djtehch34t Joined: May 7, 2016
• Posts: 42
March 15th, 2020 at 4:37:17 PM permalink
Posting my PM'd answer at the Wizard's request:

First, note that if we know the expected time T_N for the first of N lightbulbs to burn out, then we can reduce the problem to T_N +T_{N-1}, since the exponential distribution is memoryless. If there are N lightbulbs, since each expires independently, then they are N times as likely to burn out as 1 lightbulb. So, the expected time T_N is equal to 1/N. IIRC, it's also true that it follows an exponential distribution with mean 1/N, but we only really need the expected time for the first lightbulb for this problem. Then, we can sum over all 10 lightbulbs, giving 1/10 + 1/9... + 1 = the 10th Harmonic number H_10 = 7381/2520.
Wizard Joined: Oct 14, 2009
• Posts: 21932
March 15th, 2020 at 4:45:55 PM permalink
Quote: djtehch34t

Posting my PM'd answer at the Wizard's request:

First, note that if we know the expected time T_N for the first of N lightbulbs to burn out, then we can reduce the problem to T_N +T_{N-1}, since the exponential distribution is memoryless. If there are N lightbulbs, since each expires independently, then they are N times as likely to burn out as 1 lightbulb. So, the expected time T_N is equal to 1/N. IIRC, it's also true that it follows an exponential distribution with mean 1/N, but we only really need the expected time for the first lightbulb for this problem. Then, we can sum over all 10 lightbulbs, giving 1/10 + 1/9... + 1 = the 10th Harmonic number H_10 = 7381/2520.

Congratulations! This correct. I owe you another beer.
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ksdjdj Joined: Oct 20, 2013
• Posts: 894
March 15th, 2020 at 5:11:48 PM permalink
I will have a go, the worst that will happen, is that I won't "win/be owed a beer" (which would be the same result as not answering at all)
Note: some of this may be "irrelevant to the question", but I am also trying to learn.

If the average life(mean) for 1 light bulb is 1 year, then 50% of the light bulbs last ~0.69*** years (or less)

*** = LN(0.5)/-1

So, this got me thinking, for 10 light bulbs to = 0.5, you need to do this

0.5√10 = 0.93....
then 1-0.93.... = 0.0669...

Therefore. the average life of the last bulb is : 2.70...^^^ years

^^^ = LN(0.0669670084631926)/-1

(Edit/update): Since the answer by djtehch34t above is correct, I didn't find the average (mean), instead I think I found the "median" (but I could easily be wrong, since I am new at this).

Note: In this case, the mean should be bigger than the median, going by the articles I have read on the internet.

Note 2: I hope I am using the term median correctly (that is why i put " " around it, in case I am wrong in my interpretation).

Note 3: Again, I know the question was "....mean waiting time for the LAST bulb to burn out" , but I am mainly trying to learn, so I will be happy if I am correct with just getting the "median"

----
Edit/update: Congrats to djtehch34t (the response from the Wiz wasn't posted, when I first started writing my post).
Last edited by: ksdjdj on Mar 15, 2020
Ace2 Joined: Oct 2, 2017
• Posts: 680
March 17th, 2020 at 10:04:08 AM permalink
Quote: Wizard

You have 10 light bulbs going. Each has an average life of 1 year. The life expectancy of each bulb follows an exponential distribution. What this means is the bulbs are never overdue to burn out. Regardless of how long a bulb has been burning, it is no closer to burning out. The average remaining time is always one year per bulb.

Extra credit:

Same problem except each light bulb will be instantly* replaced once and only once after it burns out the first time. On average, how long until the last bulb burns out permanently (after being replaced once)?

No beer rules. First correct answer gets a valuable stock market tip.

*assume instantly means zero time needed to replace
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Wizard Joined: Oct 14, 2009
• Posts: 21932
March 17th, 2020 at 4:43:50 PM permalink
Quote: Ace2

Same problem except each light bulb will be instantly* replaced once and only once after it burns out the first time. On average, how long until the last bulb burns out permanently (after being replaced once)?

Good question. I hope to post an answer (hopefully a correct one) later today.
It's not whether you win or lose; it's whether or not you had a good bet.
djtehch34t Joined: May 7, 2016
• Posts: 42
March 17th, 2020 at 8:08:33 PM permalink
Assuming I programmed it right, using the recurrence T_{i, j} = ((i-j)/i T_{i-1, j} + j/i T_{i, j-1}) + 1/i, where i is the current number of bulbs and j is the number of those with a backup, calculating T_{10, 10} via dynamic programming I get ~4.6229.
Ace2 Joined: Oct 2, 2017
• Posts: 680
March 17th, 2020 at 8:39:53 PM permalink
Quote: djtehch34t

Assuming I programmed it right, using the recurrence T_{i, j} = ((i-j)/i T_{i-1, j} + j/i T_{i, j-1}) + 1/i, where i is the current number of bulbs and j is the number of those with a backup, calculating T_{10, 10} via dynamic programming I get ~4.6229.

That is a correct approximation (all digits shown are correct), but there is a formulaic solution with the answer expressed as an integer.
Last edited by: Ace2 on Mar 17, 2020
It�s all about making that GTA
Ace2 Joined: Oct 2, 2017
• Posts: 680
March 18th, 2020 at 9:24:40 AM permalink
I�ll post the solution a bit later today unless someone wants more time?
It�s all about making that GTA
djtehch34t Joined: May 7, 2016
• Posts: 42
March 18th, 2020 at 9:46:58 AM permalink
Quote: Ace2

I�ll post the solution a bit later today unless someone wants more time?

I'd like to work on it tonight, if you don't mind.
Wizard Joined: Oct 14, 2009