## Poll

19 votes (45.23%) | |||

14 votes (33.33%) | |||

6 votes (14.28%) | |||

2 votes (4.76%) | |||

12 votes (28.57%) | |||

3 votes (7.14%) | |||

6 votes (14.28%) | |||

5 votes (11.9%) | |||

12 votes (28.57%) | |||

9 votes (21.42%) |

**42 members have voted**

See: OEIS A103277

"Smallest i such that there exists j such that i = x + y + z, j = x*y*z has exactly n solutions in positive integers x <= y <= z".

3, 13, 39, 118, 185, 400, 511, 1022, 1287, 2574, 4279, 8558, 11777, 24377, 23554, 46111, 99085, 165490

Edit: Interesting that it does not monotonically increase. the 15th term, 23554 is smaller than the 14th term.

Quote:GialmereQuote:WizardHappy Thanksgiving to you too and all the math geniuses of the forum!

I think everyone has an equal chance to be last.

My solution is I answered this for the 3, 4, and 5-person case and all of them showed, hopefully correctly, that everyone had an equal chance of being last. I can explain how I did those cases on request, but I thought of it like a coin flipping game where the goal was to keep flipping until you won $x or lost $y.

Extending the logic for those three cases, I say regardless of the number of mathematicians, everyone will have an equal chance to be last.

Correct!!

Very good.

Yes. It's a surprisingly fair way to determine who gets stuck with the last helping (if you don't mind spending hours at the table passing food around).

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I am not getting this. What I get is:

If the first person is included in having "touched" the cranberries at the start, then yes, everybody but that person has an equal chance.

If the first person is not included until the cranberries are passed to that person by someone else, then the two people adjacent to the first person are half as likely as the others.

Quote:GialmereHere is a link to the Riddler page containing the official solve (at the bottom). I would just post it here but the rather fun pie chart animations from Twitter will not embed.

link to original post

Thank you. At first I thought that was a "hand waving" kind of proof, but now I can see the logic in it. Here is how I might word it.

Let the starting position be x and y be any other specific person you wish to find the probability of being last.

Eventually, the cranberry sauce will find it's way to someone to either side of y. Then, for y to be last, the sauce must immediately move away from y and eventually get all the way to y's other neighbor, without ever getting to y from the other side. That is 19 steps it must move.

If you have one unit to gamble with on a 50/50 game, your chances of winning n units before losing the one unit is 1/(1+n). I can prove this upon request, but I hope it's obvious.

So, in the case of the sauce moving 18 steps before moving one from the starting point is 1/19.

You can use this logic for anyone at the table.

A man had a 10-gallon keg of wine and a jug. One day, he drew a jugful of wine and then topped off the keg with water. Later on, when the wine and water had gotten thoroughly mixed, he drew another jugful and again topped off the keg with water. The keg then contained equal quantities of wine and water.

What was the capacity of the jug?

Quote:Gialmere...What was the capacity of the jug?

^{2}+x). Since the keg now has equal quantities of wine and water, x

^{2}+x=1/2. Thus x = ( -1 +/- SQRT(1+2) ) / 2. This gives x=(SQRT(3)-1)/2, i.e. the keg is about 3.660 gallons.

Quote:Gialmere

A man had a 10-gallon keg of wine and a jug. One day, he drew a jugful of wine and then topped off the keg with water. Later on, when the wine and water had gotten thoroughly mixed, he drew another jugful and again topped off the keg with water. The keg then contained equal quantities of wine and water.

What was the capacity of the jug?

link to original post

Let J = volume of the jug.

Now, follow the water. Initially, the volume of water in the wine keg is 0. The first step (removing J wine from the keg) does not change the volume of water in the keg.

The second step adds a volume of J water to the keg.

The third step removes a fraction (J/10) of the previously added water. So (J/10)*J water has been removed.

The last step adds J water to the keg.

So, the water in the jug is now 0 + 0 + J - J

^{2}/10 + J which should equal 5 if the volume of water in the keg equals the volume of wine in the keg.

Solve this equation for J:

J

^{2}- 20J + 50 = 0

Add 50 to both sides and manipulating:

J

^{2}- 20J + 100 = 50

(J - 10)

^{2}= 50

J - 10 = +/- 50

^{1/2}

J = 10 - 50

^{1/2}or J = 10 + 50

^{1/2}

J = 2.92893... or J = 17.0711...

Of course J cannot be greater than 10, so the volume of the jug must be about 2.93 gallons.

Quote:GialmereA man had a 10-gallon keg of wine and a jug. One day, he drew a jugful of wine and then topped off the keg with water. Later on, when the wine and water had gotten thoroughly mixed, he drew another jugful and again topped off the keg with water. The keg then contained equal quantities of wine and water.

What was the capacity of the jug?

link to original post

Let J be the capacity of the jug

At the start, the keg has 10 (gallons) wine and no water

Draw a jug of wine, and replace it with a jug of water:

The keg has 10-J wine and J water

Draw another jug of wine, and replace it with water:

The amount drawn is J (10 - J)/10 = (10 J - J^2) / 10 wine

The amount of wine remaining in the keg is 10 - J - (10 J - J^2) / 10

Since the amount of wine remaining is half the total, 10 - J - (10 J - J^2) / 10 = 5

100 - 10 J - (10 J - J^2) = 50

50 - 20 J + J^2 = 0

J = 10 +/- sqrt(400 - 200) / 2 = 10 +/- 5 sqrt(2)

Since 10 + 5 sqrt(2) is greater than the capacity of the keg, the only solution is 10 - 5 sqrt(2) gallons

Quote:Gialmere...What was the capacity of the jug?

^{2}=1/2. Thus x = ( -2 +/- SQRT(4-2) ) / 2. This gives x=1-SQRT(2)/2, i.e. the keg is about 2.929 gallons.

Let j = volume of jug.

Wine after second dipping = wine after first dipping * ratio of wine after first dipping

5 = (10-j)*((10-j)/10)

j^2 - 20j + 50 = 0

j = 10 - 5*sqrt(2)