# Logic Puzzles

I have a short attention span for these kinds of debates, debates that have no resolution.

I declare you the winner.

BTW, I leaned logic reading Sherlock Holmes.

If you don't like people deriding your posts as simplistic, then don't post such simplistic dreck, and definitely don't then assume (out loud) that the lack of interest from other members is due to their lack of ability.

Is it also simplistic dreck?

Twenty minutes of invigorating fun!! (Cheaper than pleasuring yourself on Floating Fortune Road.)

If you don't like people deriding your posts as simplistic, then don't post such simplistic dreck, and definitely don't then assume (out loud) that the lack of interest from other members is due to their lack of ability.

Is it also simplistic dreck?

Twenty minutes of invigorating fun!! (Cheaper than pleasuring yourself on Floating Fortune Road.)
Clearly the butler did it. Why else would he have destroyed the evidence?

Sorry if my posts are off-topic. I took the thread title "Logic Puzzles" as a general injunction to post Logic Puzzles and/or their solutions.

Here's one of the Greats. It didn't make the rounds till circa 21st century so some haven't heard it:

Four prisoners are brought to the warden's throne room and presented with a challenge:
On the morrow, a game will be played. Each of the four prisoners will be assigned a color randomly, Red or Green. Each prisoner will be informed of his fellow prisoners' colors, but not his own. Each prisoner will write one of three words on his ballot: Red, Green, or Abstain.
• If at least one prisoner guesses his color correctly, and none of the four guesses his color wrong, then all four will be released, given a million silver shekels, and married to one of the warden's four beautiful daughters.
• If any Red prisoner writes Green then all four will be executed.
• If any Green prisoner writes Red then all four will be executed.
• If all four write Abstain they will all be sent back to the Dungeon and perhaps given another opportunity after another 10 years.
The prisoners are allowed to discuss with each other on the eve of the game. What strategy should they adopt to maximize their chance at the marriages and silver shekels?​

~ ~ ~ ~ ~ ~ ~ ~

If you don't like people deriding your posts as simplistic, then don't post such simplistic dreck, and definitely don't then assume (out loud) that the lack of interest from other members is due to their lack of ability.

Is it also simplistic dreck?

Twenty minutes of invigorating fun!! (Cheaper than pleasuring yourself on Floating Fortune Road.)
Clearly the butler did it. Why else would he have destroyed the evidence?

BRILLIANT!! Still, one is faced with the stated problem -- naming the butler's accomplices.

I have a short attention span for these kinds of debates, debates that have no resolution.
It's not a debate but a set of logic puzzles all of which have "resolutions."
I declare you the winner.
No. In a sense you're the winner. You've honestly admitted that you didn't know the answer to this puzzle and have been willing to learn the answer not denying that knowledge. "Losers" are those who pretend to know what they don't know and who deny that those who have knowledge do have that knowledge.
BTW, I leaned logic reading Sherlock Holmes.
That's a start. I like Schaum's Outline of Logic.

Keping with degree of difficulty in the OP.

A plabe crashes on the border between two countries. All are killed. Where are the survivers burried?

Hopefully in the countries they came from.
uhhhh...if they all died so how could there be surivors to be buried?
Everybody dies.
Says who?
Uh, isn't the punchline supposed to be that we don't want to bury any survivors?

Keping with degree of difficulty in the OP.

A plabe crashes on the border between two countries. All are killed. Where are the survivers burried?

Hopefully in the countries they came from.
uhhhh...if they all died so how could there be surivors to be buried?
Everybody dies.
Says who?
Uh, isn't the punchline supposed to be that we don't want to bury any survivors?
Nobody survives forever.

### Puzzle: Who murdered Fyodor Karamazov?​

I've composed a wide variety of logic puzzles off-and-on for several decades. Mostly for fun, but Dell Logic Puzzles paid me $100-$125 for very difficult puzzles. Here's one I composed two years ago for this message board! ... Getting only one solution and minimal applause I'll repost it.

In a certain district of Russia, many of the men are members of either the Cult of Knights or the Cult of Knaves. The Knights utter only true statements; the Knaves only false. Even when guilty of a crime they are, strangely, bound by this code.

A few years ago, Chief Inspector Plato was called upon to solve some crimes in that province, and usually was able to test guilt by asking
"If Someone of your cult was witness to this crime and I asked him if you were guilty, would he say Yes?"

