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I saw a version with two colors, where 99.5 people are saved. Can't be bothered to think about three, but maybe if you're not too lazy.
And in general, I log into HeroesWorld to degrade.

I saw a simpler version, but it was more focused on the **PSYCHOLOGY** that we love.

The first respondent knows the number of hats of each color for 99 people. The second respondent knows the number of hats of each color for 98 people + the answer of the first + whether or not the first was executed. I'll figure out how to put this into a formula. Or maybe I won't.

Out of 4, only 3 can be saved because, based on the first person's answer, 2 won't be able to distinguish between "red, red, red" and "blue, red, red". So, unfortunately, 3 out of 2 is the maximum.

Oops... I tricked everyone...:D
The first person looks at 2, 3, and 4.
If all 3 colors are the same, they name that color. If 3 and 4 have the same color, and 2 is different, they name the remaining color (which none of the other 3 have). If all 3 are different, they name the color of the second person. If 2 and 4 have the same color, and 3 is different, they name the color of the third person. If 2 and 3 have the same color, and 4 is different, they name the color of the fourth person.

The second person listens to the first person's answer and looks at the remaining ones.
If 3 and 4 have colors that match the first person's answer, they repeat the answer. If 3 and 4 have the same color, but it's different from the first person's answer, they name the remaining color. If 3 and 4 have different colors and they are different from the first person's answer, they repeat the first person's answer. If 3 and 4 have different colors and one of them matches the first person's answer, they name the other color.

The third person listens to the first two answers and looks at 4.
If the first and second answers match the color of 4, they repeat the answer. If the first and second answers match, but are different from the color of 4, they name the remaining color. If the first and second answers are different and they are different from the color of 4, they name the color of 4. If the first and second answers are different and one of them matches 4, they name the other color.

The fourth person listens to the first three answers.
If all 3 answers are the same, they repeat the answer. If the first and second answers match, but are different from the third answer, they name the remaining color. If all 3 answers are different, they repeat the third answer. If 2 answers are the same (except for the case where 1 and 2 are the same!), and the third is different, they repeat the answer that is different.

2, 3, and 4 are 100% safe! Probably...

Added 7 minutes later

I have a feeling that the answer is that all but one can be saved, and that one has a chance to be saved.

It seems that this is indeed the case, but I'm too lazy to go through 100 samples.:D

Come on, Hermit, give us your answer. We give up...:smile32:

Maybe I'll wait a bit longer before providing the voiceover and just say that the closest correct answer is what Uranium235 wrote.

I have a feeling that the answer is that 99 people can be saved, but only one has a chance to be saved.
This is based on my memory of the problem; I don't remember the solution.

Yes. 99 people are guaranteed to be saved, and one has only a chance!

The first person to answer knows the number of hats of each color for 99 people.
The second person to answer knows the number of hats of each color for 98 people, plus the first person's answer, and whether or not the first person was executed.
I'll figure out how to put this into a formula, or maybe I won't.

This contains everything that's needed and even some extra information. Maybe you can try to figure out how each person can determine the color of their hat based on what the wise men said before them and the colors of the hats of those standing behind them?

I tried to derive a formula based on the difference between the number of hats of different colors, but I couldn't. So, I'll write it down; maybe it will give someone an idea.
The idea was to come up with a color for each value of the function of the differences (the problem is with the function itself), which the first person to answer would name.
For example, if the value is 0, it's green; if it's 1, it's red; if it's 2, it's blue; if it's 3, it's green again... In general, as an option, N (where N is a natural number and 0) means green, N+1 means red, and N+2 means blue.

As for the function itself, I tried adding all 3 differences (taking the absolute value of the difference, of course), as well as any 2 of them. Something doesn't work.
That is, I had this: let's say there were 36 red hats, 32 blue, and 31 green in front of the first person to answer.
He adds, for example, all 3 differences: 4+5+1. He gets 10. 10 is the color of the hat, and he names it.
The second person counts the hats in front of him (and he heard "red," so he understands that the sum of the differences in front of the last person was 1, 4, 7, 10, 13...
He counts the differences in front of him (let's say he has a red hat, but he doesn't know it. Then there should be 35 red, 32 blue, 31 green. The sum of the differences is 8.
He figures that the difference couldn't have changed much because his hat wasn't taken into account; for the first person, it was 7 or 10.
Then, if I have a green hat, the second person thinks, the first person would have: 3+3+0. That doesn't work.
If I have a blue hat, he would have: 2+2+4=8. That doesn't work.
With a red hat, it obviously works.
But now let's consider another case, if the second person actually has a blue hat.
Then he figures (he sees 36 red, 31 blue, 31 green): the difference is 10.
And then it turns out that a red hat doesn't suit him (because then the difference would be 12 for the first person, but that's not red, it's green).
But he can't choose between blue and green because the difference turns out to be the same because they are now equal.

