# Why am I expected to take n/m attempts to complete something with m/n odds?

I assume there are various ways of explaining it, but what's the most intuitive (yes I understand that is a relative concept)

Edit due to request:

Obvious example: There is a 1/6 chance of rolling a 6 on a 6-sided die. Expected number of rolls is 6/1.

Another example: You have 4/6 chance of rolling 1-4 on a 6 sided die. Expected number of rolls is 6/4.

Another example: There is a 2/52 chance of picking a red queen from a deck of cards. Expected number of draws (with replacement) is 52/2.

edit: Please replace "odds" in my title to "probability". As pointed out in a reply, I did not use the right word.

• Could you please describe a little more precisely what "complete something" means? What is your experiment, data collection process, or model? – whuber Mar 8 at 17:04
• Are my examples enough or do you want something different/additional? – lewikee Mar 8 at 17:48

Let $$p$$ be the probability of the event (so for you, $$\frac{m}{n}$$), and $$C$$ be the expected number of tries,
Then $$C = p + (1-p)(C+1)$$, since with probability $$p$$ you get it on the first try, and with probability $$1-p$$ you don't, and it takes $$C+1$$ total tries in expectation.
Then we solve for $$C$$, to get $$C = \frac{1}{p}$$