He had trained the local police about this technique, and was seldom called to help anymore. He could spend much of his time contemplating how to attract interest from the lovely Anastasia. But today he was called in urgently. One of the notables had died in suspicious circumstances.

The local citizenry had adopted a counter-measure to inquisitive police: they would ignore the questions asked and just make arbitrary statements that happened to be True (if they were Knights) or False (if they were Knaves). This created a lot of extra work for the detectives but they had workarounds. For starters they often water-boarded the suspects for a while, then asked whether they wanted the torture to stop. This separated the Knights from the Knaves very easily! But they couldn't do that with the Chief Inspector in town!

Anyway, C.I. Plato was surprised that he was called on to attend an inquest. And he had mixed feelings when he learned that it was the death of Fyodor Karamazov that was being investigated.

You see, Fyodor was neither a Knight nor a Knave, but a Knucklehead: He always spoke two sentences at a time, with exactly one of the sentences (either the 1st or the 2nd) true and the other sentence false. Although it was disrespectful to say so aloud, Plato was somewhat relieved that Fyodor was the deceased: the times he had questioned this old man got very confusing. But now it was Fyodor's murder that was under investigation.

Or was he even murdered? The local police had determined that Fyodor's three sons were all present at the death and all knew exactly what had happened. Either one, or two, or all three had participated in the murder of their father. Or maybe it was an accident, and all three were innocent. There was nothing whatsoever to go on; the butler had blundered and cleaned up all the material evidence.

Plato prepared to question the three brothers. "Which are knaves and which are knights?", he asked. The local police became apologetic. "Nobody knows, they kept to themselves all these years. And — sorry about this Chief Inspector — it's very possible that one or more of them are Knuckleheads like their father. At least we're sure that each brother is in one of the three cults. And of course the brothers know which cult each of their brothers are in."

What The F**k, Plato thought to himself in Russian, as he prepared his questions.
"Did you kill your father?", Plato suddenly yelled at Dmitri, hoping to take him by surprise.

• (1) Exactly one of us three brothers is guilty of the old man's murder.
• (2) Exactly one of us three brothers is a Knucklehead.
Plato couldn't make much of this. How many Knuckleheads were there? What if they are all Knuckleheads, he groaned. Great, thought Plato, just great. He tried again, shouting the same question at Ivan.

• (3) Exactly two of us three brothers are guilty of the old man's murder.
• (4) Exactly one of us three brothers is a Knight.
Without being prompted, Alexei chimed in.
• (5) I am not guilty of my father's death.
• (6) None of us three are Knuckleheads.
It was Dmitri's turn again. He said
• (7) None of us three are Knights.
• (8) Papa had a mole on his left ankle.
"Go check the body for moles," said Plato decisively. "Fyodor has already been cremated" was the sheepish reply. On inquiry it turned out that nobody except for the sons had ever seen the old man with his boots off, even when he was taking comfort at the local brothel.

"Are you guilty of your father's death, Ivan? A simple yes or no, please."
• (9) No, I am not guilty.
• (10) Papa had a mole on his right ankle.
By now the Chief Inspector was whimpering. "Can you help me, Alexei?" That brother answered
• (11) Exactly one of us three brothers is a Knave.
• (12) There are exactly two true statements among (1), (7) and (10).
Now we might be getting somewhere, thought Plato. They're beginning to slip. But just then the Karamazov attorney showed up, telling his clients to shut up. And they'd all be leaving the next morning on a train to Paris if C.I. Plato couldn't crack the case by dawn.

Can you help? Which, if any, of the brothers killed their father? Which cults are they in, anyway?

Well, if I were writing a program to solve this I would probably try all possible combinations until I encountered the first combination that satisfied all of the requirements. We have 3 brothers, Dmitri, Ivan, and Alexei. Each of them may be one of three cults, Knight, Knave, or Knucklehead. Each of them makes 4 statements. Each statement may be either true or false. So that's 3x3x4x2 = 72 possible combinations. This would explain why the puzzle is not attempted.

Ideally, there are simplifications available. But one would only learn those by prior experience with similar puzzles. One simplification might be to assume that all three are knuckleheads, like their father, because all three
always spoke two sentences at a time

Nobody survives forever.

But hopefully they will not be buried while they continue to survive.

### Puzzle: Who murdered Fyodor Karamazov?​

I've composed a wide variety of logic puzzles off-and-on for several decades. Mostly for fun, but Dell Logic Puzzles paid me $100-$125 for very difficult puzzles. Here's one I composed two years ago for this message board! ... Getting only one solution and minimal applause I'll repost it.

