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# I am a member of a golf club that runs a random monthly draw.

I am a member of a golf club that runs a random monthly draw. here are 750 members in the club BUT you need to be at the monthly meeting to win the draw. How do the odds of me winning vary by the number of attendees at the meeting?
It would depend on the depend on how the drawings are conducted.

If only those who are present at the meeting are included in the drawing; you would have a 1 in X chance (where X is the number of members present). If there were 10 members present; you would have a 1 in 10 or 10% chance of winning.

The other method that could be used, would be to conduct the drawing with all member's names included, and the prize would only be given if that person was present. In this case; if there were 10 members present; each of those would have a 1 in 750 (or 0.133%) chance of winning, and there would only be a 10 in 750 (1.33%) chance that the prize would be awarded. This type of drawing often is done is such a way that the "jackpot" is increased each month until the prize is eventually awarded.
Customer: replied 5 years ago.

It is the latter of your scenarios where, if the drawn name isXXXXX then it is a so-called roll-over.

I am interpreting you reply as the less that are there, the greater the chance of a given member winning.

Technically, is it 0.133 x 0.1 if there are 10 present compared to 0.133 x 0.01 if there are 100 present?

In the rollover, the chance of an individual winning is NOT influenced by the number of members present.

In the situation; all 750 names are XXXXX XXXXX the drawing, but the prize is awarded only if that person is there. If one person is present they have a 1 in 750 chance of winning, and if all 750 are present they still have the 1 in 750 chance of winning.

What does change is the chance that someone will take home the prize. The fewer people present, the higher the chance that the prize will roll-over till the next month.