If you are offered to play a game three times, and the chance of winning each time is 4%, what is the overall probability of you winning? This assumes that you can win only once, e.g. if you win on your first or second attempt then play stops.
Intuitively I know (or at least think) it can't be 12% overall chance, as this would imply tossing a coin at least twice would guarantee you get what you want. So after some research I discovered binomial distribution. I don't know if it's even the correct method to solve this but here was my attempt:
If $w$ is a win and $l$ is a loss, then we have:
$1.\quad w$
$2.\quad lw$
$3.\quad llw$
$4.\quad lll$
as our 4 combinations. If $k$ is then the number of times we won (either one or zero) and $n$ is the number of times we lost:
$1.\quad (0.04)^1(0.96)^0 = 0.04$
$2.\quad (0.04)^1(0.96)^1 = 0.0384$
$3.\quad (0.04)^1(0.96)^2 = 0.036864$
$4.\quad (0.04)^0(0.96)^3 = 0.884736$
After adding these probabilities I got an overall chance of winning to be… exactly 100%. Which presumably is just the chance that any of these events occurs, and means I've done something wrong. If I add just the first three probabilities I get 11.5% chance, which seems more sensible, but I don't know if it's correct.
Any help would be much appreciated!
Jack
Best Answer
Your item 4 does not result in the player winning. Adding the first three does give the chance of winning. If you got to play three times you would expect to win $0.12$ but occasionally you would win more than once in a group of three, so the chance of at least one win must be a little less.