Well, if I were writing a program to solve this I would probably try all possible combinations until I encountered the first combination that satisfied all of the requirements. We have 3 brothers, Dmitri, Ivan, and Alexei. Each of them may be one of three cults, Knight, Knave, or Knucklehead. Each of them makes 4 statements. Each statement may be either true or false. So that's 3x3x4x2 = 72 possible combinations. This would explain why the puzzle is not attempted.
I don't know about your 3x3x4x2. Did you intend 33·24·3? Or perhaps multiply by 32 again to account for the murderers' identities and the two moles. This is still trivially small for a computer program. But manual solution is MUCH easier. Right off the bat:
. . . . Dmitri said: (7) None of us three are Knights.
It is impossible for Dmitri to be a Knight.

It is trivial to make arbitrarily complex logic puzzles. The puzzles that Dell paid \$125 for were ridiculously difficult, but evidently there's a market for them. Best I think, are puzzles that are not too difficult but not too easy. But I guess solving logic puzzles is just NOT a common hobby.

Composing logic puzzles is FUN. This was one of the first "Knucklehead" puzzles I composed, and I didn't bother with more because I decided this one was "perfect"! The parameters (3 suspects, 12 statements) seemed near the "Goldilocks" value. Most of the statements are VERY straightforward, along with the two throwaways involving moles. The straightforward statements fall into place as deductions are made. This "nifty" statement
. . . . Alexei said (12) There are exactly two true statements among (1), (7) and (10).
becomes key near the end of the solution process. But having more than one statement of that type might be too sadistic! My friend solved this puzzle in 20 minutes. . . . But that's 20 minutes wasted if solving them isn't fun.

I do use a computer program to help me design such puzzles. Not only does this double-check that there is exactly one solution, but I can tune it to eliminate unnecessary clues, and so on.

I've composed a wide variety of puzzles other than this Dell Logic genre, but to say much more might lead Googlers to my Real Name, which I am (half-heartedly?) trying to keep concealed.

Here's one of the Greats. It didn't make the rounds till circa 21st century so some haven't heard it:

Four Three prisoners are brought to the warden's throne room and presented with a challenge:​
On the morrow, a game will be played. Each of the four three prisoners will be assigned a color randomly, Red or Green. Each prisoner will be informed of his fellow prisoners' colors, but not his own. Each prisoner will write one of three words on his ballot: Red, Green, or Abstain.​
• If at least one prisoner guesses his color correctly, and none of the four three guesses his color wrong, then all four three will be released, given a million silver shekels, and married to one of the warden's four three beautiful daughters.
• If any Red prisoner writes Green then all four three will be executed.
• If any Green prisoner writes Red then all four three will be executed.
• If all four three write Abstain they will all be sent back to the Dungeon and perhaps given another opportunity after another 10 years.
The prisoners are allowed to discuss with each other on the eve of the game. What strategy should they adopt to maximize their chance at the marriages and silver shekels?​

This is a WONDERFUL puzzle which should appeal to fans of combinatorics and information theory. Even very smart people usually give up and think 50% is the best that can be achieved.

I did NOT compose this one, except for replacing the number of prisoners (3) with 4: a "red herring." I have reverted that change above.

Well, if I were writing a program to solve this I would probably try all possible combinations until I encountered the first combination that satisfied all of the requirements. We have 3 brothers, Dmitri, Ivan, and Alexei. Each of them may be one of three cults, Knight, Knave, or Knucklehead. Each of them makes 4 statements. Each statement may be either true or false. So that's 3x3x4x2 = 72 possible combinations. This would explain why the puzzle is not attempted.
I don't know about your 3x3x4x2. Did you intend 33·24·3? Or perhaps multiply by 32 again to account for the murderers' identities and the two moles.

It's like a pair of dice. Each has six sides. For any given side of dice1, dice2 can have 6 possibilities. So, 6x6 gives the total number of combinations. 1:1 1:2 1:3 1:4 1:5 1:6, 2:1 2:2 2:3 ... etc. If we had 3 dice, it would be 6x6x6, as in 1:1:1 1:1:2 1:1:3 etc.

This is still trivially small for a computer program. But manual solution is MUCH easier. Right off the bat:
. . . . Dmitri said: (7) None of us three are Knights.
It is impossible for Dmitri to be a Knight.

