it as though it is only one name: Robinhood. With the emphasis on the first syllable.

The weird thing is that Americans don't do that with other names with the same syllable count. Charlie Brown, for example, is not pronounced Charliebrown. Donald Trump is not Donaldtrump.

Okay, carry on. I might have more to say when I've listened to the episode.

From the Wikipedia entry on Robin Hood, under “Early References” comes this quote:

From 1261 onward, the names 'Robinhood', 'Robehod' or 'Robbehod' occur in the rolls of several English Justices as nicknames or descriptions of malefactors.

So there is some precedent for Robinhood as a single name.

That's a good point. But I think by the time of Errol Flynn, it was pretty much locked in as two names.

[bachfiend]

To see this more easily, consider a planet where there are only three days per year, and try to find the probability that, in a group of three people, at least two people will share a birthday.

bachfiend's reasoning would lead to a result of 1-(2/3)^3=19/27.

The correct answer is 1-(2/3)*(1/3)=7/9.

The key error is the following: Just because there are three pairs, and a pair has probability 2/3 of not sharing a birthday, it doesn't mean you get to say that the probability of all three pairs not sharing birthdays is (2/3)^3. That multiplicative property of probabilities is only valid for independent events.

No, it wouldn’t.  My reasoning would go along the line that if there’s three people One, Two and Three, and that if there’s only three days in the year, then One could have a birthday on any of the three days, and the chance is therefore 3/3, Two could have a birthday on either of the other two days, so as to not share a birthday, so the chance is therefore 2/3, and Three must have a birthday on the remaining, again not to share a birthday with One and Two, with a chance of 1/3.  So the odds are 3/3 x 2/3 x 1/3 or 6/27 or 2/9.  And hence the chance of at least two sharing a birthday
is 1 - 2/9 ie 7/9.  Which is the answer I’m supposed to have got.

Oh, wait.  I now see it.  My original explanation was apparently the correct one - it’s the product of 364/365, 363/365, ...  I shouldn’t have taken any notice of the simplified method of calculation.

I'm glad that you were able to identify the error in your explanation. Everyone makes mistakes; the key is admitting when one is wrong.

My error was accepting brilligtrove’s answer as to the solution (reply #5).  I was overjoyed to find an easier method of calculating the probabilities that I didn’t examine it to ensure that it was actually correct.  My method, described in reply #2 is laborious with having to multiply 364/365, 363/365, 362/365, etc, etc, etc.  There’s no easy way of calculating it.  You have to multiple 22 different numbers if there’s 23 people.  364!/342! divided by 365^22 won’t work, at least not easily.  171! can’t be calculated on my Scientific calculator, let alone 364!

[bachfiend]

...
I went to a staging of ‘Hamlet’ by the Royal Shakespeare Company, and I’m very ashamed to admit that it was the first Shakespeare play I’d ever seen.  It was very long (I was extremely grateful for the interval) and ‘interesting.’  It was set in a modern African country with the actors armed with AK-47s, but talking in Elizabethan English.

I’m sometimes bemused by directors feeling the need to update the classics.  I went to a Wagner Ring Cycle in Leipzig last year.  In ‘Siegfried,’. Mime was conspiring to get Siegfried to kill Fafner with his sword, despite having his own AK-47, and could have done it himself.

Interesting. I saw a performance of Romeo and Juliette by the Royal Shakespeare Company, in which, at the beginning of the play, the guards have machine guns and shoot them off to break up the fight. But the rest of the play was as the Bard wrote it. Juliette was white, and played by an actress much older than the character’s 13 years, and Romeo was a Rastafarian. But none of the language was changed. It was an amazing performance. I could have done without the machine guns, but after that first scene they didn’t get in the way of the play.

There are semi-mythological figures whose historicity may be questioned, and Steve would do well to get a historian on the show. But I think that we know enough about Shakespeare to regard him as a real person. BTW, at the time he was writing, English spelling had not yet been standardized. There’s nothing odd, in the context of the time, in the variability of the spelling of his name. We’ve standardized it now, but it’s a mistake to make a big deal about the different spellings used then, even by the man himself.

I didn’t know whether I should have mentioned it, but, well, all the actors were heavily pigmented - but they were very good.    After the first few minutes, it wasn’t noticeable.  The text was exactly the same as the original.

Not only weren’t spellings standardised, but the spelling of many words has changed.  The same in German.  I’m going to a performance of ‘die Zauberflöte’ in two weeks, and in preparation I bought the German libretto on my Kindle to avoid having to read the surtitles.  Mozart (or at least his librettist) spelt ‘drei’ (three) as ‘drey,’ and Mozart is only late 18th century.  Wagner spelt many words differently to Germans today.  Neanderthal (as in Neanderthal Man) nowadays would be Neandertal.  Bach’s Cantata 147 was originally ‘Herz und Mund und That (instead of Tat) und Leben’ (it still is, in some recordings).

[fuzzyMarmot]

To see this more easily, consider a planet where there are only three days per year, and try to find the probability that, in a group of three people, at least two people will share a birthday.

bachfiend's reasoning would lead to a result of 1-(2/3)^3=19/27.

The correct answer is 1-(2/3)*(1/3)=7/9.

The key error is the following: Just because there are three pairs, and a pair has probability 2/3 of not sharing a birthday, it doesn't mean you get to say that the probability of all three pairs not sharing birthdays is (2/3)^3. That multiplicative property of probabilities is only valid for independent events.

No, it wouldn’t.  My reasoning would go along the line that if there’s three people One, Two and Three, and that if there’s only three days in the year, then One could have a birthday on any of the three days, and the chance is therefore 3/3, Two could have a birthday on either of the other two days, so as to not share a birthday, so the chance is therefore 2/3, and Three must have a birthday on the remaining, again not to share a birthday with One and Two, with a chance of 1/3.  So the odds are 3/3 x 2/3 x 1/3 or 6/27 or 2/9.  And hence the chance of at least two sharing a birthday
is 1 - 2/9 ie 7/9.  Which is the answer I’m supposed to have got.

Oh, wait.  I now see it.  My original explanation was apparently the correct one - it’s the product of 364/365, 363/365, ...  I shouldn’t have taken any notice of the simplified method of calculation.

I'm glad that you were able to identify the error in your explanation. Everyone makes mistakes; the key is admitting when one is wrong.

My error was accepting brilligtrove’s answer as to the solution (reply #5).  I was overjoyed to find an easier method of calculating the probabilities that I didn’t examine it to ensure that it was actually correct.  My method, described in reply #2 is laborious with having to multiply 364/365, 363/365, 362/365, etc, etc, etc.  There’s no easy way of calculating it.  You have to multiple 22 different numbers if there’s 23 people.  364!/342! divided by 365^22 won’t work, at least not easily.  171! can’t be calculated on my Scientific calculator, let alone 364!

Many calculators have a built-in binomial coefficient function, so that you won't need to calculate 365! directly. You can also use an online calculator like Wolfram Alpha if you don't want to be limited by display digits. For example, simply enter 1-23!*Binomial[365,23]/(365^23)

You made an error in not recognizing that your second solution yielded a different formula than your first. Not a big deal, but it is good to recognize the mistake and learn from it. I've made more than my share over the years, that is for sure