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I enter a contest every week with $850$ others that I have a random chance of being selected as the winner. There is only one winner. If I play for $850$ weeks, should I have a $100\%$ chance of being a winner?

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  • $\begingroup$ If you flip a coin twice, is there a 100% chance that it comes up heads at least once? $\endgroup$ – Gabriel Burns Oct 18 '16 at 16:30
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    $\begingroup$ What are your chances of losing? What are your chances of losing every time after $850$ weeks? $\endgroup$ – Kevin Long Oct 18 '16 at 16:31
  • $\begingroup$ Only if the rules state that there can be no new entrants, and once you have won once, you are ineligible to win a second time. $\endgroup$ – Doug M Oct 18 '16 at 16:31
  • $\begingroup$ You are expected to win one time, but that is the average of the of the times you win $0$ times in 850 days, the times you win one time in 850 days, the times you win 2 times in 850 days, etc. $\endgroup$ – Thomas Andrews Oct 18 '16 at 16:41
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The probability that you win at least once is:

$$1-\left(1-\frac{1}{850}\right)^{850}$$

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    $\begingroup$ Assuming that you meant $850$ people including yourself (if you meant $851$ people altogether, then replace the inner "$850$" with "$851$")... $\endgroup$ – barak manos Oct 18 '16 at 16:42
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    $\begingroup$ +1, and good catch on the potential off-by-one error. :-) $\endgroup$ – Brian Tung Oct 18 '16 at 16:55
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The probability of beeing selected at least once is One minus the probability of beeing never selected (converse probability)

$P(X\geq 1)=1-P(X=0)$,

where $X$ is the random variable of beeing selected in 850 weeks. It is binomial distributed:

$X\sim Bin(850,p)$

$P(X=0)=\binom{850}{0}\cdot p^0\cdot (1-p)^{850}=(1-p)^{850}$

p is the probability of beeing selected in one week.

$1-P(X=0)$ is $1$ if $P(X=0)=0$.

We can assume that $p<1$. It is possible that $(1-p)^{850}$ becomes $0$ ?

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The winning chances are independent of each other. So, for every match you have chance of 1 for 850 members. Even if you play for infinite times, there is no guarantee for 100% winning. So for 850 matches you have probability of winning as

1/(850)^ 850

The above case is probability for winning in all the 850 matches.

The probability for winning atleast once is binomial. It is

1 - ( losing all the times ).

Losing all the times is (no. of players - 1) / no. of players raised to power of no. of matches played

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  • $\begingroup$ So, the more times you play, the less likely you are to ever win? $\endgroup$ – Thomas Andrews Oct 18 '16 at 16:41
  • $\begingroup$ No, the probability is less likely. It is independent event. It doesn't depend on previous plays. So, it is multiplied. $\endgroup$ – Sameer Oct 18 '16 at 16:43
  • $\begingroup$ This gives probability of winning every time- the question asked for probability of winning at least once. $\endgroup$ – Kevin Long Oct 18 '16 at 16:51
  • $\begingroup$ Yes, that's correct, I gave for winning all the times. $\endgroup$ – Sameer Oct 18 '16 at 16:52
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    $\begingroup$ @Sameer: That is not what the question asks for. The question asks for the probability that the OP wins at least once. $\endgroup$ – Brian Tung Oct 18 '16 at 16:55

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