That's not clear to me. The sentence "At least we're sure that each brother is in one of the three cults" does not rule out the possibility that they are all in the same cult! If each brother is a Knucklehead, then "each brother is in one of the three cults" is satisfied. We would need to add a qualification like "and each brother is in a different cult".

... I do use a computer program to help me design such puzzles. Not only does this double-check that there is exactly one solution, but I can tune it to eliminate unnecessary clues, and so on. ...

Cool. I've often thought it would be easy to write a program to generate (and solve) SODUKU puzzles. I would rather solve the programming problem than to solve a puzzle. The only puzzles I bother with these days are the ones over on lichess.org. And I get a lot of those wrong.

Well, if I were writing a program to solve this I would probably try all possible combinations until I encountered the first combination that satisfied all of the requirements. We have 3 brothers, Dmitri, Ivan, and Alexei. Each of them may be one of three cults, Knight, Knave, or Knucklehead. Each of them makes 4 statements. Each statement may be either true or false. So that's 3x3x4x2 = 72 possible combinations. This would explain why the puzzle is not attempted.
I don't know about your 3x3x4x2. Did you intend 33·24·3? Or perhaps multiply by 32 again to account for the murderers' identities and the two moles.

It's like a pair of dice. Each has six sides. For any given side of dice1, dice2 can have 6 possibilities. So, 6x6 gives the total number of combinations. 1:1 1:2 1:3 1:4 1:5 1:6, 2:1 2:2 2:3 ... etc. If we had 3 dice, it would be 6x6x6, as in 1:1:1 1:1:2 1:1:3 etc.

I DO know how to count combinations. Your "3x3x4x2" led me to wonder if YOU know. Your dice example does not explain your "3x3x4x2."

This is still trivially small for a computer program. But manual solution is MUCH easier. Right off the bat:
. . . . Dmitri said: (7) None of us three are Knights.
It is impossible for Dmitri to be a Knight.

That's not clear to me. The sentence "At least we're sure that each brother is in one of the three cults" does not rule out the possibility that they are all in the same cult! If each brother is a Knucklehead, then "each brother is in one of the three cults" is satisfied. We would need to add a qualification like "and each brother is in a different cult".
We DO agree on
. . . The sentence "At least we're sure that each brother is in one of the three cults" does not rule out the possibility that they are all in the same cult!

The point is that if Dmitri is a Knight "None of us three are Knights" is false. But Knights speak only true statements.

So, we are having a review of Probability An Statistics 101?

Given a continuous normal probability distribution from 1 to 3 what is the probability 0f 2.0?

When we make make measurements we take the arithmetic average of repeated measurements as an estimate of a true value. Why is this valid?

I have three rods cumming off a continuous production line. The mean and standard deviations are
3 0.4
2.5 0 .2
4 0.8.
When put end to end what are the max and min lengths? What is probability of three rods together equaling 9.5 meters?

Let Marvin give it a try first.

I DO know how to count combinations. Your "3x3x4x2" led me to wonder if YOU know. Your dice example does not explain your "3x3x4x2."

3 brothers, 3 cults per brother, 4 statements per brother, 2 truth values (T or F) per statement. That was my assessment of the possible combinations. Brother 1 could be in cult 1, or cult 2 or cult 3. Brother 1 also made 4 statements, each of which could be true or false. Etc. So, we could have B1:C1:S1:T, B1:C1:S1:F, B1:C1:S2:T, B1:C1:S2:F etc.

As you rotated through each combination, you would test whether the statements were consistently true or consistently false, and finish when all of the statements are consistent.

The point is that if Dmitri is a Knight "None of us three are Knights" is false. But Knights speak only true statements.

So, that would eliminate all combinations in which B1:C1 is the case (B1=Dmitri and C1=Knight). If the puzzle is sound then there will only be one specific combination in which each statement is either true and false.

So, we are having a review of Probability An Statistics 101?

Given a continuous normal probability distribution from 1 to 3 what is the probability 0f 2.0?

When we make make measurements we take the arithmetic average of repeated measurements as an estimate of a true value. Why is this valid?

I have three rods cumming off a continuous production line. The mean and standard deviations are
3 0.4
2.5 0 .2
4 0.8.
When put end to end what are the max and min lengths? What is probability of three rods together equaling 9.5 meters?

Let Marvin give it a try first.

I was a Psych major and we covered statistics in the Tests and Measurements course. But that was long ago and I'd have to look up stuff, so I'll skip this one.

