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?
Alexander Howard
1/infinity^2
Adrian Moore
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?
Dylan Price
[math]\sum_{k=0}^\infty \frac{1}{2^{2^k}}[/math]
Gavin Hall
i concur
Leo Williams
...
Hunter Baker
How can we put this in closed form?
Chase Long
100%
Mason Campbell
I see the intuition. Care to explain the formal path?
Brayden Scott
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]
Nathan King
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.
Justin Garcia
Right of course.
Jackson Long
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%.
Joseph Perry
If you remove the option to give up, is there any way to actually lose this game?
Chase Walker
you could die
Jose Cruz
Woah Woah Calm Down Mate
Logan Williams
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.
Ryan Brooks
Does that give us a closed form?
Jonathan Hughes
no but it shows its less than 1
Chase Murphy
its aproximately 0,65
Tyler Reed
it's one half. Either you eventually win or you don't.
Eli Barnes
Approximately 62%
Leo Young
[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
Dominic Sullivan
100% if it was predetermined that you would win, 0% if it was predetermined that you would lose.
Eli Lee
GET OUT OF MY FUCKING THREAD I WILL SLAUGHTER YOUR FAMILY
Logan King
Why? There is no such thing as probability save in instances of ignorance.
Hunter Carter
I SAID GET OUT YOU FUCKING CULTIST GET OUT GET OUT GET OOOOOUUUUUUUTTTTTTTTTT
Ryder Williams
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.
Camden Allen
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.
Brandon Hall
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.