Today's problem reviews a topic you covered in your school daze.
Let's test your memory -

In King Midas' realm, gold is the only valid currency
(those were the days). His minter stamps coins,
100 per box. But he's a shady character, who plugs
one coin in each box.

The king's treasurer, suspicious, samples one randomly
selected coin from each box. Consider a run of 100
boxes. What's the chance the minter gets away, scot free?

No need for an exact number, a formula will do.

In the land of Asymptopia, they mint n coins per box.
Again, one plugged coin per box. And one sample
tested from each box. Again, we consider n boxes tested.

Now, what's the chance the minter gets the last laugh,
as n --> infinity?

--
Rich

On Saturday, July 17, 2021 at 8:50:10 PM UTC-4, RichD wrote:
> Today's problem reviews a topic you covered in your school daze.
> Let's test your memory -
>
> In King Midas' realm, gold is the only valid currency
> (those were the days). His minter stamps coins,
> 100 per box. But he's a shady character, who plugs
> one coin in each box.
>
> The king's treasurer, suspicious, samples one randomly
> selected coin from each box. Consider a run of 100
> boxes. What's the chance the minter gets away, scot free?
>
> No need for an exact number, a formula will do.
>
> In the land of Asymptopia, they mint n coins per box.
> Again, one plugged coin per box. And one sample
> tested from each box. Again, we consider n boxes tested.
>
> Now, what's the chance the minter gets the last laugh,
> as n --> infinity?

The chance of getting away with each box is independent of the other boxes and is 99/100. (99/100)^n gives you the probability of "getting away with it" for n boxes, so 0.3660323412732 (approximately) for 100 boxes.

The probability tends to 0 as the number of boxes tends to infinity unless he cuts the treasurer in on the deal.

--

Rick C.

- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209

On 18/07/2021 01:50, RichD wrote:
> Today's problem reviews a topic you covered in your school daze.
> Let's test your memory -
>
> In King Midas' realm, gold is the only valid currency
> (those were the days). His minter stamps coins,
> 100 per box. But he's a shady character, who plugs
> one coin in each box.
>
> The king's treasurer, suspicious, samples one randomly
> selected coin from each box. Consider a run of 100
> boxes. What's the chance the minter gets away, scot free?
>
> No need for an exact number, a formula will do.
>
> In the land of Asymptopia, they mint n coins per box.
> Again, one plugged coin per box. And one sample
> tested from each box. Again, we consider n boxes tested.
>
> Now, what's the chance the minter gets the last laugh,
> as n --> infinity?

What does "plugs" mean in this context?

--
Regards,
Martin Brown

On Sunday, July 18, 2021 at 5:26:03 AM UTC-4, Martin Brown wrote:
> On 18/07/2021 01:50, RichD wrote:
> > Today's problem reviews a topic you covered in your school daze.
> > Let's test your memory -
> >
> > In King Midas' realm, gold is the only valid currency
> > (those were the days). His minter stamps coins,
> > 100 per box. But he's a shady character, who plugs
> > one coin in each box.
> >
> > The king's treasurer, suspicious, samples one randomly
> > selected coin from each box. Consider a run of 100
> > boxes. What's the chance the minter gets away, scot free?
> >
> > No need for an exact number, a formula will do.
> >
> > In the land of Asymptopia, they mint n coins per box.
> > Again, one plugged coin per box. And one sample
> > tested from each box. Again, we consider n boxes tested.
> >
> > Now, what's the chance the minter gets the last laugh,
> > as n --> infinity?
> What does "plugs" mean in this context?

Fake coins.

--

Rick C.

+ Get 1,000 miles of free Supercharging
+ Tesla referral code - https://ts.la/richard11209

On 18/07/2021 01:50, RichD wrote:
> Today's problem reviews a topic you covered in your school daze.
> Let's test your memory -
>
> In King Midas' realm, gold is the only valid currency
> (those were the days). His minter stamps coins,
> 100 per box. But he's a shady character, who plugs
> one coin in each box.
>
> The king's treasurer, suspicious, samples one randomly
> selected coin from each box. Consider a run of 100
> boxes. What's the chance the minter gets away, scot free?

The King's treasurer is unfortunately far too dim to solve the problem
efficiently. All you need is a balance and a bit of patience and a
binary chop procedure to isolate the one coin with the wrong mass.

It is highly unlikely that the plug is the same mass as a gold coin of
those dimensions although one made of tungsten might pass the density
test it wouldn't pass the bite test...

> No need for an exact number, a formula will do.
>
> In the land of Asymptopia, they mint n coins per box.
> Again, one plugged coin per box. And one sample
> tested from each box. Again, we consider n boxes tested.
>
> Now, what's the chance the minter gets the last laugh,
> as n --> infinity?

Too boring to be worth bothering to answer.

--
Regards,
Martin Brown

On gnuarm.del...@gmail.com wrote:
>> Today's problem reviews a topic you covered in your school daze.
>> Let's test your memory -
>> In King Midas' realm, gold is the only valid currency
>> Hiss minter stamps coins, 100 per box. But he's a shady character,
>> who plugs one coin in each box.
>> The king's treasurer, suspicious, samples one randomly
>> selected coin from each box. Consider a run of 100
>> boxes. What's the chance the minter gets away, scot free?
> > In the land of Asymptopia, they mint n coins per box.
> > Again, one plugged coin per box. And one sample
> > tested from each box. Again, we consider n boxes tested.
>> Now, what's the chance the minter gets the last laugh,
>> as n --> infinity?
>
> The chance of getting away with each box is independent of the other boxes
> and is 99/100.
> (99/100)^n gives you the probability of "getting away with it" for n boxes,
so 0.366032 (approximately) for 100 boxes.

