The paradox goes like this:
Let’s say there is a crazy, all-powerful man who one day kidnaps one person and then rolls a fair die. If a six comes up, he kills the person, and if any other number comes up, he releases him and takes a number of people double what he took before and does the same thing to them. He rolls the die, and if a six comes up he kills the people he kidnapped, and if any other number comes up, he releases them and takes double the number of people he took. So on the second day there are two people, on the third four people, and so on.

One day you wake up and realize that you’ve been kidnapped by him, but you don’t know how many people were kidnapped with you and you don’t know when he began taking people—whether you’re on the first day or some later day. Still, he gives you the option of pressing a button that gives you a fifty-fifty lottery—either to die or to be released. What should you do? Press the fifty-fifty button, or go with the one-in-six die roll?

The paradox says that on the one hand, a one-in-six chance is a better chance, but on the other hand it’s likely that you’re not on the first day, and on every later day there’s a higher chance that you’ll die (because if no one has died until now, it’s more likely that it’ll happen now). In addition, if you look at it in terms of how many people are supposed to die, then on the second day two out of three can die, and on the third day four out of seven, and so on. That is, on every day there’s a greater than half chance that you’re in the group that will die, and therefore it’s better to press the button and turn your odds into fifty-fifty.

There is no connection at all between the days. Each day an independent die roll is made, and therefore your chance of dying on any given day is 1/6. It doesn’t matter what day you’re on. So I don’t understand what the paradox is here.
Your description at the end is just an unclear jumble of words. What does it mean that on the second day 2 out of 3 will die? Why are you accumulating the people from all the days? In short, this looks like pure nonsense to me. Neither a paradox nor anything of the sort.

Dvir,
First, you explained the paradox badly.
Second, like many pseudo-paradoxes, it isn’t really a paradox but an exercise meant to show that your intuition about a certain thing is wrong.

The 50/50 option is basically an escape attempt—and then if you succeed (50%) you leave the die statistics.
The idea is that the last group (the one that gets killed) is larger than the total number of all the other kidnapped people in all the earlier groups put together (in the original paradox, the number of captives increases by multiplying itself), and therefore even though statistical intuition says to prefer the die option (1/6), you should prefer the 50/50 option.
Because if you take a random person from all those ever kidnapped, the probability that he is in the last group that gets killed is greater than 50%, and therefore you should assume that you are in the group that will be executed. Very unintuitive, but that’s the answer considered correct. I have to say that I always feel uncomfortable with exercises like this, as well as with the anthropic principle. As is well known, we could also say: what are the odds that we were born now and not in the future, when the world population reaches one hundred trillion people.
But on the face of it, the calculation is correct.

A really confusing thing you wrote, which has nothing at all to do with solving the “paradox,” and which completely threw me off when I read your explanation, is: “if no one has died until now, it’s more likely that it will happen now.”
That has absolutely nothing to do with the paradox, and it’s a silly statistical mistake; the probability is of course reset on every roll.

I hope I clarified the paradox above. If not, I’d be happy to explain it briefly.
What does the Rabbi think about the paradox?
Apparently, according to Bayes’ principle, one really should calculate the probability that I was kidnapped in the first place and which group I was kidnapped with. But it’s hard to digest the solution. At the end of the day, the die still hasn’t been rolled, and it hasn’t yet been decided which group is the last one. Right now we’re facing a 1/6 roll that will determine the group’s fate, and it’s not clear how the additional information affects the die.

You didn’t clarify it.

The crazy murderer works like this:
He kidnaps one person and rolls a die. If a 6 comes up he kills him; if not, he releases him and kidnaps 2 people. He rolls a die again; if a 6 does not come up, he releases them and kidnaps 4, and so on.
You find yourself tied up and blindfolded, without knowing how many people were kidnapped with you. You have the option of trying to escape, and the chance that you die in the escape attempt is 50%.
The supposed solution is that despite the intuitive scandal, you should escape.

