# Probability of a few possible outcomes with certain probability beat the original

Lets say we have a contest of exactly 5 contestants who are all competing against the Original.

The probability that some contestant will win over the Original are as follows:

• cont1 = 26.9%
• cont2 = 21.3%
• cont3 = 20.7%
• cont4 = 8.96%
• cont5 = 67.5%

Please tell me what is the formula to calculate what are the chances that any one of ( cont1, cont2, cont3, cont4) beat the Original ? (Please note I'm not taking cont5 into account).

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## 1 Answer

I'm assuming that there is independence between each of the constestants. Let $p_i$ denote the probability that the $i$'th contestant beats the Original. Then $$P(\text{none of cont1-cont4 wins})=\prod_{i=1}^4 (1-p_i)\approx 0.4153$$ and therefore $$P(\text{any of cont1-cont4 wins})=1-P(\text{none of cont1-cont4 wins})\approx 0.5847.$$

It's just elementary probability theory. Let $A_i$ denote the event that the $i$'th contestant wins, i.e. $P(A_1)=0.269$, $P(A_2)=0.213$ and so on. Then the complement $A_i^c$ is the event that the $i$'th contestant does not win and $P(A_i^c)=1-P(A_i)$ for all $i=1,\ldots,5$ (elementary property of a probability measure). Now $$P(\text{none of cont1-cont4 wins})=P\left(\;\bigcap_{i=1}^4 A_i^c\right)\stackrel{\rm indep.}{=} \prod_{i=1}^4 P(A_i^c)=\prod_{i=1}^4 \big(1-P(A_i)\big)\approx 0.4153,$$ where we have used the definition of independent sets. At last we use that the complementary set of $\{\text{any of cont1-cont4 wins}\}$ is exactly the set $\{\text{none of cont1-cont4 wins}\}$.

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thx, that's exactly it, can you please point me to the theory about this - what's this formula called, so i can google it. Thx! – Nikola Jul 10 '12 at 10:35
@Nikola: See edit. – Stefan Hansen Jul 10 '12 at 10:48
I think also implicit in the question and in Stephan's answer (but never stated is that all 5 contestants compete independently against the original in 5 distinct contests rather than there being say one contest where all six compete. – Michael Chernick Jul 10 '12 at 15:12