since I’m getting lost in the math here I though I could ask someone who knows his probability maths:
there’s a game online kind of like roulette. if you bet on the “even” they give you a payback of 2.4 and if you bet on the “odd” they give you 1.75. (kind of like betting on all black or all red in roulette).
there are 45 tiles, 18 of them are even and 25 of them are odd. so I figured that the odds of getting an even is 40% and the odds of getting an odd is 55%.
now, my system I’m trying is really simple, I just bet on even and if I win I bet on even again. if I LOSE though then I bet twice the starting amount on even. if I lose again I bet twice the previous amount and so on. which means that after 9 losses I get up to 1000 bucks bets. which is my limit here, let’s say.
now using this strategy it’s easy to win a tiny bit every try. but sooner or later there are going to be 10 odds in a row and since I can’t bet twice that since I’m over my limit I’ve lost.
SO, it’s obvious that you CAN win in this game, but it depends on how long you play it.
MY QUESTION:
what are the odds that I’ll be able to play using this strategy without exeeding my 1024 cash limit up until I get 2048 in winning? I want to know if I have a chance at all or if the amount of bets I would have to do to slowly win a thousand bucks will be so many that the odds of a 10 times odd-streak occurs.
if you understood all that and can figure it out, then please tell me. this method worked up to 10000 once but then I kept playing and lost it all on a freak 14 odds in a row-happening.
hm, ok. but still…what are the odds of 9 losses coming in a row? how do you calculate that?