Here's one of the Greats. It didn't make the rounds till circa 21st century so some haven't heard it:

Four Three prisoners are brought to the warden's throne room and presented with a challenge:​
On the morrow, a game will be played. Each of the four three prisoners will be assigned a color randomly, Red or Green. Each prisoner will be informed of his fellow prisoners' colors, but not his own. Each prisoner will write one of three words on his ballot: Red, Green, or Abstain.​
• If at least one prisoner guesses his color correctly, and none of the four three guesses his color wrong, then all four three will be released, given a million silver shekels, and married to one of the warden's four three beautiful daughters.
• If any Red prisoner writes Green then all four three will be executed.
• If any Green prisoner writes Red then all four three will be executed.
• If all four three write Abstain they will all be sent back to the Dungeon and perhaps given another opportunity after another 10 years.
The prisoners are allowed to discuss with each other on the eve of the game. What strategy should they adopt to maximize their chance at the marriages and silver shekels?​

This is a WONDERFUL puzzle which should appeal to fans of combinatorics and information theory. Even very smart people usually give up and think 50% is the best that can be achieved.

I did NOT compose this one, except for replacing the number of prisoners (3) with 4: a "red herring." I have reverted that change above.
Not sure that's a red herring -- adding the 4th prisoner makes the problem harder to analyze. Three can get to 75% chance of marriage/silver easily:

If a prisoner is told the other two are the same color, he guesses he's the opposite. If a prisoner is told the other two are opposite colors, he abstains. 75% chance of marriage/silver, 25% chance of execution.

The best solution I've so far found to the four-prisoner puzzle achieves more than 50% but less than 75% chance of marriage/silver:

If a prisoner is told the other three are the same color, he guesses he's the opposite. If a prisoner is told the other three are not all the same color, he abstains. 50% chance of marriage/silver immediately, 12.5% chance of execution immediately, X * 37.5% chance of repeating the game in 10 years, where 0 < X < 100% is the unspecified probability of playing again. 50% < Total chance of marriage/silver < 68.75%.

(All this analysis assumes the warden is playing fair. If he intends to assign the colors randomly but not with 50-50 probability or with nonzero correlation among the prisoners' colors, all bets are off.)

Keping with degree of difficulty in the OP.

A plabe crashes on the border between two countries. All are killed. Where are the survivers burried?

Hopefully in the countries they came from.
uhhhh...if they all died so how could there be surivors to be buried?
Everybody dies.
Says who?
Everybody knows Scully is immortal.

I DO know how to count combinations. Your "3x3x4x2" led me to wonder if YOU know. Your dice example does not explain your "3x3x4x2."

3 brothers, 3 cults per brother, 4 statements per brother, 2 truth values (T or F) per statement. That was my assessment of the possible combinations. Brother 1 could be in cult 1, or cult 2 or cult 3. Brother 1 also made 4 statements, each of which could be true or false. Etc. So, we could have B1:C1:S1:T, B1:C1:S1:F, B1:C1:S2:T, B1:C1:S2:F etc.
I'm not going to wade through this ( " : ... : ... : etc." )

Instead I found it fun to whip up a program which I think implements your plan. It prints
. . . . Numcases = 884736. That's 3^3 * 2^15. That's the number I mentioned multiplied by 2x2x2 -- the possible guilts by brother.
(No need to worry about moles!)

Code:
#include        <stdlib.h>
#include        <stdio.h>

char    *Tname[3] = {
"Knight", "Knuck", "Knave",
};

char    *Murds[8] = {
"none.", "Dmit.", "Ivan.", "Dm+Iv", "Alex.", "Dm+Al", "Iv+Al", "All_3",
};

int     Tval[13];       // The 12 truth values
int     Bt[3];          // Cults per brother: 0,1,2 = Knight, Knuck, Knave
int     Popcult[3];     // Number in a cult

int     Guilt;          // !!(Guilt & 1 << brix)
#define NUMMU           (!(Guilt & 1) + !(Guilt & 2) + !(Guilt & 4))
// Brothers:  0[1] - Dmitri, 1[2] = Ivan; 2[4] = Alexei

int wrongt(int btyp, int sentix1, int sentix2) {
return btyp != !Tval[sentix1] + !Tval[sentix2];
}

void tryit(void) {
int     i;