In the end the die will eventually come up 6, and some group will be executed. One has to notice that the number of people in the group that is executed will always be larger than the number of all the people in all the previous groups together. (Suppose there were 4 rounds. In the first there was 1 person, in the second 2, in the third 4, and in the fourth 8—who were murdered. Altogether 15 people participated in the game; of them 7 were saved and 8 were murdered. This of course works for any number of rounds that occur.) According to Bayes’ principle (I’m bringing Bayes’ principle in here; I’m rusty, so I hope I’m not mistaken), you calculate the probability that you were kidnapped in one of the previous rounds versus the chance that you were kidnapped in the final round that gets executed. The calculation shows that the chance that you were kidnapped in the final round is greater than 50%, and therefore you should risk escaping.

I think the description you gave is also not precise/complete. Two points need clarification:
1. You wrote that in each round they release the people immediately after the roll. In my opinion, the earlier rounds are not released, but rather they wait blindfolded until the game ends. Their fate has already been sealed (for mercy and not for punishment), but they do not know this. Now you are sitting there with all the blindfolded people and facing the dilemma whether to escape or not. Everyone is offered this deal (to escape), and the question is whether they should accept it. The chance for each of them to die from the die is greater than half. And the chance of escape is half.
2. But here there is another point that needs clarification: if I fail to escape, do I die in any case, or am I then handed over to whatever fate came up on the die? Because if I am only handed over to the fate of the die, then obviously it’s worth trying to escape. So I assume the case is that I will be murdered in any event if I don’t succeed in escaping, even if I am from one of the earlier rounds that got life on the die.
In such a situation it is obvious that it is worth trying to escape, because your chance of dying is only half, and if you don’t escape the chance is greater than half.
In short, if you describe the case precisely, I don’t see any dilemma or any problem there. The solution is obvious. The dilemma exists only because the description is imprecise.

1. What difference does it make whether the groups that may have preceded you (and maybe you’re the very first) were released or are being held tied up?

2. I wrote explicitly: the probability of death in escape is 0.5. You can simply say that you’re given the option to press a button and receive a 50% chance of death and a 50% chance of release.

And in the solution, basically you don’t calculate the die—you calculate the groups. What is the probability that you are a captive in one of the groups that were released, versus the probability that you are in the group that gets killed.
The calculation is simple. In four rounds:
8/15 > 0.5

The solution really is simple, and I already wrote that this is a pseudo-paradox. But its unintuitiveness is really troubling.
If we make it more extreme, and say that the group grows by multiplying itself, you are kidnapped, and you face the dilemma—can you stand before the die roll and say: “I am 90% sure that a 6 will come up”?
And that is an absolute scandal!

If I didn’t explain it properly, I’d appreciate it if you used the principle of charity.

If every group that was spared by the lottery is released immediately, then the deal is offered only to the last group and not to everyone. In that case it is obvious that it is better to try to escape, because otherwise you are certain to die (because if they offered it to you, then you are necessarily in the last group). So obviously the deal is offered to all the groups, and the earlier ones are not released.
However, one can now think of the following policy of offer: indeed the offer is made to everyone, but not all at once. They simply offer the deal to each group after it has been kidnapped (we saw that this must be the case); whoever tried to escape and succeeded escaped, and whoever didn’t, they kill him (even though a 6 did not come up for him). Is that the case description you are proposing?
Notice that in such a case, when I receive the offer to escape, it is not at all clear that we have already reached the last group. It may be that all the groups until now, including mine, are supposed to be released. In that situation it seems better not to escape. Though on second thought, it still seems one must look at the complete game that will continue afterward until the end. This requires further analysis.
Think about a situation where they offer me the deal before they roll the die. Now I consider whether to accept the deal, in which case there is a half chance I die, or wait for the die roll, in which case there is a one-sixth chance I die. What difference does it make which group I’m in? Here it seems simple that one should not escape but rather wait for the roll.
However, if they have already rolled the die and only then offer me the deal, I can make my own calculation as follows: in the lottery they made about me, either a 6 came up or it didn’t. The probability that a 6 came up there is 1/6, so it is better not to escape. The question is whether to look at it this way, or to look at the probability that I belong to the last group. It seems to me that at least if they offer me the deal before the die is rolled for my group, it is incorrect to look at the whole game rather than my own situation.
In any case, that is why I thought the game had already ended and everyone was sitting there blindfolded, and only now they are being offered a lottery, as I explained above. That is a clearer case for discussion. But if the previous case is equivalent to it, then I understand why they talked about it so that the answer would not be trivial. But as I said, it still needs to be clarified exactly when the offer to escape is made.

