If a number has a 25% chance of being selected... but you get 4 chances to make a selection...

[TheWarmestHole]

Okay guys and gals, I'm still having a difficult time wrapping my head around some probability questions that have arisen around the lottery game of Keno.

In Keno, the player will see 20 numbers selected at random from the number 1 to the number 80.

This seems to tell me that there would be a 1 in 4 chance of a chosen number being one of the selections for any given round, right?

Now what if we make four separate guesses... what are the chances that at least one of my four selections is one of the chosen ones?

 

[Steven G]

Now what if we make four separate guesses... what are the chances that at least one of my four selections is one of the chosen ones?

1 - chances that none of my four selections is one of the chosen ones.

 

[Romsek]

[MATH]1- \dfrac{\dbinom{60}{4}}{\dbinom{80}{4}} = \dfrac{218789}{316316} \approx 0.691679[/MATH]

 

[Dr.Peterson]

[MATH]1 - P(\text{none of the four}) = 1- \dfrac{\dbinom{76}{20}}{\dbinom{80}{20}} = 1 - 0.30832 = 0.691679[/MATH]
Looks like we have the same answer two ways, so it must be right!

 

[TheWarmestHole]

[MATH]1 - P(\text{none of the four}) = 1- \dfrac{\dbinom{76}{20}}{\dbinom{80}{20}} = 1 - 0.30832 = 0.691679[/MATH]
Looks like we have the same answer two ways, so it must be right!

Okay that's great! Thanks to you I finally understand this. I was hung up on the math behind it, and stuck in the false understanding that if something that had a 25 % of occurring were to be taken 4 times, then this was somehow supposed to mean that there was a better chance than not that for a given round, at least one of my numbers would make the selected list.

But now I see that while this was, in fact, correct, it was actually correct even at 3 numbers... as in

---