cool math problem

$1

Is there some trick here? This seems simply ev = payoff*probability = (2^n)*(1/2^n) = 1

Raul Oly ...?

EV = 1 yes

:O. how do u get a number from an equation with n in it and n isn't defined. It isn't 1.

there is no trick here but most people find the answer to be quite unexpected.

Oh yes 1 is a silly answer when the player wins a minimum of 2!

So if you sum the ev over all possible outcomes (n=1 to infinity) you get the same cancelling and end up with the divergent series 1+1+1+... which implies you should pay any amount asked! If that’s right this time that is indeed a very cool problem!

On October 30 2019 21:34 Nitewin wrote: Raul Oly ...? EV = 1 yes

Not him. I used to post here a fair bit a long time ago but I stopped playing professionally years ago but still occasionally surf liquipoker as the habit seems deeply ingrained. I’m so old school I played the original Party Poker Monster days. 3betting JJ+ was cutting edge.

On October 31 2019 11:03 Oly wrote: Oh yes 1 is a silly answer when the player wins a minimum of 2! So if you sum the ev over all possible outcomes (n=1 to infinity) you get the same cancelling and end up with the divergent series 1+1+1+... which implies you should pay any amount asked! If that’s right this time that is indeed a very cool problem!

yup, exactly right. It's called the St petersburg paradox

pay me 100 bucks and you flip the coin and ill pay the $2 for every non tails

$2 for every non tails? Then the summation converges on $2, lol. You would be +$98 EV.

Change the problem to the following and it becomes something fun to solve (this way is trivial as the answer is 2*.5+4*.25+8*.125+...=1+1+1+1...= infinity):

You play a game where a fair coin is flipped until it comes up tails the first time. After each heads flip, you win back $n, where n is the number of times the coin was flipped thus far. How much should one be willing to pay for each game to still be +EV?

ex: H +1, H +2, H +3, H +4, T end and receive $10

nvm, read it wrong... edit.

response to Daut:

1/2(2) +1/4(3) +1/8(6) +1/16(10)+...

=sum 0 to n: 1/2^n* n(n+1)/2=4

the EV is $4? is this correct?

Yes, but prove it, no mathematica! I had some trouble, but found an interesting proof that it's 4

+ Show Spoiler +

nice proof, can use that same kind of argument to prove that 1+2+3+4+5...=-1/12

isn't the paradox caused because infinity becomes ergodic?

I mean the paradox as why it seems non-intuitive since its time probability and not ensamble probability

edit: nevermind its just a math glitch

This centrifuge problem applies to seating for positional neutrality in poker. (no god seating)

On November 01 2019 04:43 Stroggoz wrote: nice proof, can use that same kind of argument to prove that 1+2+3+4+5...=-1/12

Been about 15years since I tutored Calc 2, but I believe you can only use that method for series that converge (otherwise S doesn't have a value that we can multiply). And I want to say the smallest same sign series that diverges is 1/n, so any monotone series that shrinks faster than 1/n converges. Gets much trickier with series that alternate signs.

Strogg, where are you taking game theory classes?

im not. i'm just reading through this book. https://www.amazon.com/Game-Theory-Introduction-Steven-Tadelis/dp/0691129088