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[QUOTE=Ganelon;860696]Any chance you could post that equation? I’m kinda curious[/QUOTE]

[SPOILER=Note that we will use the following formulae:]

[probability of winning round n]=2^-n

[probability of losing round n]=1-2^-n

[probability of independent events A and B both occuring]=[probability of A occuring][probability of B occuring]

Informally speaking, events are independent if the probability of one occuring is irrelevant to the other's probability. Suppose you roll a die. Whether you rolled an even or odd number is independent of the probability that a number less than 3 was rolled. i.e. being informed the roll is odd has no effect on whether the second condition is satisfied, and vice versa. However, whether you rolled an even or odd number is not independent of the probability that a number less than 4 was rolled. Being told your roll is even makes it rather less likely a number less than 4 was rolled. Write out all the possibilities and see why.

[probability of one of two mutually exclusive events A or B occuring]=[probability of A occuring]+[probability of B occuring]

Two events are mutually exclusive if they cannot occur together. Back to the die, rolling a 5 is mutually exclusive from rolling a 6. Hence, the probability of rolling a 5 or a 6 is simply 1/6+1/6.
[/SPOILER]

[SPOILER=final formula][probability that you win] = [probability that you win on exactly round 1] + [probability that you win on exactly round 2] + [probability that you win exactly on round 3]...

= [probability of winning round 1] + [probability of losing round 1][probability of winning round 2] + [probability of losing round 1][probability of losing round 2][probability of winning round 3]...

= [2^-1]+[(1-2^-1)(2^-2)]+[(1-2^-1)(1-2^-2)(2^-3)]+[(1-2^-1)(1-2^-2)(1-2^-3)(2^-4)]...
[/SPOILER]

[SPOILER=possibility of losing]
[2^-1]+[(1-2^-1)(2^-2)]+[(1-2^-1)(1-2^-2)(2^-3)]+[(1-2^-1)(1-2^-2)(1-2^-3)(2^-4)]...

< [2^-1]+[2^-2]+[2^-3]+[2^-4]...

= 1

Hence, the probability of the game being won is less than 1, and it's possible to never win the game.
[/SPOILER]

I neglected to include all this to make it a little less intimidating lol. Also, idk how to make forum latex work with words so you have to suffer the horrible notation lol.

Originally Posted by Ganelon

Any chance you could post that equation? I’m kinda curious

Spoiler : Note that we will use the following formulae: :

[probability of winning round n]=2^-n

[probability of losing round n]=1-2^-n

[probability of independent events A and B both occuring]=[probability of A occuring][probability of B occuring]

Informally speaking, events are independent if the probability of one occuring is irrelevant to the other's probability. Suppose you roll a die. Whether you rolled an even or odd number is independent of the probability that a number less than 3 was rolled. i.e. being informed the roll is odd has no effect on whether the second condition is satisfied, and vice versa. However, whether you rolled an even or odd number is not independent of the probability that a number less than 4 was rolled. Being told your roll is even makes it rather less likely a number less than 4 was rolled. Write out all the possibilities and see why.

[probability of one of two mutually exclusive events A or B occuring]=[probability of A occuring]+[probability of B occuring]

Two events are mutually exclusive if they cannot occur together. Back to the die, rolling a 5 is mutually exclusive from rolling a 6. Hence, the probability of rolling a 5 or a 6 is simply 1/6+1/6.

Spoiler : final formula :

[probability that you win] = [probability that you win on exactly round 1] + [probability that you win on exactly round 2] + [probability that you win exactly on round 3]...

= [probability of winning round 1] + [probability of losing round 1][probability of winning round 2] + [probability of losing round 1][probability of losing round 2][probability of winning round 3]...

= [2^-1]+[(1-2^-1)(2^-2)]+[(1-2^-1)(1-2^-2)(2^-3)]+[(1-2^-1)(1-2^-2)(1-2^-3)(2^-4)]...

Spoiler : possibility of losing :

[2^-1]+[(1-2^-1)(2^-2)]+[(1-2^-1)(1-2^-2)(2^-3)]+[(1-2^-1)(1-2^-2)(1-2^-3)(2^-4)]...

< [2^-1]+[2^-2]+[2^-3]+[2^-4]...

= 1

Hence, the probability of the game being won is less than 1, and it's possible to never win the game.

I neglected to include all this to make it a little less intimidating lol. Also, idk how to make forum latex work with words so you have to suffer the horrible notation lol.

I remember hearing the law from Interstellar. The law states that "anything that can happen, will happen" or "anything that can go wrong, will go wrong".

This law makes the most sense when applied to, say a die. If you roll a die once, you only have a 1 in 6 chance of rolling a 6. However, if you repeatedly roll a die, you necessarily have a 100% chance of eventually rolling a 6. This can be computed directly. The probability that a die does not roll a 6 n times is given by the following formula:

[LATEX](5/6)^n[/LATEX]

If we take the limit as n tends to infinity, the probability dwindles to 0. Put differently, that means that "It's impossible for one to roll a die over and over until the end of time and never get a 6".

