Author |
Topic: The Spider-Man of Paris (Read 176 times) |
|
alien2
Uberpuzzler
Gender: 
Posts: 6988
|
 |
The Spider-Man of Paris
« on: Jan 11th, 2024, 11:10am » |
 Quote  Modify
|
There are a dozen most valuable paintings on the walls of Musée d'Art Moderne de Paris: Rembrandt, Picasso, Monet etc. A thief knows this. He is also aware of a guard patrol in intervals of about an hour and a half. Each painting is worth 50.000 euros for his cut. But, even though he discovered flaws in security, that the alarm is faulted, he also finds out, a few days back from a reliable source, that one painting has the only working alarm. There is no way for him to know which painting has the hidden alarm, not without setting the alarm off. If the alarm goes off, there is no chance for him to be saved by the bell and he will rot in a prison cell for ten years. If you were in the shoes of this skillful thief, how many paintings would you steal? And what are the chances, statistically and mathematically speaking, for a relatively safe margin to steal minimum paintings, acquire a substantial compensation for your effort measured in tens even hundreds of thousands of euros, and feel with greater certainty that you will, hopefully, not be caught? I’d steal none.
|
| « Last Edit: Jan 11th, 2024, 11:17am by alien2 » |
 IP Logged |
|
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2873
|
|
Re: The Spider-Man of Paris
« Reply #1 on: Jan 12th, 2024, 10:15am » |
Quote Modify
|
There isn't a well-defined answer here. It depends on the relative value of 10 years imprisonment and large sums of money.
If you don't care about imprisonment at all, then you get the best expected payout by stealing half the paintings (with the expected value being a quarter of the total value of all paintings). That's also where the marginal value of stealing another painting becomes negative.
More generally, if you plan to steal a proportion, p, of the paintings, you have an expected monetary return of p(1-p) times the total value of all the paintings, and a probability of p of being arrested and imprisoned.
So in this case, if you value a decade's imprisonment at -V, you have an expected payout of 600,000*p(1-p) - Vp, so the break-even point in terms of value is at p=1 - V/600000.
Obviously, that treats p as continuous, while, with paintings being discrete objects, the actual proportion you steal must be one of only thirteen possibilities. If you value avoiding prison more than the value of all but one of the paintings (550,000 in this case) then you shouldn't attempt stealing any.
|
|
IP Logged |
|
|
|
alien2
Uberpuzzler
Gender:
Posts: 6988
|
|
Re: The Spider-Man of Paris
« Reply #2 on: Jan 16th, 2024, 1:31am » |
Quote Modify
|
Thank You for the reply and fine input, rmsgrey.
The Spider-Man of Paris II
Everything is the same as it is in the original riddle, except that this time there is the following addendum to the tip: “A reliable source assured the thief that there is one working alarm, hidden behind one painting out of dozen paintings lined up in a line on a single wall. However, the snitch offered another twist, that is to say, he claimed that, from what he found out, the hidden alarm is behind one of every either third, fourth or fifth painting, starting from either the left or from the right of the series of paintings.” What would be Your strategy (if you were the thief), hopefully best strategy, depending on amount of courage as well, especially 1) if you were aiming at becoming rich, ready to gamble, 2) or if you wanted desperately to avoid prison sentence, 3) or if you consider your personal strategy, possibly your current financial situation?
|
| « Last Edit: Jan 16th, 2024, 10:58am by alien2 » |
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: The Spider-Man of Paris
« Reply #3 on: Jan 16th, 2024, 8:41am » |
Quote Modify
|
on Jan 16th, 2024, 1:31am, alien2 wrote:| the hidden alarm is behind one of every third, fourth or fifth painting, starting from the left or from the right of the series of paintings. |
|
So, the dangerous paintings are numbers 3, 4, 5, 6, 8, 9, 10 and 12, counting from either end, so number 1 is number 12 from the other end, and number 7 is number 6 from the other end, leaving only numbers 2 and 11 guaranteed safe.
