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I were playing Hearthstone the other day (a card game).

I got a card that gave me 50% chance to draw a extra card, when it where to be my turn.

Then i got another one of those. It made me wonder what my chances then would be to draw a extra card, if i had two of them on the battleground.

LateChicken
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  • I do not know the complete rules, but two independent $50-50$-chances give a probability of $\frac{3}{4}$ of at least one success. Maybe I miss the point of the question. – Peter Aug 14 '16 at 19:08
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    You got the point, 3/4/75% were also my guess. But how did you come up to that number? – LateChicken Aug 14 '16 at 19:13

2 Answers2

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The probability that both $50-50$- chances fail is $$\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$ because the events are indepenent.

So, the probability of at least one success is $$\ 1-\frac{1}{4}=\frac{3}{4}$$

Another way is to consider the possible outcomes

$$SS,SF,FS,FF$$

where $S$ is a success and $F$ is a failure. Each of the sequences has the same probability , namely $\frac{1}{4}\ $ , and in $3$ of the $4$ cases, you have at least one success.

Peter
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If you are curious and want to learn more how this works you can read about binomial distribution, which basically is this sum:

$$(p+q)^n = \sum_{k=0}^n{n \choose k} p^kq^{n-k}$$

for probabilities, which add up to 100%, we have: $p+q=1$ so $q=1-p$ so it becomes

$$(p+q)^n = \sum_{k=0}^n{n \choose k} p^k(1-p)^{n-k}$$

The probability you are looking for is 1 minus one of the end points in this sum: $$1-\left({1 \over 2}\right)^2=1-{1\over 4}= {3\over 4}$$

mathreadler
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