Okay, gotcha. And yep, agreed.
I guess that’s really the difference. If the number of rounds is unbounded, then players in the last round have 0% chance to survive instead of the normal 35 / 36 chance. (Speaking of the Type A infinite variant.)
To be honest, I’m not quite sure if Little_Nemo was describing a Type A or Type B game (above, I was assuming Type B). Basically: is the last part where everyone dies a round of the game or not? If not, then it’s just a Type A game, and the fact that everyone dies if it makes it past the last round is totally irrelevant. They aren’t players and so don’t count in the odds. On the other hand, if it’s considered a legitimate round (except that the odds go to 100%), then it’s a Type B game.
This is the key point you still don’t seem to be getting. The sizes of the group matters more than the dice rolls.
It’s like asking which state a random American is from. There are fifty states but that doesn’t mean you have a 50-1 chance of a random American being from any one of them. There’s actually a 12% chance that a random American will be from California and only a 0.2% chance they will be from Wyoming.
The same is true in the deadly sixes game. The group that loses will be nine times larger than all of the winning groups combined. Which means that a random player will have a 90% chance of being in the losing group.
Only the dice rolls matter. The group sizes are irrelevant.
Every player has a 1 / 36 chance to be in the losing group, full stop. Every single one.
The only way that’s not true is if the final group is guaranteed to lose. Which changes the premise. The only way to change (invalidate) the premise in that way is if the number of groups is unbounded. Infinity breaks things like simple arithmetic and the deadly sixes game.
EDIT: It might be worth mentioning that zero groups also invalidates the premise, just like unbounded groups invalidates it. With zero groups, your chance of losing is zero. With unbounded groups, your chance of losing is 90%. But if the number of groups is any positive integer, your chance of losing is 1 / 36.
And you don’t seem to be internalizing what that means in the context of infinity. Because no matter what round you pick, the next one after that is ~10x as important. And the one after that is 100x, and so on. You can pick the trillionth, the googolplexth, the Graham’s numberth, the BB(BB(1000))th number, or whatever–everything you have learned so far is completely irrelevant, because the rounds after that are more important yet. You simply cannot conclude things about the infinite case that you can about smaller cases.
Little Nemo has convinced me.
There is no paradox. The chance of winning is 35/36.
With a finite pool of players the chance of winning is 35/36, this is unarguable.
With an infinite pool of players there is no necessary last round, there is an infinite progression of ever more unlikely next rounds, each with a 35/36 chance of winning.
The chance of being a loser, if you are playing in the subset of games that lose is 90%. .
However, this doesn’t mean anything except after the game has ended, or the outcome has been predetermined. The insistence from the OP on the present/future tense of “if you’re playing” is contradicted by the randomness of the rolls.
Once you’re in a room, we can all agree that you have a 35/36 chance of winning.
Once the game has ended, we can all agree that out of those who were subjected to a roll, 9/10ths of them lost.
What we can’t say is that you will be playing the game, and that you will be subjected to a roll. You can’t know that ahead of time.
So a more precise way of putting it would be “The chance of being a loser, if you were playing in the subset of games that lose, is 90%.”
Not sure if you’re being sarcastic in your first line. But obviously I have not convinced you if you think a player can have both a 97% chance of winning and a 90% chance of losing and not see a paradox in that.
Right? I was all “Oh shit, is that what Little Nemo has been arguing? I haven’t been reading this right at all!” heh.
No one is claiming there is a paradox. There are different answers depending on the question you are asking, the rules of the game, and how you define certain terms.
Is a “player” someone who enters the room at some point, or is it someone in the pool who will potentially be picked to enter the room? Is there a maximum number of rounds? What happens in the last round?
And on preview, maybe Little Nemo is still claiming there is a paradox. His statement demonstrates how being imprecise in wording and definitions can lead to an unclear answer. There is a 97% chance of winning if you are part of the pool of players who could potentially enter the room. There was a 90% chance of losing if you were one of the players who did enter the room and the game ended.
I’m going to take another shot at this. This time I’m leaving the dice out of it.
You’re in a group of eleven people. You’re told the group will be randomly divided up and placed in two rooms. One person will go into one room and ten people will go into another room. Then the ten people in the second room will be killed.
Do you think “This isn’t so bad. The people in one room will live and the people in the other room will die. So I have a 50/50 chance of surviving.”
Or do you think “Uh oh. Ten of us are going to die and only one of us will live. The odds against me are 10-1.”
Now pretend there’s a larger group of one hundred and eleven people. This time you’re told you’ll be randomly divided up into three rooms. One person in the first, ten people in the second, and one hundred people in the third. And then the one hundred people in the third room will be killed.
Do you think “This isn’t so bad. The people in two rooms will live and the people in the only one room will die. So I have a two out of three chance of surviving.”
Or do you think “Uh oh. One hundred of us are going to die and only eleven of us will live. The odds against me are a little bit over 9-1.”
Now pretend there’s an even larger group of eleven hundred and eleven people. This time you’re told you’ll be randomly divided up into four rooms. One person in the first, ten people in the second, one hundred people in the third, and one thousand in the fourth. And then the one thousand people in the fourth room will be killed.
Do you think “This isn’t so bad. The people in three rooms will live and the people in the only one room will die. So I have a three out of four chance of surviving.”
Or do you think “Uh oh. One thousand of us are going to die and only one hundred and eleven of us will live. The odds against me are a little bit over 9-1.”
Are you starting to see how this works?
No, I am not claiming there is a paradox. I have been saying that the players have a ninety percent chance of losing. And that’s it. There is no 35/36 chance of winning.
That’s not how odds work.
You do have to be clear about what you mean by a “random American.” The most natural interpretation is one individual, chosen at random from among all residents of the USA, in such a way that each resident has an equal chance of being chosen.
In the case of a “random player” of the “deadly sixes” game, it’s not quite so clear what that means, and I think that’s the source of some of the confusion. What is the sample space for this “random player”?
The odds in this scenario are indeed grimmer than 1/36, but that is because you know that 9/10 or so of the people in the group will be killed, and you know that you will be in the group that may be part of the 9/10.
You can’t know that beforehand about the deadly sixes game unless the rolls have been predetermined. If the rolls have been predetermined, and then you are told that you will be subjected to a roll and that the game will end in double sixes, then your odds seem like they should be 1/10 for survival.
If the rolls have not been predetermined, you can’t be assured that you will be subjected to a roll at all if the game only happens once, even if there are an infinite number of people.
I agree that the 90% chance of losing is true when you know you will be one of the people chosen to go in the room, and you know the last roll will be double-sixes. Do you dispute that qualification? If yes, under what other conditions do you think you have a 90% of losing?
I said players have a ninety percent chance of losing. I assumed most people would understand that a player is a person who plays the game. If somebody does not play, they are not a player.
Players had a ninety percent chance of losing. They don’t have a ninety percent chance of losing. Whether or not they were a player can only be determined after the end of the game.
To clarify: Little Nemo’s obtuseness has driven me to the opposite conclusion on principle.
If a player, chooses to play the game, and is selected to enter a room they have a 35/36 chance of winning.
The end.
This is factually incorrect. It is the most wrong possible. Maximum wrong.
There is only a 90% chance of losing once you introduce infinity. That is just as wrong as saying there’s a 0% chance of losing when there are zero rounds so that makes a paradox because zero comes up with a different result.