IQ (moved subtopic)

I'm not buying your book and then searching through it to find this exact scenario. You can post the relevant section here or I can go and find the calculation myself.

But yes, taking that calculation as given, I think this is what I was lacking when thinking about this problem, hence struggling. I need to think about it some more with this new found knowledge in hand now.

do it with the 1 coin vs 2 coins playing coinflips for a coin.

then tell me why it shouldn't generalize

Are you ever going to direct me to this post that you are recalling?

I answered the hypothetical you posed. You explicitly stated that I knew the person would vote for Trump or someone like him. Of course I was taking how the person would vote into account.

I don't understand. I thought you were arguing that I am a slave to principles without regard for the consequences. What are you arguing now?

Also, since when did giving someone a ride to a polling station amount to using "artificial means" to win an election?

This is an absurd hypothetical. I honestly don't know what I would do. You certainly could find a candidate who was bad enough that I would push the button. Generally speaking, I would be far less likely to push the button if I were the opposing candidate because I would be worried that my judgment was badly clouded by self-interest.

ooh, i would actually like that, i have no math education so most of the academic papers i read to learn how to tackle a problem just go straight into notation so it may as well be written in egyptian hieroglyphs as far as i'm concerned

do you guys have a physical store, deliveries not good right now since i'm on x-country tour that will continue for a few months and will be passing through vegas before i'm somewhere permanently

Because there is superseding logic. If we play this game many times our long range results must approach being even, percentagewise, because all bets are fair. That can only happen if the chances are the ratio of the two stacks. The argument that the game could go on forever is silly. The chances of that "approach" zero. You know that. You are the epsilon fan.

I dont want to live in a world where there's no shops in vegas selling DS's books

You would "like" the book or like the chance to make $100? Anyway the best option is Amazon.

Right. All I am saying is that we need to do work to actually show this: "That can only happen if the chances are the ratio of the two stacks." You are saying it's obvious regardless of any other variables like if we e.g. use some progressive betting strategy etc. etc. I'm saying that it's an appealing statement but so are a lot of statements (especially in probability) that end up being totally wrong. I would not want to reach this conclusion without doing the work.

if that book has what i hope is in it then it's worth far more than $100 to me so purely interested in the book

guess book will have to wait until i am done with the trip then

Better to die with dignity.

lets say i'm trying to improve how i factor wind into my golf calculations

a bunch of times it's in accessible language which i can follow and understand

the above is rare, most lean heavily into jargon :(

such as this one below - i have to then avoid this kind of golf wind paper because it would require me to go back to school and take a math class for the first time just to understand all the stuff they don't explain and just take for granted that anyone reading it must know (i foolishly never took any math classes in college as i'd had the ap credits to not need them)

so hence the genuine interest in your book because i've looked at a bunch of math books for dummies before but it's often not focused on the kind of applications i would use it for - because again, most people in my situation actually studied math

You sound like Melkerson, who once said I need to stop trying to come to a conclusions via "first principles." But he was wrong for being such a stickler. Ergo 17th.

The book is not that advanced.

Especially when it comes to probabilitym, I tend to adhere to the adage that every problem has a solution which is simple, elegant, intuitive, and wrong. I am sure you are familiar with many of the more famous examples of this.

Actually, I am not. Give me an example. In any case I am simply saying, for example, that you don't need to find a flaw in a dice system to prove that there is one. You need only to know that adding up only negative numbers can't result in a positive one. (In spite of that negative 1/12 stuff.)

I'm thinking stuff like Monty Hall, or "Mrs Robinson has two children. At least one is a boy born on a Monday. What’s the chance that both are boys?" There are plenty more where those came from.

By "dice system" I assume you mean "a wagering strategy that involves altering bet sizes in order to beat a negative expectation game". Yes, once you have shown once that this is impossible, you can then use this principle for all progressive betting systems. But you do need to do the work to show that it is impossible initially to establish the "first principle". And in this case, the question is not about beating the odds, the question is "what are our odds of winning x units before we lose y units in a fair game", and it's not self evident, at least to me, that the same principle applies to all possible betting patterns in this case, since variance may or may nor play a factor in the solution..

word, will buy it once i am no longer on the road

50% you lose.

50% game is flipped.

50% I lose again (you win), or we are back to the original game.

so every two move sets, 1 game repeats, one you win, 2 I win.

so just by analyzing two moves we have you winning more than 25%, me winning more than 50%.

every two move I win 50% of that again, you win 25% of that again.

so your total wins are (game is the two move sets that ends it, recursive) 1/4 + 1/16 + 1/64 etc.

so your win total is the geometric series (1/4)^n which converges to 1/3 isn't that obvious?

Ok, I mean that's the bit I would have needed a pen and paper for. It's obviously correct, but it was not "obvious" to me as I would have needed a pen and paper for it. I'm not great at just seeing stuff like that in my head instantly.

You asked why this wouldn't generalise. It wouldn't generalise because you've taken advantage of the symmetry of the game after each flip that allows it to continue to arrive at the answer, I think. If I start with 1 and you start with 3, you wouldn't be able to do that.

Btw it's obvious without having quick geometric series convergence memorized.

When every 4 game one you win, two you lose, one is a redraw and you start from the same position, it's already 1/3 intuitively right?

Idea is the intuition step to bracket in 2 steps games.

The 1 v 3 coins is even easier.

Half you lose, half you reach 2-2 where equity is obviously split so you gain half, so you gain 2 halves and lose 2 every 4 games, 1/4.

All even coin amounts, you can model the minigame to reach equity and solve them.

All odd coin amounts you have the symmetry of redraws

I can sum the geometric series in my head lol (and an additional shortcut is that the sum of 1/x^n for n 1 to infinity is 1/(x-1)), it was identifying that was the correct series to sum. I mean, since I know that the sums of the reciprocals of powers of 4 is 1/3 I knew that would be the answer, but that's working ass backwards.

You say "every 4 games" but you can lose at any point in that 4 game stretch at which time the game stops. There might not even be 4 games in total. Looks like that doesn't affect the result, but again, that this is the case is not immediately obvious to me.

I'm not going to deny that my intuition for problems of this type is terrible by the way, which is why I prefer to do the calculations.

What do you mean by "symmetry of redraws"? As in when the number of coins A and B have is swapped? Is that easier than solving the original problem?

? I say every 4 complete games, 4 games of recurring coin flipping.

2 you lose (just lose first flip), 1 you win (win 2 consecutive flips), one you are back to the beginning (you have one coin vs 2).