In the first case, the third wise man would probably be able to calculate his difference and compare it with the answers of the first and second. But again, in the case of a tie, he falls asleep.
I tried to modify the formula, but I didn't succeed.

For example, if there were 2 colors of hats, only blue and red, it would be easier. There are a total of 99 hats in front of the first person. Let's say 51 are red and 48 are blue. And agree in advance that if the number of red hats is even, you name the color red; if it's odd, you name the color blue. So, he names blue.
The second person looks (let's say he has a red hat). In front of him, there are 50 red and 48 blue. And the first person said that there was an odd number of red hats in front of him. Therefore, seeing an even number of red hats, the second person understands that he has a red hat.
Similarly, if he had a blue hat, the number of red hats would still be odd, and the second person would understand this.
The third person reasons in the same way: there was an odd number of red hats in front of the first person, and the second person named red. Therefore, if the third person has a red hat, the number of red hats in front of him is odd; if he has a blue hat, it's even.
But I can't figure out how to apply a similar formula for three colors.

For example, you can look at the divisibility by certain numbers or the divisibility between the differences, but I can't derive a formula, although it probably exists.

There was another idea: to calculate the sum of the hats with coefficients: 1*number of red + 2*number of blue + 3*number of green. (with the same principle - the sum is divisible by three - the first person names green, divisible by three + 1 red, three + 2 blue). And the second person counts his sum and compares it, but something went wrong again.

Okay, here's the solution.

It is indeed possible to save 99 wise men.

The wise men agree to assign a numerical value to each color, for example, red - 0, white - 1, blue - 2.

The first wise man calculates the sum modulo 3 of the color values of all the wise men standing in front of him:
(0+0=0
0+1=1
0+2=2
1+1=2
1+2=0
2+2=1) and says the color corresponding to this number. (Hopefully, it will match the color of his hat.) The second wise man calculates the sum modulo 3 of the color values of all the wise men standing behind him and subtracts it modulo 3 from the number said by the first wise man:

(0-1=2
0-2=1
1-0=1
1-1=0
1-2=2
2-0=2
2-1=1
2-2=0). He then determines the color of his hat. The third wise man subtracts from the number said by the first wise man the sum modulo 3 of all the wise men standing behind him and the number said by the second wise man (the color of the second wise man's hat). The fourth wise man subtracts modulo 3 from the number said by the first wise man the sum of the answers given by the second and third wise men and the sum modulo 3 of the colors of all the wise men standing behind him, and so on. That is, the J-th wise man determines his color:

c_j=a₁-(a₂+a₃+...+a_j-1)-(c_j+1+c_j+2+...+c₁₀₀)

c - colors, and a - the answers of the wise men.
Starting with the second wise man, the answers will definitely match the colors.
All arithmetic operations must be performed modulo 3.

Uranium came closest to the answer, so let him be the one to solve it, although Yaschur also did a great job and probably guaranteed to save 75 wise men.

Added 10 minutes later
P. S. And I also have a thought that it is generally possible to save any number of wise men with any number of hat colors. You just need to add modulo a larger number - the number of hat colors.

Well, well... This solution doesn't seem to match the "examples" mentioned in the question. It states that if colors 2 and 3 are the same, then player 1 should name that specific color. But according to the "algorithm," if the colors are white, then player 1 should name... blue.

Added 8 minutes later

From the second wise man onwards, the answers will definitely match the colors.

Based on what article of the Criminal Code

Based on which article of the Criminal Code, theorem, axiom, etc., is such a conclusion made?

Because from x+y=z it follows that mod(x)+mod(y)=mod(z).
If the wise men have a hash sum of all the hats, then, by successively subtracting from this sum, they should be able to figure out their own hat color.

It is more profitable to travel on the second one than on the first. For example, for the same money, you can travel an extra 40 centimeters from Omsk to Moscow.
Name the first and second ones.

Car and rail roads.

A train and a horse? )
That's a rather unconventional comparison. 40 cm is a very small distance; if it were about transportation, any error in cost calculation (tickets, fuel consumption, and horse feed) would result in distances much greater than these forty centimeters. I think the point lies elsewhere.

No, a commuter train and a horse are not the correct answer.

Something seems a bit unconventional. 40 cm is a very small distance; if it were transportation, any error in cost (tickets, consumption, and fuel price, horse feed) would result in distances much greater than these forty centimeters. I think the essence lies in something else.

The upper shelf, due to the Earth's sphericity, describes a longer trajectory than the lower one. This is noticeable, of course, only at large distances. That's why they travel to Moscow from Omsk, and not from some Ryazan... :D And discounts on upper berths are indeed not uncommon.