The paradox proceeds exactly as you described. Sorry that I didn’t describe it well enough. Of course it is possible that no group has yet been killed.
I was referring to the “paradox” where they ask you before the die is rolled. That surely seems to be the intended meaning. But even then, mathematically I don’t understand why not to calculate the entire game. In my humble opinion, that is how it should be calculated according to Bayes’ principle. Except that then one gets a blatantly absurd result.
Am I missing something? What prevents us from calculating the whole game? It shouldn’t be relevant to me whether it has ended yet or not, or whether the die has already been rolled—that is exactly the calculation: whether I appear at the end of the game (and am destined to be murdered) or in the middle of it.

It shouldn’t make any difference at all whether the die has already been rolled or not. It’s like rolling the die, having nobody look at it, and covering it with a cup. Still, according to the calculation, if you arrived in such a game that ends when a six is rolled, then most likely the die is showing six under the cup. And that is a completely paradoxical result.
Because you calculate: what is the probability that I arrived in the game at a stage where the die came up six, versus the probability that I arrived at a stage where it came up some other number. And it makes no difference at all whether it was rolled and covered with a cup, or whether it has not yet been rolled at all.It shouldn’t make any difference at all whether the die has already been rolled or not. It’s like rolling the die, having nobody look at it, and covering it with a cup. Still, according to the calculation, if you arrived in such a game that ends when a six is rolled, then most likely the die is showing six under the cup. And that is a completely paradoxical result.
Because you calculate: what is the probability that I arrived in the game at a stage where the die came up six, versus the probability that I arrived at a stage where it came up some other number. And it makes no difference at all whether it was rolled and covered with a cup, or whether it has not yet been rolled at all.

Sorry for the length and lack of clarity. I was using voice typing while driving. It would have been better if I had restrained myself 😀.
Bottom line, it seems to me that my remarks already explained the obvious long ago. All that remains is the paradox, which now that I’ve dived into it feels to me a bit more paradoxical. From an anthropic-Bayesian calculation, it seems simple that my chances of being in the group that will be killed are greater than half, because I should ask myself not only what the chance on the die is, but whether it is more likely that I was kidnapped along with the minority in every non-final group, or with the majority in the final group.
But intuition and common sense say that as long as the die hasn’t been rolled, there is a 5/6 chance that the game will continue one more round, and I will go to life and peace.
Do you have a clear-cut solution?
If we are supposed to ignore such an anthropic calculation, the question is: why?

I don’t have time right now to delve into it and spell it out. Usually, to understand such a calculation, one has to formulate the experiment that would test the result of the calculation—that is, what exactly should we do to check, by expected value, whether the decision is correct or not.
Briefly, it seems simple to me that if I am faced with two possibilities—either to conduct a lottery in which the probability that I die is 1/6, or to conduct a lottery in which the probability is 1/2—then obviously the first option is preferable. This does not depend on the size of my group, what happened to other groups, when I entered the game, and so on.
Notice that I am not asking which group I am in or at what stage I entered the game. None of that interests me. I am being asked, in the given situation, which of the two lotteries I want to choose. So right now I have two possibilities before me, and the comparison between the probabilities is what determines the answer.