IMPOSSIBLE TO SOLVE

Let's say you're flipping a coin, but with some special rules:

- if you flip /heads/ the first time, you win, and the game is over.
- if you flip /tails/, then you can either give up, or try to flip /heads/ twice in a row. Then:
- if you flip /heads/ twice, you win.
- if you flip /tails/ again, before flipping /heads/ twice, then you can either give up, or try to flip /heads/ four times in a row. Then:
[...]
- if you flip /tails/ [math]n[/math] times without winning, you can either give up or try to flip /heads/ [math]2^n[/math] times in a row.

Hypothetically, let's say you can flip the coin infinite times. What is the probability that you will win?

1/infinity^2

No, man, there's a definite 50% probability, since you can win if you flip heads the first time. After that it just converges to some value. But which one?

[math]\sum_{k=0}^\infty \frac{1}{2^{2^k}}[/math]

i concur

...

How can we put this in closed form?

100%

I see the intuition. Care to explain the formal path?

samefag here, I think I change my mind. I was just adding up the probabilities that you win in the kth round. But there needs to be some probability of NOT having won the previous rounds. I think it should be
[eqn]\sum_{k = 0}^{\infty} \frac{1}{2^{2^k}}\prod_{j=0}^{k-1}\left( 1 - \frac{1}{2^{2^j}}\right)[/eqn]

now that I've changed my answer to this: ,
here's my explanation. We write the event "you eventually win" as a disjoint union of the events "you win in the kth round, after losing the first k-1 rounds". (the part about specifying that you lost the first k-1 rounds is important as it makes these events disjoint). Then it's just a matter of multiplying the probabilities of events that are independent, and then adding the probabilities of events that are disjoint.

Right of course.

You know you messed up there

>if you flip /tails/ n times without winning, you can either give up or try to flip /heads/ 2n times in a row.

you are flipping for rows of 2,4,16,256, you know that right?

The probability you will win is 1 = 100%.

If you remove the option to give up, is there any way to actually lose this game?

you could die

Woah Woah Calm Down Mate

In this sum, 1/2^n will appear once for each n>0 with + if n has odd number of 1s in binary and - if it has even number of 1s.

Does that give us a closed form?

no but it shows its less than 1

its aproximately 0,65

it's one half. Either you eventually win or you don't.

Approximately 62%

[math]
\frac{1}{2}+
\frac{1}{4}+
\frac{1}{16}+
\frac{1}{256}+
\frac{1}{4096}+
\frac{1}{65536}=
0.8166656494140625
[/math]
49/60 for some reason

100% if it was predetermined that you would win, 0% if it was predetermined that you would lose.

GET OUT OF MY FUCKING THREAD I WILL SLAUGHTER YOUR FAMILY

Why? There is no such thing as probability save in instances of ignorance.

I SAID GET OUT YOU FUCKING CULTIST GET OUT GET OUT GET OOOOOUUUUUUUTTTTTTTTTT

Nope. This is not it. It averages to little more than 0.75, but my experimental code says that the result should be around 0.64~0.65.

How do you get 0.75?
The sum of the first 10 parts is 0.6498... and the rest is too small to have a 0.10 effect.

His logic checks out, I'm inclined to believe you haven't done enough simulations to encounter the tail probabilities (large k) often enough.
As a rough gauge, to encounter the scenario k=3 (win by flipping 8 heads) once you'd expect to require a number of simulations of the order of 1 over
1/2^(2^8)*(1/2)*(3/4)*(15/16) ~ 3 e -78.

[eqn]\frac{1}{2}\sum_{n=1}^\infty\prod_{k=2}^n\left(\frac{1}{2}-\frac{1}{2^k}\right)=0.7112119049133975787211002\dots[/eqn]