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18 votes (51.42%)
13 votes (37.14%)
5 votes (14.28%)
2 votes (5.71%)
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3 votes (8.57%)
6 votes (17.14%)
5 votes (14.28%)
10 votes (28.57%)
8 votes (22.85%)

35 members have voted

gordonm888
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gordonm888
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November 26th, 2021 at 8:21:27 AM permalink
Eureka!

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.
Last edited by: gordonm888 on Nov 26, 2021
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy
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November 27th, 2021 at 8:30:28 AM permalink
Quote: Gialmere

Quote: Wizard

Happy 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.
Gialmere
Gialmere
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November 27th, 2021 at 12:15:54 PM permalink
Yes, you serve yourself first. The question is worded poorly. Here 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.
Have you tried 22 tonight? I said 22.
Wizard
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Wizard
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November 27th, 2021 at 6:03:37 PM permalink
Quote: Gialmere

Here 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.
Last edited by: Wizard on Dec 27, 2021
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
Gialmere
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November 30th, 2021 at 8:33:24 AM permalink


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?
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
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November 30th, 2021 at 8:55:33 AM permalink
Quote: Gialmere

...What was the capacity of the jug?

Let x be the ratio of wine removed into the jug. Then the keg now holds 10*(1-x) wine and 10*x water. So the keg now has a ratio of x water. In the second drawing the ratio of the keg removed is x, so the amount of water is 10*x * x. Thus the total water removed is 10 * (x2+x). Since the keg now has equal quantities of wine and water, x2+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.
Edit: I can see my mistake was using the water removed in the second phase, please ignore this answer!
Last edited by: charliepatrick on Nov 30, 2021
ChesterDog
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Gialmere
November 30th, 2021 at 10:11:35 AM permalink
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 - J2/10 + 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:
J2 - 20J + 50 = 0

Add 50 to both sides and manipulating:
J2 - 20J + 100 = 50
(J - 10)2 = 50
J - 10 = +/- 501/2
J = 10 - 501/2 or J = 10 + 501/2
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.
ThatDonGuy
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Gialmere
November 30th, 2021 at 10:19:13 AM permalink
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 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

charliepatrick
charliepatrick
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Gialmere
November 30th, 2021 at 11:42:13 AM permalink
Quote: Gialmere

...What was the capacity of the jug?

Let x be the ratio of wine removed into the jug. Then the keg now holds 10*(1-x) wine and 10*x water. So the keg now has a ratio of x water and (1-x) of wine. In the second drawing the ratio of the keg removed is x, so the amount of wine removed is 10*x*(1-x). Thus the total water removed is 10* (x + x*(1-x) ). Since the keg now has equal quantities of wine and water, 2x-x2=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.
Wizard
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Wizard
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Gialmere
November 30th, 2021 at 12:49:19 PM permalink
I also get 10-5*sqrt(2) =~ 2.9289 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)
It's not whether you win or lose; it's whether or not you had a good bet.

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