[bonanova]

According to folklore, marital fortunes were once told by a simple method.

A girl would hold six long blades of grass in her closed hand, the blades protruding from each side.

Another girl would tie the ends in pairs, at random, first on one side, then on the other.

The first girl would then open her hand and inspect how the blades were connected.

The possible outcomes are three small loops; one small loop and one medium loop; and one large loop.

If the blades became one large loop, the girl doing the tying would be married within the year!

The puzzle is in two parts:

.

1. What is the probability of obtaining one large loop?
2. How many blades of grass must the first girl hold for the probability to fall to 1/8?

.

As always, enjoy!

[Guest]

Consider the 6 blades on one side, ABCDEF, to be paired A-B, C-D, E-F. On the other side, the first blade chosen has a 4/5 chance of allowing a single loop by avoiding its partner on the other side. Then the second blade only has a 2/4 chance of allowing the large loop. This makes the probability of one large loop 2/5. Is this correct, or am I oversimplifying it?

[Guest]

I just figured out all the combinations for one side, lets say the right side

Then on the left side I paired (1,2) (3,4) (5,6)

Then I determined the proportions of each case with the left side as I said with every single combination on the right

This proportion will be the same as the proportion of the cases if you did every single combination on the left, with every single combination on the right

Because if you have (1,3) (2,4) (5,6) on the left, and then pair that up with every combination on the right you could just rearrange the left back to (1,2) (3,4) (5,6) (while they are still tied, so it changes the order on the right) and have every single one correspond to exactly one combination as if you started with (1,2) (3,4) (5,6) on the left

The numbers I got were

1/15 are 3 smalls

6/15 are 1 small, 1 medium

8/15 are 1 large

And I'm working on part two...

Edited July 18, 2009 by K4D

[superprismatic]

According to folklore, marital fortunes were once told by a simple method.

A girl would hold six long blades of grass in her closed hand, the blades protruding from each side.

Another girl would tie the ends in pairs, at random, first on one side, then on the other.

The first girl would then open her hand and inspect how the blades were connected.

The possible outcomes are three small loops; one small loop and one medium loop; and one large loop.

If the blades became one large loop, the girl doing the tying would be married within the year!

The puzzle is in two parts:

.

1. What is the probability of obtaining one large loop?
2. How many blades of grass must the first girl hold for the probability to fall to 1/8?

.

As always, enjoy!

8/15

It never will! For 2n+2 blades, the probablility of a loop is Product{i=1 to n}(2i/(2i+1)). This product has all odd factors in its denominator. Since there are never any 2s in the factorization of the denominator, the fraction could never simplify to 1/8.

[superprismatic]

8/15

It never will! For 2n+2 blades, the probablility of a loop is Product{i=1 to n}(2i/(2i+1)). This product has all odd factors in its denominator. Since there are never any 2s in the factorization of the denominator, the fraction could never simplify to 1/8.

The closest that the probability gets to 1/8 is with 110 blades, having a loop probability of approximately .11977.

[bushindo]

The closest that the probability gets to 1/8 is with 110 blades, having a loop probability of approximately .11977.

Shouldn't it be a bit less? You're probably thinking that 1/8 = .12000

1/8 = .125. If you have 100 or 102 blades, you'll have a probability of 0.1256451 or 0.1244011, respectively. Still seems an awful lot to me, but induction said so, so it must be so.

[superprismatic]

Shouldn't it be a bit less? You're probably thinking that 1/8 = .12000

1/8 = .125. If you have 100 or 102 blades, you'll have a probability of 0.1256451 or 0.1244011, respectively. Still seems an awful lot to me, but induction said so, so it must be so.

Oops, you'r right! Stupid slip up on my part. Thanks!

[Guest]

Consider the 6 blades on one side, ABCDEF, to be paired A-B, C-D, E-F. On the other side, the first blade chosen has a 4/5 chance of allowing a single loop by avoiding its partner on the other side. Then the second blade only has a 2/3 chance of allowing the large loop. This makes the probability of one large loop 8/15. Is this correct, or am I oversimplifying it?

Ooops, I corrected a mistake.

[Guest]

As I understand the problem, the girl is holding six blades of grass.

The ends on the left would be A,B,C,D,E,F and the ends on the right would be a,b,c,d,e,f.