So you steal those two paintings, giving you 100,000 to lose in addition to the cost of being imprisoned. The previous analysis then applies but with slight changes. If p is now the proportion of the unsafe paintings you steal, and V remains your cost for being imprisoned, the expected gain in payout from stealing the rest becomes:
(500000p)(1-p) - (100000+V)p
with break-even at:
(500000p)(1-p) - (100000+V)p = 0
=> p = 0.8 - V/500000
In particular, if you rate avoiding prison over 350,000, you should only take the safe paintings.
However, for picking the correct proportion to steal, rather than looking for the break-even point, you should be looking for the maximum value, which is at:
p = 0.4 - V/1,000,000
If you are indifferent to being imprisoned, you should steal 4 of the unsafe paintings, giving you 6 in total, with a 60% chance of getting away with it.
In fact, for any number of safe paintings, a thief who doesn't care about prison but does want to maximise their expected value will walk away with half of them, or all the safe ones, whichever is more. Meanwhile, the more safe paintings there are, the less imprisonment needs to cost in order for a cautious thief to avoid all gambling.
|
|
IP Logged |
|
|
|
alien2
Uberpuzzler
Gender:
Posts: 6988
|
|
Re: The Spider-Man of Paris
« Reply #4 on: Jan 16th, 2024, 11:02am » |
Quote Modify
|
It's always good to hear from You, rmsgrey.
I modified the reply and added "either" two times in the variant of the riddle, just so there is no confusion, if there was one to begin with in the 1st place.
|
|
IP Logged |
|
|
|
alien2
Uberpuzzler
Gender:
Posts: 6988
|
|
Re: The Spider-Man of Paris
« Reply #5 on: Jan 16th, 2024, 4:12pm » |
Quote Modify
|
The Spider-Man of Paris III
There are a dozen valuable paintings lined up on a single wall of Musée d'Art Moderne de Paris. As it happens, because of a strange case of time travel and temporal paradox several months back, they happen to be all one and the same painting: “Impression, Sunrise” by Claude Monet. However, six paintings are the original, and six are fugazies. A skillful thief who broke in knows this, that is to say, a reliable source assured him of this a few months back. Still, he is not an art expert, so he can’t tell the difference. The snitch also assured him that either even or odd paintings, starting from left to right, happen to be the multilocated original. The thief is also aware of a guard patrol in intervals of about an hour and a half, and that each painting is worth 50.000 euros for his cut. But, even though he discovered flaws in security, that the alarm is faulted, he is up against the clock because of guard patrol. He can steal exactly six paintings, or else he’ll surely be caught and he will rot in a prison cell for maximum ten years. Which paintings should the thief steal, that is to say, what would be the best strategy for a partial success, so that he doesn’t flee entirely empty-handed? Or, for that matter, do You think that he should gamble in order to possibly maximize his success and become as rich as he can be, considering the win/loss ratio?
P.S. This variant could be intriguing and a lot more complex. Alas, I can't think of anything else at the moment.
|
| « Last Edit: Jan 16th, 2024, 4:21pm by alien2 » |
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: The Spider-Man of Paris
« Reply #6 on: Jan 17th, 2024, 7:54am » |
Quote Modify
|
For the third variation, your expected return from taking 6 paintings is always the same, so, if you're just concerned with the monetary return, the only question is how risk-averse or risk-seeking you are. If you're risk-averse, taking 3 of each type guarantees a 3-painting payout; if you're risk-seeking, taking 6 and 0 gives you either all or nothing.
You can get more interesting by delving into utility curves - if, rather than 2 paintings being twice as valuable to you as 1, 3 paintings being triple, and so on, if you instead have some other relative values for different numbers of paintings, then that changes your expected value for various strategies.
You have four options, each of which gives you a 50% chance of each of two possible outcomes - taking all the paintings from one set gives you 0 or 6; taking five from one and one from the other gives you 1 or 5; taking four and two gives you 2 or 4; and taking three of each gives you 3 or 3, which is just a guaranteed 3. Whichever pair gives you the highest total utility across the two possibilities would be your personal best pick.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print