Easy math puzzles
Poll
 | 18 votes (50%) | ||
| | 13 votes (36.11%) | ||
| | 5 votes (13.88%) | ||
| | 2 votes (5.55%) | ||
| | 11 votes (30.55%) | ||
| | 3 votes (8.33%) | ||
| | 6 votes (16.66%) | ||
| | 5 votes (13.88%) | ||
| | 11 votes (30.55%) | ||
| | 8 votes (22.22%) |
36 members have voted
To celebrate Thanksgiving, you and 19 mathematicians are seated at a circular table. Everyone at the table would like a helping of cranberry sauce, which happens to be in front of you at the moment.
Instead of passing the sauce around in a circle, you decide to pass it randomly to the person seated directly to your left or to your right. They then do the same, passing it randomly either to the person to their left or right. This continues until everyone has, at some point, received the cranberry sauce.
Of the 20 people in the circle, who has the greatest chance of being the last to receive the cranberry sauce?
Quote: ThatDonGuy
Rephrase the problem: you are looking for positive values p and q (where p is the total number of marbles and q the probability) such that a + b + c = p and 6 abc / (p (p-1) (p-2)) = q for multiple unordered triples (a, b, c).
The second equation can be rewritten as abc = p (p-1) (p-2) q / 6.
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Thank you for your generous answer.
I am running off to my Thanksgiving feast and don't have the time to study your results closely. I am wondering what the properties are of 91 that it (or its multiples) are in the denominator of every q.
Quote: gordonm888I am running off to my Thanksgiving feast and don't have the time to study your results closely. I am wondering what the properties are of 91 that it (or its multiples) are in the denominator of every q.
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That was my mistake - I was assuming p = 14 for each one when I calculated them.
The probabilities have been corrected in my earlier post.
I also just realized that those may not be all of the answers where the probability < 1.
Quote: ThatDonGuy
Yes.
Change 100 to 14 and 0.2 to 18/91:
6 * (8 * 3 * 3) / (14 * 13 * 12) = 18/91
6 * (6 * 6 * 2) / (14 * 13 * 12) = 18/91
For that matter, change 100 to 14 and 0.2 to 10/91:
6 * (10 * 2 * 2) / (14 * 13 * 12) = 10/91
6 * (8 * 5 * 1) / (14 * 13 * 12) = 10/91
Rephrase the problem: you are looking for positive values p and q (where p is the total number of marbles and q the probability) such that a + b + c = p and 6 abc / (p (p-1) (p-2)) = q for multiple unordered triples (a, b, c).
The second equation can be rewritten as abc = p (p-1) (p-2) q / 6.
One factor that limits the solutions in this particular case is, q is a probability, so q < 1 is a boundary condition.
Here are the only pairs of (total number of balls, probability of getting one of each) that have more than one solution:
Edit - probabilities have been corrected
13, 18/143 has (9, 2, 2) and (6, 6, 1)
14, 10/91 has (10, 2, 2) and (8, 5, 1)
14, 18/91 has (8, 3, 3) and (6, 6, 2)
16, 9/56 has (10, 3, 3) and (9, 5, 2)
17, 18/85 has (9, 4, 4) and (8, 6, 3)
19, 48/323 has (12, 4, 3) and (9, 8, 2)
20, 3/38 has (15, 3, 2) and (10, 9, 1)
21, 48/665 has (16, 3, 2) and (12, 8, 1)
21, 12/95 has (14, 4, 3) and (12, 7, 2)
21, 24/133 has (12, 5, 4) and (10, 8, 3)
22, 18/77 has (10, 6, 6) and (9, 8, 5)
23, 360/1771 has (12, 6, 5) and (10, 9, 4)
24, 30/253 has (16, 5, 3) and (12, 10, 2)
25, 18/115 has (15, 6, 4) and (12, 10, 3)
26, 63/1300 has (21, 3, 2) and (18, 7, 1)
26, 27/260 has (18, 5, 3) and (15, 9, 2)
26, 36/325 has (18, 4, 4) and (12, 12, 2)
27, 56/975 has (21, 4, 2) and (14, 12, 1)
28, 44/819 has (22, 4, 2) and (16, 11, 1)
28, 80/819 has (20, 4, 4) and (16, 10, 2)
