weary saffron
"if n is defined as the round number and your chances of winning are defined as (1/3)^n what are the chances you will win as n approaches infinity"
I came up this question one day and tried to solve the question but I had conflicting answers between 100% and 50%. Which is correct?
tardy mantleBOT
weary saffron
to get 100% i thought as you play infinite times, the chances never reach 0 therefore it must be certain
to get to 50% i used the geometric series "1/3 + 1/9 + 1/27...." which resulted in 1/2
toxic orbit
probably worth clarifying:
chances to win on round n, or chances of winning at least once after n rounds?
if it's the former one quickly realizes that the chances of winning rather rapidly approaches 0 as more rounds are played.
(or maybe there's another interpretation you're using here? in which case it's worth clarifying.)
weary saffron
I’ll try rephrase the question. The game starts as a 1/3 chance of winning and if you win, the game is over. Otherwise, the game continues with a chance of 1/9 of winning and each round gets thrice as unlikely to win. What are the chances you will win eventually
Sorry if this is a bit confusing this is the best I could do
toxic orbit
ah, right.
so I think it might be easier to ask the complementary question of "what is the probability of you never winning the game as n approaches infinity".
because then you can subtract what you get from 1 (or 100%), and I believe it may simplify calculations a fair bit.
toxic orbit
it's either .close or .solved.