This is the same principle underlying the monkey-typewriter principle. It's very unlikely a monkey hits a random selection of keys and writes out the works of Shakespeare. But the probability that it doesn't happen, when powered by large enough n, will eventually dwindle to 0 too. This means the monkey will necessarily eventually write the full works of Shakespeare. Though it's very difficult, the monkey has literally all of eternity to get it done.

However, Murphy's Law does have its limitations. You can make up games where at every stage you could win but there's still a chance you'll never win. For example:

Suppose you play a game where, in the first round, you just need to flip a heads to win. If you lose, you decide to make it harder for yourself - in the second round, you demand you need to flip two heads in a row to win. If you fail this, then in the third round you'll demand you need three heads, and so on. It turns out that it's possible to never win this game, even though the probability of winning in just a couple rounds is greater than 60%.

How come? In the previous cases, though the probability of winning at any particular moment may have been challenging, it was at least fixed. This game gets harder the longer we play. Thus, after writing out the equation and taking the limit as n tends to infinity, one sees that even all of eternity may not be enough to win, if you're unlucky enough.

But is that really the case? What does that statement even mean - "it's possible to never win"? The statement about the previous games "you'll eventually win" was at least something that was possible to tangibly construct. You could get out a die, roll it over and over and verify that you do eventually get a 6 every time. The task of seeking a win will take finite time, even if you're not sure how much time exactly.

On the other hand, this new statement is another kind of monster. You can't verify that there exist games you'll never be able to win. You play this game for a hundred rounds and sure, you're probably in a game you'll lose no matter what because now you need a hundred coinflips in a row to win. But you may nevertheless still win the next round. And if not the next round, then possibly the one after that. It may be unlikely you'll win soon, but you can literally keep going until you win.

That's kind of where I'm going with this. This mathematical statement is, as an equation, at the very least logically consistent. But we will literally never be able to verify or falsify the statement in the real world. Hence, can the statement truly be trusted to reflect our reality? Perhaps our reality does respect our pretty little formula and will allow some games to last forever - or perhaps it doesn't. Perhaps it simply doesn't care.

If your instinct is to say "our reality will respect the formula because math is math and math is intrinsically correct", I ask you to question where this trust in maths comes from. Indeed, we trust maths so much precisely because we can verify that our calculations reflect our reality's behaviour through experimentation - without the experimentation, why should we still trust math?

I remember hearing the law from Interstellar. The law states that "anything that can happen, will happen" or "anything that can go wrong, will go wrong".

This law makes the most sense when applied to, say a die. If you roll a die once, you only have a 1 in 6 chance of rolling a 6. However, if you repeatedly roll a die, you necessarily have a 100% chance of eventually rolling a 6. This can be computed directly. The probability that a die does not roll a 6 n times is given by the following formula:



If we take the limit as n tends to infinity, the probability dwindles to 0. Put differently, that means that "It's impossible for one to roll a die over and over until the end of time and never get a 6".

This is the same principle underlying the monkey-typewriter principle. It's very unlikely a monkey hits a random selection of keys and writes out the works of Shakespeare. But the probability that it doesn't happen, when powered by large enough n, will eventually dwindle to 0 too. This means the monkey will necessarily eventually write the full works of Shakespeare. Though it's very difficult, the monkey has literally all of eternity to get it done.

However, Murphy's Law does have its limitations. You can make up games where at every stage you could win but there's still a chance you'll never win. For example:

Suppose you play a game where, in the first round, you just need to flip a heads to win. If you lose, you decide to make it harder for yourself - in the second round, you demand you need to flip two heads in a row to win. If you fail this, then in the third round you'll demand you need three heads, and so on. It turns out that it's possible to never win this game, even though the probability of winning in just a couple rounds is greater than 60%.

How come? In the previous cases, though the probability of winning at any particular moment may have been challenging, it was at least fixed. This game gets harder the longer we play. Thus, after writing out the equation and taking the limit as n tends to infinity, one sees that even all of eternity may not be enough to win, if you're unlucky enough.

But is that really the case? What does that statement even mean - "it's possible to never win"? The statement about the previous games "you'll eventually win" was at least something that was possible to tangibly construct. You could get out a die, roll it over and over and verify that you do eventually get a 6 every time. The task of seeking a win will take finite time, even if you're not sure how much time exactly.

On the other hand, this new statement is another kind of monster. You can't verify that there exist games you'll never be able to win. You play this game for a hundred rounds and sure, you're probably in a game you'll lose no matter what because now you need a hundred coinflips in a row to win. But you may nevertheless still win the next round. And if not the next round, then possibly the one after that. It may be unlikely you'll win soon, but you can literally keep going until you win.

That's kind of where I'm going with this. This mathematical statement is, as an equation, at the very least logically consistent. But we will literally never be able to verify or falsify the statement in the real world. Hence, can the statement truly be trusted to reflect our reality? Perhaps our reality does respect our pretty little formula and will allow some games to last forever - or perhaps it doesn't. Perhaps it simply doesn't care.

If your instinct is to say "our reality will respect the formula because math is math and math is intrinsically correct", I ask you to question where this trust in maths comes from. Indeed, we trust maths so much precisely because we can verify that our calculations reflect our reality's behaviour through experimentation - without the experimentation, why should we still trust math?