Fun Probabilty Problem.......

[ToraToraTora]

You have a bag of 10 keys, one of which opens a lock. You pick a key at random and try to open the lock. If the key you pick does not open the lock you must put it back in the bag shake them up and start again. On which try are you most likely to get the correct key?

[JJ109]

The First you have a 1/10 chance

For it to be second, ywo things must happen. Fail on first = 9/10 and succeedon second = 1/10. Therefore 9/10 * 1/10 = 9/100 < 1/10

For third 9/10 * 9/10 * 1/0 = 81/1000

probabilty is getting worse all the time!

[JJ109]

Apologies for my typing, but I think you can understand
1. 1/10
2. 9/100
3.81/1000
and getting worse as you go further along

[ToraToraTora]

Too easy I guess!

[bobbinwales]

Depends on how you look at it. I can see jj109s point and its right in away.
But maybe TTT's point is that theres always a 1 in 10 chance. If you always start again that is

[Zacs-Master]

Why is it not exactly the same chance on each try?

[bobbinwales]

Ok too slow to enter...

[bobbinwales]

Zacs...'cos it assumes if you get it you dont start again...

[Zacs-Master]

Oh, right.

[ToraToraTora]

bob jj109 got the answer I was looking for, I am making no "Point" - it's just one of those problems that can lead people in the wrong direction.

[ClareTGold]

The first try is the one with the largest probability, but isn't it going to be more interesting to work out the expected number of tries? Which is

(1/10)*(1+2*9/10+3*(9/10)^2+...)

and this sum equals exactly ten, as it's the same thing as

(1/10)*(d/dx) Sum x^n from n=1 to infinity, evaluated at x = 9/10

[ToraToraTora]

ZM: "Why is it not exactly the same chance on each try? " - that's the non intuitive part. In order to get to the second try you must fail on the first try so 9/10 * 1/10 for the second go. So that's longer odds than the chance of getting it on the first attempt and it gets worse each attempt.

[ClareTGold]

Clare*. Or ClaretGold.

[bobbinwales]

It's clare not Jim tora although your probably not alone in that 'mistake'

[ClareTGold]

She*... Her*.

But do go on, please :)

[bobbinwales]

Clares expected figure of 10 go's is what I get but without all their advanced. maths...its just an extension of needing 1 go on average (everytime) to pick a black ball from a bag containing just one ball (black) ...or 2 go's on average to get a head with a fair coin toss....or 6 go's on average to throw a 6 on a fair dice

[Prudie]

Just shows how you can use maths to fiddle stats. The chance of getting the right key is always 1 in 10 irrelevant of how many times you've tried - otherwise you could use the incorrect argument that if you've thrown 10 consecutive heads on a perfectly legitimate coin then the 11th toss is more likely to be another head...

[jno]

Prudie, the question isn't quite what are the chances on each attempt but what are they in a series of attempts.

[bobbinwales]

Yes but that's a different question prudie not quite what toratoratora asked