[bonanova]

Three switches, numbered 1, 2 and 3, and wired in series, control a single light bulb, which is not lighted.

For the bulb to light, all three switches must be On.

The switches' On positions are not labeled.

You flip the switches, one at a time, in any sequence you choose.

1. After how many flips [call this number N] can you be guaranteed to light the bulb?
   
2. What is the probability of lighting the bulb after each flip: 1, 2, ... N?
   
3. What is N if there are S switches [instead of 3]?
   

[Guest]

Binary number (since the switches are simple on/off) x2 (since each switch must be returned to its original position to reach the next number) 111 which is 20 flips (10 x 2) will do it, but may not be the least number of times that switches must be flipped.

[Guest]

1- since it's already in one preset combination, and u have a total of 8 combinations (000-111) then u should get it in 7 tries at most.

2- cant be bothered to do probabilty calc

3- N= (2^S)-1 ....assuming the initialy condition counts as a try, otherwise, its 2^S

Edited June 20, 2009 by bonanova
spoiler added

[Guest]

1, you have to be careful though order can be important, basically you want something like this

000

100

110

010

011

001

101

111

so yes 2^n -1, but make sure you sort such that only 1 switch is changing at a time.

2. with each switch you eliminate one possibility,

so first switch, 1/7 chance. second switch 1/6. etc.

Edited June 20, 2009 by bonanova
spoiler added

[bonanova]

phillip1882 and alizawi have N.

Care to give the sequence?

Probability is still up for grabs: phillip1882 is close, but not quite.