That's easy.
> The probability tends to 0 as the number of boxes tends to infinity
Massive fail.
There are n coins per box, and n boxes get are sampled.

--
Rich

On Monday, July 19, 2021 at 7:47:04 PM UTC-4, RichD wrote:
> On gnuarm.del...@gmail.com wrote:
> >> Today's problem reviews a topic you covered in your school daze.
> >> Let's test your memory -
> >> In King Midas' realm, gold is the only valid currency
> >> Hiss minter stamps coins, 100 per box. But he's a shady character,
> >> who plugs one coin in each box.
> >> The king's treasurer, suspicious, samples one randomly
> >> selected coin from each box. Consider a run of 100
> >> boxes. What's the chance the minter gets away, scot free?
> > > In the land of Asymptopia, they mint n coins per box.
> > > Again, one plugged coin per box. And one sample
> > > tested from each box. Again, we consider n boxes tested.
> >> Now, what's the chance the minter gets the last laugh,
> >> as n --> infinity?
> >
> > The chance of getting away with each box is independent of the other boxes
> > and is 99/100.
> > (99/100)^n gives you the probability of "getting away with it" for n boxes,
> so 0.366032 (approximately) for 100 boxes.
>
> That's easy.
> > The probability tends to 0 as the number of boxes tends to infinity
> Massive fail.
> There are n coins per box, and n boxes get are sampled.

Yeah, I didn't notice that n was also increasing the size of the box. So it would be ((n-1) / n)^n or

(n-1)^n / n^n

Yeah, not sure. The ratio itself tends to one as n increases, but taking the power of n makes it smaller as n increases. So not sure if it goes to 1 or 0. I'm wondering if (n-1)^n forms a well known power series. Not in a math mood at the moment.

--

Rick C.

--- Get 1,000 miles of free Supercharging
--- Tesla referral code - https://ts.la/richard11209

On 20/07/2021 02:26, Rick C wrote:
> On Monday, July 19, 2021 at 7:47:04 PM UTC-4, RichD wrote:
>> On gnuarm.del...@gmail.com wrote:
>>>> Today's problem reviews a topic you covered in your school daze.
>>>> Let's test your memory -
>>>> In King Midas' realm, gold is the only valid currency
>>>> Hiss minter stamps coins, 100 per box. But he's a shady character,
>>>> who plugs one coin in each box.
>>>> The king's treasurer, suspicious, samples one randomly
>>>> selected coin from each box. Consider a run of 100
>>>> boxes. What's the chance the minter gets away, scot free?
>>>> In the land of Asymptopia, they mint n coins per box.
>>>> Again, one plugged coin per box. And one sample
>>>> tested from each box. Again, we consider n boxes tested.
>>>> Now, what's the chance the minter gets the last laugh,
>>>> as n --> infinity?
>>>
>>> The chance of getting away with each box is independent of the other boxes
>>> and is 99/100.
>>> (99/100)^n gives you the probability of "getting away with it" for n boxes,
>> so 0.366032 (approximately) for 100 boxes.
>>
>> That's easy.
>>> The probability tends to 0 as the number of boxes tends to infinity
>> Massive fail.
>> There are n coins per box, and n boxes get are sampled.
>
> Yeah, I didn't notice that n was also increasing the size of the box. So it would be ((n-1) / n)^n or
>
> (n-1)^n / n^n
>
> Yeah, not sure. The ratio itself tends to one as n increases, but taking the power of n makes it smaller as n increases. So not sure if it goes to 1 or 0. I'm wondering if (n-1)^n forms a well known power series. Not in a math mood at the moment.

You chose a very unfortunate way to write it down.
You should recognise this classic standard form.

lim n-> inf (1-1/n)^n

(although it is usually written with the opposite sign)

It is a variant of expanding (1+x)^n with x = -1/n in this case

--
Regards,
Martin Brown

On 2021-07-19, RichD <r_delaney2001@yahoo.com> wrote:
> On gnuarm.del...@gmail.com wrote:
>>> Today's problem reviews a topic you covered in your school daze.
>>> Let's test your memory -
>>> In King Midas' realm, gold is the only valid currency
>>> Hiss minter stamps coins, 100 per box. But he's a shady character,
>>> who plugs one coin in each box.
>>> The king's treasurer, suspicious, samples one randomly
>>> selected coin from each box. Consider a run of 100
>>> boxes. What's the chance the minter gets away, scot free?
>> > In the land of Asymptopia, they mint n coins per box.
>> > Again, one plugged coin per box. And one sample
>> > tested from each box. Again, we consider n boxes tested.
>>> Now, what's the chance the minter gets the last laugh,
>>> as n --> infinity?
>>
>> The chance of getting away with each box is independent of the other boxes
>> and is 99/100.
>> (99/100)^n gives you the probability of "getting away with it" for n boxes,
> so 0.366032 (approximately) for 100 boxes.
>
> That's easy.
>
>> The probability tends to 0 as the number of boxes tends to infinity
>
> Massive fail.
> There are n coins per box, and n boxes get are sampled.

the probability of winning ( not being dioscovered) is (1-(1/n))^n

or / n-1 \ n
| --- |
\ n /


or (n-1)^n
-------
n^n

That tends asymptotically to 1/e

--
Jasen.