// case (1): // D: Exactly one of us three brothers is guilty.
if (Tval[1] != (NUMMU == 1))
return;
// case (2): //  D: Exactly one of us three brothers is a Knucklehead.
if (Tval[2] != (Popcult[1] == 1))
return;
// case (3): //  I: Exactly two of us three brothers are guilty.
if (Tval[3] != (NUMMU == 2))
return;
// case (4): //  I: Exactly one of us three brothers is a Knight.
if (Tval[4] != (Popcult[0] == 1))
return;
// case (5): //  A: I am not guilty of my father's death.
if (Tval[5] != !(Guilt & 4))
return;
// case (6): //  A: None of us three are Knuckleheads.
if (Tval[6] != (Popcult[1] == 0))
return;
// case (7): //  D: None of us are Knights.
if (Tval[7] != (Popcult[0] == 0))
return;
// case (8): //  D: Papa had a mole on his left ankle.
// case (9): //  I: No, I am not guilty.
if (Tval[9] != !(Guilt & 2))
return;
// case (10): //  I: Papa had a mole on his right ankle.
// case (11): //  A: Exactly one of us three brothers is a Knave.
if (Tval[11] != (Popcult[2] == 1))
return;
// case (12): //  A: Exactly two true among (1), (7) and (10).
if (Tval[12] != (2 == 0
+ Tval[1]
+ Tval[7]
+ Tval[10]))
return;
printf("Solution: Dmitri is %s Ivan is %s Alexei is %s Murderers are %s\n",
Tname[Bt[0]], Tname[Bt[1]], Tname[Bt[2]], Murds[Guilt]);
for (i = 1; i < 13; i++) printf("%s%c%s",
i == 1 ? " Truthvals: " : "",
Tval[i]["FT"],
i == 12 ? "\n" : "");
}

int main(int argc, char **argv)
{
long     Numcases = 0;

Popcult[1] = Popcult[2] = Popcult[3] = 0;
for (Bt[0] = 0; Popcult[Bt[0]]++, Bt[0] < 3; Popcult[Bt[0]]--, Bt[0]++)
for (Bt[1] = 0; Popcult[Bt[1]]++, Bt[1] < 3; Popcult[Bt[1]]--, Bt[1]++)
for (Bt[2] = 0; Popcult[Bt[2]]++, Bt[2] < 3; Popcult[Bt[2]]--, Bt[2]++)
for (Guilt = 0; Guilt < 1 << 3; Guilt++) {

#define         TV(ix)       for (Tval[ix] = 0; Tval[ix] < 2; Tval[ix]++)

TV(1) TV(2) TV(3) TV(4)
TV(5) TV(6) TV(7) TV(8)
TV(9) TV(10) TV(11) TV(12) {
++Numcases;
if (wrongt(Bt[0], 1, 2)) continue;
if (wrongt(Bt[1], 3, 4)) continue;
if (wrongt(Bt[2], 5, 6)) continue;
if (wrongt(Bt[0], 7, 8)) continue;
if (wrongt(Bt[1], 9, 10)) continue;
if (wrongt(Bt[2], 11, 12)) continue;
tryit();
}
}
printf("%ld cases\n", Numcases);
exit(0);
}

Doubtless several of you will find parts of this C code unpleasant. While criticizing please post your proposed replacement.

If the puzzle is sound ...
IF my puzzle is sound? Same to you, buddy!

Not sure that's a red herring -- adding the 4th prisoner makes the problem harder to analyze.

Three can get to 75% chance of marriage/silver easily:

If a prisoner is told the other two are the same color, he guesses he's the opposite. If a prisoner is told the other two are opposite colors, he abstains. 75% chance of marriage/silver, 25% chance of execution.

BRAVO! Did you find this yourself just now? I always thought you were a very smart guy. More than most puzzles, this one stumps many of the best solvers. It's just so "obvious" that you can't outperform 50%. Do you have an INTUITIVE explanation of how you can beat 50%?
The best solution I've so far found to the four-prisoner puzzle achieves more than 50% but less than 75% chance of marriage/silver:
Aha! But my red herring was indeed just a herring ... but a tricky herring if I do say so myself. You still get 75% easily:

Pick one of the players. Call him George.

George always Abstains, and the other three players all ignore George.

(All this analysis assumes the warden is playing fair. If he intends to assign the colors randomly but not with 50-50 probability . . .
Pedant. "Randomly" used informally like this always means "Randomly with a uniform distribution."
If "Randomly" is the intent, "Randomly but not necessarily uniformly" would be written.