Poll

 Both are too hard. No votes (0%) One is I can handle but no clue on question 2. 1 vote (16.66%) Both are too easy. 1 vote (16.66%) I'm learning Korean in my spare time. 1 vote (16.66%) I'm wasting my spare time here. 1 vote (16.66%) This lock down is making me fat. 1 vote (16.66%) I want to help somehow, but laziness prevailed. 1 vote (16.66%) Total eclipse reminder -- 04/08/2024 4 votes (66.66%) What's the next TV show I should binge on? 1 vote (16.66%) Any WoV CV cases yet? 1 vote (16.66%)

6 members have voted

Wizard
Joined: Oct 14, 2009
• Posts: 22054
March 20th, 2020 at 2:24:35 PM permalink
I know this one is similar to the Ace2 variant of the light bulb puzzle, but I'm finding what Ace2 brought to the table very interesting.

You have a single fair six-sided die. You keep rolling it until you've rolled each side at least twice.

Question 1: What is the average number of rolls needed?
Question 2: Put the solution in a closed form expression (warning -- difficult -- I think).

Usual rules apply. If you're already a member of the beer club and can't restrain yourself for 24 hours, feel free to PM me.

Last edited by: Wizard on Mar 20, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
Joined: Oct 16, 2011
• Posts: 453
March 20th, 2020 at 5:50:15 PM permalink
I think you should change it to ..... until you've rolled each side AT LEAST twice
Wizard
Joined: Oct 14, 2009
• Posts: 22054
March 20th, 2020 at 8:24:25 PM permalink
Quote: ssho88

I think you should change it to ..... until you've rolled each side AT LEAST twice

Good point. I just made that edit. I think we need to let the 24-hour clock run out before the fun can begin.
It's not whether you win or lose; it's whether or not you had a good bet.
EdCollins
Joined: Oct 21, 2011
• Posts: 1248
March 20th, 2020 at 10:02:40 PM permalink
I'm not looking for a beer, but for Question 1, I think the average number of rolls needed is 28.165.
I have no idea regarding Question 2.
Suited89
Joined: Dec 23, 2019
• Posts: 93
March 20th, 2020 at 10:24:10 PM permalink
If at least twice I estimated the puzzle as

28 + 1/6 as n*((2n+1)/n)^2 where n is 6 sides of the die.
some people need to reimagine their thinking
Klopp
Joined: Mar 10, 2020
• Posts: 16
March 21st, 2020 at 3:27:31 AM permalink
I am not looking for a beer either, but I think the answer to question 1 is 24.134
ssho88
Joined: Oct 16, 2011
• Posts: 453
March 21st, 2020 at 3:59:23 AM permalink
Quote: Klopp

I am not looking for a beer either, but I think the answer to question 1 is 24.134

Agreed. I got the same answer( through a not so intellectual method)
Wizard
Joined: Oct 14, 2009
• Posts: 22054
March 21st, 2020 at 5:31:27 AM permalink
Quote: EdCollins

I'm not looking for a beer, but for Question 1, I think the average number of rolls needed is 28.165.
I have no idea regarding Question 2.

I disagree on question 1, but you're in the ballpark
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Joined: Oct 14, 2009
• Posts: 22054
March 21st, 2020 at 5:33:54 AM permalink
Quote: Suited89

If at least twice I estimated the puzzle as

28 + 1/6 as n*((2n+1)/n)^2 where n is 6 sides of the die.

Let's see, where's my calculator...

That comes out to 56.166666, which is way too many.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Joined: Oct 14, 2009
• Posts: 22054
March 21st, 2020 at 5:35:21 AM permalink
Quote: Klopp

I am not looking for a beer either, but I think the answer to question 1 is 24.134

I agree.

To get the beer, one must show his work. I'll award two beers for this one, one for each part.