Odds of drawing

Question on a drawing?10 names are called you have 5% of total entries which are one million. what are the odds that you don’t get called? Once someone is called they can’t get called again. Also same as above but what if 15 names were called. What if you only had 2% of the entries? Thanks for any help.

There's no way to calculate this accurately unless you know how many people have one entry, how many have two, how many have three, and so on, because of the "once someone is called they can't get called again" rule.

Quote: Hunterhill

Question on a drawing?10 names are called you have 5% of total entries which are one million. what are the odds that you don’t get called? Once someone is called they can’t get called again. Also same as above but what if 15 names were called. What if you only had 2% of the entries? Thanks for any help.
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Unfortunately, the solution cannot be determined without knowing precisely how many entries (if different) that everyone else has. The reason why is because someone else being called at anytime (starting with the first person called) effectively eliminates all of their other entries from the pool...but we don't know how many entries those are. That person being called could eliminate 100 entries or 100,000, so you really don't know.

Simply put: Assuming you aren't called the first time, the person with the greatest number of entries being called helps you WAY more than the person with the fewest number of entries being called.

The answer would be easy if you want to assume that everyone has the same number of entries. In this case, we'll just say that you have 5 of 100 (as does everyone else) because that would be the same thing as saying 50,000 of 1,000,000:

Your probability of being called is:

(5/100) + (5/95) + (5/90) + (5/85) + (5/80) = 0.27951066391

Or, roughly 27.951066391%

As you can see, we are simply eliminating five entries for each person called. Why what I suggested above matters is that we can look at the same thing, except for the fourth call, we will assume that the person called either had TEN entries or ONE entry and see what happens to the probabilities based on that---everyone else still has five entires or as close as possible---it really doesn't matter in this example:

(5/100) + (5/95) + (5/90) + (5/85) + (5/84) = 0.27653447343

(5/100) + (5/95) + (5/90) + (5/85) + (5/75) = 0.28367733058

In the most extreme case, only five individual people would have entries at all, in which event, the probability that you would be drawn sooner or later would be 100%.

Thanks guys.It was all sort of a best guess scenario.
I don’t really know how many entries in total there were.If I were to continue guessing I would say the average person had about 300 while a group of 10 players had 5000 each and not one of those ten players were called in the drawing.