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I am working at the Paris motor show. Everyday there is a draw where you can play for free and win a video camera if your name is picked. We are a team of 10 employees. We are all playing. The chance of winning is one in 200 but there are 10 of us playing which makes it a 5% chance of winning the draw (I know it is not exactly 5% it is slightly more but let's say that it is 5%). Am I right in concluding that if we play 13 times (that is everyday till the end of the motor show) the probability of one of us winning is around 48% ? My reasonning is as follows: probability of winning is 5% that is 0.05, therefore probability of not winning is 0.95 probability of not winning any of the 13 draws 0.95^13 that is: 0.51334208327951 probability of wining at least one of the draws: 1-0.51334208327951≈0.48 Let me know if I am wrong. My math skills are poor. I know that probability can be really misleading, but I guess that I followed the right method.Thanks in advance for your help.

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makes sense to me :) – gt6989b Oct 3 '12 at 18:41
why do you think it's not 5% for one draw (10/200)? Your second reasoning is correct that at least one of you winning once is 48% in 13 draws. – karakfa Oct 3 '12 at 18:46
Also you might want to take into account that by the time the 13th drawing comes around, the chances are 10 in 188, assuming the previous winners' names are not put back into the pool. – dadinck Oct 3 '12 at 18:49
@ Karafka you are right, I got confused but actually they put the winners name back in the pool so it is 5% (by my estimate, I don't know exactly how many people are playing everyday but on average it is 5ù chance for the team to win a video camera). Thank you all for your help. – user1119429 Oct 3 '12 at 19:45

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