False. It is not 90% if it’s capped at only 10 rounds, or 20 rounds, or 80 rounds, or a googol rounds. It’s only 90% if there are unlimited rounds.
As another example of how infinity invalidates the premise, consider the sum of multiple unique integers.
1 + 2 = 3
1 + 2 + 3 + 4 + 5 = 15
etc…
In all cases, the sum is greater than the number of elements. For the above two examples, the sum of 3 is greater than 2, and the sum 15 is greater than 5.
But if you say an unlimited amount of integers, you now have an infinite number of elements summing to infinity. Those are both the same number: Infinity. The sum is no longer greater than the number of elements. A paradox! But it’s not a paradox; it’s just a matter of infinity invalidating the premise.
But as I showed in a previous post, there will never be a final group that has an infinite number of players. The number of players in the final group may be very large - but not infinite. So if the only value of X that shows a paradox is infinity then there is no paradox.
Exactly right. There is no paradox.
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Can you explain the paradox of the sum of counting numbers being equal to the number of elements summed given an unlimited number of them? How is it possible that the number of elements and their some are equal?
What is the maximum number of players that can be in the final group? If it’s not infinite, give us a number.
To further elaborate: I think your confusion is that you are thinking of infinity as a number. “The game must end sometime, and that final group must have a number of players you can count, so it’s not infinite.”
But that’s not what infinity is. It’s a concept. No matter what number you give for the final group, I can give one bigger. And the bigger number is still a number, not “infinity”, but the fact that I can’t define that biggest number means it’s infinite.
Hey look, I can prove that the sum of all integers is finite. It’s obviously finite in the case of just the first integer 1. And if the sum is finite for the first n integers, then it’s clearly also finite in the case of the first n+1 integers. By mathematical induction, then, the statement is true for all integers and I can say the sum of all integers is finite.
We haven’t had any math experts weigh in regarding my question yet, but Google is confirming that infinity + 1 = infinity. So let’s use your own example to explain:
I bolded the incorrect part.
2 + 3 + X - X = 5
Unless X = infinity, in which case the answer is 0.
You see how infinity changes everything? You didn’t include parentheses, which means simple addition and subtraction go left to right. That gives us:
2 + 3 = 5
5 + Infinity = Infinity
Infinity - Infinity = 0
If you had added parentheses, the answer would be different. (2 + 3) + (Infinity - Infinity) = 5.
Which means that once we introduce infinity as a valid number, we have just proved that:
2 + 3 + X - X <> (2 + 3) + (X - X)
You see how introducing infinity invalidates the premise? Your own example disproves simple arithmetic if we allow infinity.
I do not know if I am any kind of “math expert”, but I can confirm that “infinity” is not a positive integer. However, I am not convinced that there is any kind of apparent paradox like a violation of the Law of Large Numbers, any more than with the fact that a gambler can always make money betting a martingale provided he or she has an infinite bankroll.
When I did a simulation, I got different answers depending on how I calculated the results
One way of calculating always got 35/36 (or 1/36), while the other way of calculating got closer and closer to 90%/10% the more rounds that occurred. Weird.
I think the problem there is that the limit of increasing the number of rounds is not equal to the overall limit, because the later rounds are far more influential than the earlier ones. In fact, for whatever value you can come up with for the first n rounds, that figure has zero influence on the overall answer. Since the expected number of players is infinite, rounds n+1 onward are the only ones that contribute to the answer, regardless of the value of n.
That makes sense
(a) there is no guarantee that there will be a final round
(b) the possible sizes of potential final rounds is unbounded, which is the same as being “infinite”, in informal language.
How big is the largest possible integer? Unbounded, there are infinite possible integers, but each one is finite. How big is the largest possible “final group”? Unbounded, even though each one is finite.
That said, I’m not sure quite what your position is at this point. Here’s my claim:
-For any finite population of potential players, the chances of a player winning are 35/36, no matter how you calculate it
-There’s a convincing-seeming argument that, for infinite players, the chances of winning are 1/10
-There’s also a convincing-seeming argument that, for infinite players, the chances are 35/36
-That’s clearly an apparent contradiction/paradox
-And the resolution of that paradox is that the 1/10 argument is flawed, because it tries to manipulate infinity incorrectly
Do you disagree with any of that?
Okay, let me try to reframe this to eliminate any possibility of infinity being a factor.
Here’s how the game works. A group of people enters a room (starting with a single person). A pair of dice is rolled.
If anything other than a pair of sixes is rolled, everyone in the room is given a million dollars and leaves. A new group is brought into room and the game is played again. An important point is that each new group is ten times as large as the previous group.
If the result is a pair of sixes, poison gas is released into the room and everyone dies. The game is then stopped.
There is a third possibility, if the dice are rolled a trillion times with producing a double six, the game is declared over and everybody is the multiverse is killed.
For purposes of this scenario, assume there is a large enough supply of people, money, poison gas, and space in the room in order to play a trillion rounds. Also nobody plays the game more than once. Surviving and getting a million dollars is winning; dying is losing.
So what are the odds if you’re playing?
The odds of dying are a bit over 90%. They’re a bit over 90% for games of length 1-999,999,999. And then they’re 100% for the final 1,000,000,000,000-round game, which is ~9x bigger than every other game combined. So you have a >90% chance of being in the last round, which has a 100% death rate.
If are lucky enough to be picked for rounds 1-999,999,999,999, then your odds of dying are only 1/36. But >90% are unlucky enough to be picked for that last round, where there is a 100% death rate. So the overall odds shoot up dramatically.
Including the people who survived the earlier rounds?
Just a quick note, by the way, that 10trillion people in the final round dwarfs what google says is the number of planck volumes that fit in the observable universe, which is on the order of 10185.
I think the odds are 1/36, not 90%.
Yeah, that.
We’ve established two variations on the finite version of the OP game. They’re identical to the original in all rounds prior to the limit. They differ only in the last round:
Type A: Players in the last round are subject to the dice roll. They’re killed on a double-6, and set free otherwise.
Type B: Players in the last round are killed unconditionally.
Type A finite games give the expected 1/36 odds of dying. Type B players have a ~90% death rate. That Type B is worse than the dice roll should be obvious, since the players in the giant last round have much greater odds of dying than in the Type A or infinite versions (100% vs. 3%).