# Odds?

• January 4th 2011, 09:50 PM
vooch
Odds?
Hello,
I am new to the forum, and quite frankly have been out of the "classroom" for a number of years, thus I just can't seem to remember how to solve my question....so I felt compelled to get advice/answers from the folks who are well versed.

My question: If the original odds of drawing a particular item are stated as 1 in 17, with one chance taken......what are the odds of drawing if 2 chances are taken? 4 chances? 6 chances? and lastly 8 chances? Also if you folks would be so kind and also include a percentile (%) for each answer I would certainly appreciate it.
Thank each of you for your time, and I look forward to spending time on your wonderful/informative website!
• January 4th 2011, 10:22 PM
Random Variable
The probability if you're drawing twice is the probability of getting the item on your first selection, or not getting the item on your first selection but then getting it on your second. That probability would be $\frac{1}{17} + (1- \frac{1}{17})(\frac{1}{17})$, where $(1 - \frac{1}{17})$ is the probability of not getting the item on any given selection.

If you're drawing 4 times, then the probability would be $\frac{1}{17} + (1-\frac{1}{17}) \frac{1}{17} + (1-\frac{1}{17})(1-\frac{1}{17})\frac{1}{17} + (1-\frac{1}{17})(1-\frac{1}{17})(1-\frac{1}{17})\frac{1}{17}$ (i.e, getting it on your first selection, or not getting it on your first but then getting in on your second, or not getting it on your first two selection but then getting it on your third, or not getting it on your first three selections but then getting it on your fourth selection).

EDIT: Unless the probability of drawing the particular item doesn't remain constant for each selection.