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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

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    $\begingroup$ "and means I've done something wrong" Not at all. Getting $100\%$ as the answer to "What is the probability that there is an outcome?" is an indication that things are done correctly. Or at the very least that any mistakes you've made cancel out. $\endgroup$ – Arthur Mar 9 '18 at 0:33
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    $\begingroup$ You don't win in case 4 $\endgroup$ – Bram28 Mar 9 '18 at 0:33
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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.

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  • $\begingroup$ Thanks for your answer, Ross. Sorry if I'm being a bit simple here but, wouldn't that mean if you were offered 25 attempts to play, you'd be guaranteed to win? $\endgroup$ – Jack Mar 9 '18 at 0:38
  • $\begingroup$ No, you would expect one win but would get more sometimes and none sometimes. The chance of none is $0.96^{25}$ $\endgroup$ – Ross Millikan Mar 9 '18 at 0:44
  • $\begingroup$ I think what I'm not understanding is why the probability sums. If instead we can win more than once and, using the example in my question, playing 3 times would mean a 12% overall chance of winning (0.04 + 0.04 + 0.04). If you apply that principle to flipping a coin, then wouldn't you need to flip it only twice to guarantee getting either heads or tails at least once? (0.5 + 0.5) It doesn't seem intuitive to me but I'm probably overthinking it $\endgroup$ – Jack Mar 9 '18 at 1:38
  • $\begingroup$ No, flipping a coin twice does not guarantee a head. It gives you an expectation of one head because you get two heads $1/4$ of the time and one head $1/2$ of the time. In your case playing three times gives you $11.5\%$ chance of winning at least once and an expectation of $0.12$ wins, counting two or three when you win that many times. $\endgroup$ – Ross Millikan Mar 9 '18 at 2:32

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