29, 20/203 has (20, 6, 3) and (15, 12, 2)
31, 40/899 has (25, 4, 2) and (20, 10, 1)
31, 45/899 has (25, 3, 3) and (15, 15, 1)
32, 117/2480 has (26, 3, 3) and (18, 13, 1)
34, 63/1496 has (28, 3, 3) and (21, 12, 1)
34, 135/2992 has (27, 5, 2) and (18, 15, 1)
35, 8/187 has (28, 5, 2) and (20, 14, 1)
39, 297/9139 has (33, 3, 3) and (27, 11, 1)
42, 36/1435 has (36, 4, 2) and (32, 9, 1)
43, 360/12341 has (36, 5, 2) and (30, 12, 1)
link to original post
In ThatDonGuy's table, the denominator of the probability q tends to be the product of the two (or even three) largest prime factors of
p*(p-1)*(p-2)
For example, when p= 19, the value of q = 48/323. Note that 323 = p *(p-2) or 19*17. This makes perfect sense as the solutions must generally require small prime factors -with no large prime factors - for p*(p-1)*(p-2)*q/6. So the q is a fraction that has a denominator that wipes out the large prime factors and replaces them with a numerator that has many small prime factors.
Note the last row on the list, for p =43. The product p*(p-1)*(p-2) is 43*42*41 and the probability q is 360/12341 and 12341 =43*41*7. Thus, the value of q that works for p=43 is one for which the denominator eliminates the large prime factors of p*(p-1)*(p-2) and introduces a set of small prime factors with its numerator.
Quote: GialmereHappy Thanksgiving!!Of the 20 people in the circle, who has the greatest chance of being the last to receive the cranberry sauce?
link to original post
Happy Thanksgiving to you too and all the math geniuses of the forum!
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.
Quote: 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).
------------------------------------------------------------
Bad Thanksgiving jokes...
Quote: GialmereVery 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).
I'm working on a better way to state my answer. Here is what I'm trying to prove:
Consider a gambler playing a 50/50 game that pays even money. Gambler bets one unit at a time. I submit these chances are the same:
1. Gambler wins $(a+b) before losing one unit.
2. Gambler loses $a and wins $b in the same session, before exceeding either marker.
It's seems intuitive to me, but working how to put an argument in words.
Quote:Bad Thanksgiving jokes...
To get to the other sides.but some people say that’s irrational.Scared the hell out of everyone else in the grocery store.
link to original post
Those were so bad they're good!
Quote: gordonm888
In ThatDonGuy's table, the denominator of the probability q tends to be the product of the two (or even three) largest prime factors of
p*(p-1)*(p-2)
For example, when p= 19, the value of q = 48/323. Note that 323 = p *(p-2) or 19*17. This makes perfect sense as the solutions must generally require small prime factors -with no large prime factors - for p*(p-1)*(p-2)*q/6. So the q is a fraction that has a denominator that wipes out the large prime factors and replaces them with a numerator that has many small prime factors.
Note the last row on the list, for p =43. The product p*(p-1)*(p-2) is 43*42*41 and the probability q is 360/12341 and 12341 =43*41*7. Thus, the value of q that works for p=43 is one for which the denominator eliminates the large prime factors of p*(p-1)*(p-2) and introduces a set of small prime factors with its numerator.
link to original post
There's a reason they "tend to be" that way.
Let a, b, c be the three numbers, with p = a + b + c and q = abc.
The probability of drawing colors a, b, c, in that order is a / p * b / (p - 1) * c / (p - 2) = q / (p (p-1)(p-2)).
Since the order of colors is arbitrary, and there are six orders, the overall probability is 6q / (p (p-1) (p-2)).