One at a time, the blades are paired on the left, then on the right.

On the left, there are 6 ways to choose the first blade, and 5 ways to choose which blade to tie it to.

Then there are 4 ways to choose an end and then 3 ways to choose a blade to tie it to.

Then the last pair is automatically chosen.

We will assume that the blades are tied AB, CD, EF.

Because it doesn't matter what is paired on the left, only what is paired on the right.

(Sorry to take so long thinking my way through this puzzle.)

Now on the right, we have the same number of ways to choose the ends (a,b,c,d,e,f) to tie.

Assuming the first blade chosen is A's other end a, the odds of choosing a small loop on the first choice is 1/5.

This is the chance of choosing b instead of c, d, e, or f.

If this was a small loop (1/5), the remaining problem of 2 small loops or 1 medium loop is the 4 blade problem (see below).

Assuming the first choice was not a small loop (4/5), we have 1/2 chance of picking the other end of one of them.

If we pick the other end of one of them (1/2), there is a 1/3 chance of completing a small loop.

If we pick the unchosen end (1/2), there is a 1/3 chance of completing a small loop.

So, the chance of completing a small loop after not completing a small loop is 4/5 * (1/2 * 1/3 + 1/2 * 1/3) = 4/15.

The chance of NOT completing a small loop after not completing a small loop is 4/5 * (1/2 * 2/3 + 1/2 * 2/3) = 8/15.

This is the odds of completing a single large loop using 6 blades.

4 blade problem:

Assuming the first choice was a small loop, the remaining problem becomes one of 4 blades: AB, CD.

Assuming the first blade chosen is A's other end a, the odds of choosing a small loop on the first choice is 1/3.

Assuming the first choice was not a small loop (ac or ad), the other pair (bc or bd) is also not a small loop.

Assuming the first choice was a small loop (ab), the other end is automatically a small loop.

So, the odds of choosing two small loops with 4 blades is 1/3.

The odds of choosing a medium loop with 4 blades is 2/3.

The odds of choosing three small loops with 6 blades is 1/5 * 1/3 = 1/15.

The odds of choosing a single large loop with 6 blades is 4/5 * 2/3 = 8/15.

The odds of choosing a medium loop and a small loop with 6 blades is 4/5 * 1/3 + 1/5 * 2/3 = 6/15.

My head is spinning.

I will revisit this later for the 8 and 10 blade problems to see if I can see a general formula for the probabilities.

Edited July 20, 2009 by Kerrik

[Guest]

In order to get the large loop, the girl may tie the first end to any of 4 other ends (other than the one that makes a small loop)

Then pick any second end and it can be tied to any of 2 other ends (to avoid a small loop)

The remaining two are tied in 1 way

Total number of ways in which this can be done = 4.2 = 8

Consider 8 blades. These can be arranged as

4 small loops of 1 pair each

1small and 1 loop of 3 pairs

2 small and 1 loop of 2 pairs

2 loops of 2 pairs

1 large loop of 4 pairs

The number of ways of each is

1, 4x8, 6x2, 6x2 respectively

and the last one (large loop can be made as follows using the same logic as before

6x4x2 = 48

total ways = 1 + 32 + 12 + 12 + 48 = 105

This is same as number of ways in which the blades may be tied (7x5x3 = 105)

This means that for any 2n number of blades, the possible ways in which the blades may be tied on the other side are

(2n-1).(2n-3).(2n-5)…..(2n-2n+1)

The number of ways in which a large loop can be made are

(2n-2).(2n-4).(2n-6)…..(2n-2n+2)

2ⁿ.(n-1)!

Now, we need to find n such that

2ⁿ.(n-1)! / (2n-1).(2n-3).(2n-5)…..(2n-2n+1) = 1/8

For this, n = 50

So the number of blades must be 100

This gives a probability of 0.1256 which is slightly higher than 1/8 (0.125)

Exactly! Kudos also to Bushindo.

To superprismatic: I should have asked when the probability falls below 1/8.

I found it surprising the answer was that large and, for no real reason, an even power of 10.

Edited July 20, 2009 by bonanova

[bonanova]

I meant for the comments added mistakenly to DeeGee's last post

to be a reply of my own.

Moderators get confused sometimes.