IMPOSSIBLE TO SOLVE

1/infinity^2

>>9013227
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]

>>9013241
i concur

>>9013241
How can we put this in closed form?

100%

>>9013241
I see the intuition. Care to explain the formal path?

>>9013241
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]

>>9013294
now that I've changed my answer to this: >>9013321,
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.

>>9013269
Right of course.

File: dumbcunts.png (189KB, 1212x1196px) Image search: [Google]

189KB, 1212x1196px

>>9013223
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?

>>9013374
you could die

>>9013382
Woah Woah Calm Down Mate

>>9013321
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.

>>9013489
Does that give us a closed form?

>>9013683
no but it shows its less than 1

>>9013692
its aproximately 0,65

>>9013223
it's one half. Either you eventually win or you don't.

>>9013223
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

>>9013223

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

>>9014103
GET OUT OF MY FUCKING THREAD I WILL SLAUGHTER YOUR FAMILY

>>9014957
Why? There is no such thing as probability save in instances of ignorance.

File: b02.png (262KB, 600x600px) Image search: [Google]

262KB, 600x600px

>>9015002
I SAID GET OUT YOU FUCKING CULTIST GET OUT GET OUT GET OOOOOUUUUUUUTTTTTTTTTT

>>9013321
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.

>>9015152
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.

>>9015152
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.