I have also discovered the following:
39 balls and probability 1,200/9,139 has 3 solutions:
25, 8, 6
24, 10, 5
20, 15, 4
118 balls and probability 3,150/22,243 has 4 solutions:
72, 25, 21
70, 30, 18
63, 40, 15
54, 50, 14
185 balls and probability 4,158/51,911 has 5 solutions:
135, 28, 22
132, 35, 18
126, 44, 15
110, 63, 12
90, 84, 11
400 balls and probability 1,512/18,905 has 6 solutions:
288, 70, 42
280, 84, 36
270, 98, 32
252, 120, 28
245, 128, 27
196, 180, 24
511 balls and probability 56,160/631,669 has 7 solutions:
364, 75, 72
360, 91, 60
336, 130, 45
325, 144, 42
315, 156, 40
280, 195, 36
260, 216, 35
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).
How can the first person possibly be the last person - unless you are not counting the first person as having the cranberries until (and unless) they are passed to that person from someone else - i.e. the problem asks for the last person to have the cranberries passed to them?
Quote: ThatDonGuyQuote: gordonm888
In ThatDonGuy's table, the denominator of the probability q tends to be the product of the two (or even three) largest prime factors of
p*(p-1)*(p-2)
For example, when p= 19, the value of q = 48/323. Note that 323 = p *(p-2) or 19*17. This makes perfect sense as the solutions must generally require small prime factors -with no large prime factors - for p*(p-1)*(p-2)*q/6. So the q is a fraction that has a denominator that wipes out the large prime factors and replaces them with a numerator that has many small prime factors.
Note the last row on the list, for p =43. The product p*(p-1)*(p-2) is 43*42*41 and the probability q is 360/12341 and 12341 =43*41*7. Thus, the value of q that works for p=43 is one for which the denominator eliminates the large prime factors of p*(p-1)*(p-2) and introduces a set of small prime factors with its numerator.
link to original post
There's a reason they "tend to be" that way.
Let a, b, c be the three numbers, with p = a + b + c and q = abc.
The probability of drawing colors a, b, c, in that order is a / p * b / (p - 1) * c / (p - 2) = q / (p (p-1)(p-2)).
Since the order of colors is arbitrary, and there are six orders, the overall probability is 6q / (p (p-1) (p-2)).
I have also discovered the following:
39 balls and probability 1,200/9,139 has 3 solutions:
25, 8, 6
24, 10, 5
20, 15, 4
118 balls and probability 3,150/22,243 has 4 solutions:
72, 25, 21
70, 30, 18
63, 40, 15
54, 50, 14
185 balls and probability 4,158/51,911 has 5 solutions:
135, 28, 22
132, 35, 18
126, 44, 15
110, 63, 12
90, 84, 11
400 balls and probability 1,512/18,905 has 6 solutions:
288, 70, 42
280, 84, 36
270, 98, 32
252, 120, 28
245, 128, 27
196, 180, 24
511 balls and probability 56,160/631,669 has 7 solutions:
364, 75, 72
360, 91, 60
336, 130, 45
325, 144, 42
315, 156, 40
280, 195, 36
260, 216, 35
link to original post
Yes, I have been abstracting this "balls with three colors" problem and thinking of it as:
Can multiple partitions of a number, p, with the same length (in this case, length 3) have the identical set of prime factors? And if so, under what circumstances?
or
What are the characteristics of a set of prime numbers (which may include multiples) that can generate multiple partitions of the same length of the same number?
Stated like that, the problem is no longer one of probability, or balls of different colors. Thus, it was disturbing me that the existence of partitions of the same length with the same set of prime factors might depend upon the values of (p-1) and (p-2). What I was struggling to say is that the denominator of q decouples the partitions of length 3 from the prime factors of the product p*(p-1)*(p-2) and that it is the numerator of q that is central in defining the prime factor set that generates multiple partitions of p of the same length. I agree that when viewed in the framework of the original problem, my previous post was childishly obvious. I had lost myself down the rabbit hole of partitions and their sets of prime factors.
Your results continue to be fascinating. I assume those examples above are the smallest numbers that have n solutions, n = 3-7?
EDIT: A better way to say all this is that your triplets have the same sum and the same product!! Example:
25+8+ 6 = 24+10+5 = 20+15+4
and,
25*8*6 = 24*10*5 = 20*15*4
and its unnecessary to express their common products as a fraction q of p*(p-1)*(p-2)/6.
