Stonenibblers

[ibob]

In the prison of the magical world Azkaban all prisoners are kept in solitary cells and have no way of communicating. 

One day Reginald Candlenut, an auror, burst into the office of chief warden Teophilius Betelfax.
"Mr Betelfax, I fear that two prisoners have somehow found a way to communicate despite the strong measures. Thanks to the charm Amplifo Petri I heard crackling in the prison walls. I made a large number of measurements in equal intervals and found that in 45% of them there is silence, in 38% there is a single crackling sound and in the remaining 17% there are two crackling sounds."
"I wouldn't be worried, Candlenut. It seems that two stonenibblers have found refuge into our prison's walls. Stonebibblers are very rare, but as a hobbyist magizoologist I know all about them. Each stonenibbler has a personal fixed probability to be nibbling on stones at a given moment. They are exceedingly rare, but are a good explanation of your observations."

Betelfax described the stonenibblers correctly, but are they really a good explanation of Candlenut's observations?

Edited September 4, 2020 by ibob

[plasmid]

Only if they're imaginary stonenibblers.

Spoiler

CaptainEd was on the right track. Using his notation, if one stonenibbler's probability of nibbling is p and the other's is q (so the probability of the first not nibbling is 1-p and the second not nibbling is 1-q), then the observed frequency of both, one, or neither nibbling at a time implies

A) 0.17 = pq
B) 0.38 = p(1-q) + q(1-p) = p + q - 2pq
C) 0.45 = (1-p) (1-q) = 1 - p - q + pq

Since pq = 0.17 based on equation A, plug that value into equation B and you have

0.38 = p + q - 0.34

So p+q must be 0.72. Since p+q = 0.72, you know q = 0.72-p. So substitute that for q in equation A and you have

0.17 = p(0.72-p), which can be rearranged to
p² - 0.72p + 0.17 = 0

And that's a quadratic formula that can be solved as (-b ± sqrt[b²-4ac]) / 2a. But the stuff in the square root is 0.5184 – 0.68, which is a negative number, so you can’t take its square root (at least not with real numbers).

So the prisoners are up to something, and it's time for warden Betelfax to punish them with a logic puzzle that will grant them freedom if solved or instant death if flubbed.

 

[CaptainEd]

This is similar to the proofreaders problem.

Yes, here’s why.

Assume the stonenibblers’ probabilities in a time interval are p and q, and they are independent events.
Then the probability of both nibbling in the same interval is pq.

The probability of only one in a time interval is p(1-q) + q(1-p).

The probability of no sound in an interval is (1-p)(1-q).

So we have three equations

(A) pq = .17

(B) p-pq + q-qp = p+q-2pq = .38

(C) 1-q-p+pq = .45

Substitute pq in (b) and (c)

(B) p+q -2*.17 = .38 p+q = .72

(C) 1-(p+q) + .17 = .45

1-.72+.17 =.45
.45=.45

So it does look like their noises occur independently.

[\spoiler]
 

[\spoiler]

[ibob]

On 9/4/2020 at 10:06 PM, CaptainEd said:

This is similar to the proofreaders problem.

  Reveal hidden contents

Yes, here’s why.

 

  Reveal hidden contents

 

Assume the stonenibblers’ probabilities in a time interval are p and q, and they are independent events.
Then the probability of both nibbling in the same interval is pq.

The probability of only one in a time interval is p(1-q) + q(1-p).

The probability of no sound in an interval is (1-p)(1-q).

So we have three equations

(A) pq = .17

(B) p-pq + q-qp = p+q-2pq = .38

(C) 1-q-p+pq = .45

Substitute pq in (b) and (c)

(B) p+q -2*.17 = .38 p+q = .72

(C) 1-(p+q) + .17 = .45

1-.72+.17 =.45
.45=.45

So it does look like their noises occur independently.

[\spoiler]
 

[\spoiler]

You just proved that 17+38+45-(17+38) equals 45. That's not useful information