If I have 30% chance of winning a price how many times I need to win at least 1 price

Question

My basic understanding is that each time I have 30% chance of winning the prize so between 3 and 4 tries I should win it

cause .3 +.3 +.3 = 90% as I need to win only once with 3 try .3+.3+.3+.3 = 120% with 4 try

I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly

thanks for answering this noob question my math are way behind me now

Answer 1

You can't just add the probabilities (for instance, your $120\%$ can't be right...you can't have a greater than certain chance of winning!). Also, in theory there is no way to guarantee winning. You might, in theory, try a billion times and fail each time. That isn't very probable, but it is a mathematical possibility.

Say you try $n$ times. Then (under fairly reasonable assumptions) since each try fails with probability $.7$ the probability that all of them fail is $.7^n$. It follows that the probability that at least one of the tries succeeds is given by $$P_n=1-.7^n$$

How does that look numerically? Well, it's easy to compute. $P_3=0.657$ so you still have a respectable chance of not winning. $P_8=0.94235199$ so even with $8$ tries you have a greater than $5\%$ chance of not winning.

If, say, you want your chance of winning to be at least $99\%$ you have to try $13$ times! Indeed, $P_{13}=0.990311099$.

Answer 2

No, your math does not work. Indeed, there is no such thing as a $120$% chance of winning: $100$% is the best you can get, so that should immediately tell you that there is something wrong with your thinking.

The problem is that you can only add together the probabilities of two events if they are mutually exclusive, meaning that they can't both happen. That is not the case in your context: you can win the first time you try, and you can win the second time you try as well.

So, to figure out the chance of winning either the first or second time, you actually need to sibtract the probability of both events happening, since that one gets 'double-counted' when you add the idnicidual probabilities.

HEnce:

$P(1 \cup 2)=P(1)+P(2)-P(1 \cap 2) = 0.3+0.3-0.3 \cdot 0.3=0.6-0.09=0.51$

... which is of course the same as $1$ minus the probability of not winning either time, which is:

$1-P(1' \cap 2') = 1-0.7 \cdot 0.7=1-0.49=0.51$

... and that's where that general formula comes from: The chance of not winning when trying $n$ times is

$$0.7^n$$

and hence the chance of winning at least one time in $n$ tries is:

$$1-0.7^n$$

Notice that no matter how large is, this will never equal to $1$, i.e. you are never certain to win a prize, no matter how often you try. And that should intuitive sense: it is of course always possible that you never win, even if you try a gazillion times!