# Help with high school algebra problem (some probability)

• October 12th 2009, 03:40 PM
Sarah-
Help with high school algebra problem (some probability)
Keekerik is an imaginary land where the people have an interesting three-stage ritual for couples who want to get married. Wandalina and Gerik are in that situation so they go the the hope of Queen Katalana to perform this ritual. Permission for them to marry as soon as they wish depends on the outcome of the ritual.

The queen greets them and reaches into a box and pulls out six identical strings for the ritual. The queen hands the strings to Wandalina who holds them firmly in her fist. One end of each string is sticking out above and below her fists.

The queen steps to the side and Gerik is called forward. He ties two of the ends together above Wandalina's fist. Then he ties two other ends above her fist together. Finally, he ties the last two ends above her fist together. The six ends below Wandalina's fist are still hanging untied.

Now Queen katalana comes forward again. Although she was watching Gerik, she has no idea which string end below Wandalina's fist belongs to which end above. The queen does the final step and randomly picks two of the ends below and ties them together, then two more, and finally the last two. So Wandalina now has six strings in her first with three knots above and three knots below.

Whether Wandalina and Gerik will be allowed to marry right away depends on what happens when Wandalina opens her first. If the six strings form one large loop, then they will. Otherwise, they will be required to wait and reapat the ritual in six months.

Questions:

A. When Wandalina opens her first and looks at the strings, what combination of different size loops might there be?

B. What is the probability that the strings will form one big loop? In other words, what are the chances that Wandalina and Gerik will be able to marry right away?

C. What is the probability for each of the other possible combinations?

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I'm no math expert (that's why I'm asking you), but I guess this entire problem is mostly about probability. How should I go about answering these questions? I'm totally stuck. (Thinking)
• October 12th 2009, 09:47 PM
Soroban
Hello, Sarah-!

Who assigned this problem?
The solution is far too long . . .

Quote:

Keekerik is an imaginary land where the people have an interesting three-stage ritual
for couples who want to get married. Wandalina and Gerik are in that situation
so they go the the hope of Queen Katalana to perform this ritual. Permission for them
to marry as soon as they wish depends on the outcome of the ritual.

The Queen reaches into a box and pulls out six identical strings for the ritual.
The Queen hands the strings to Wandalina who holds them firmly in her fist.
One end of each string is sticking out above and below her fists.

The Queen steps to the side and Gerik is called forward.
He ties two of the ends together above Wandalina's fist.
Then he ties two other ends above her fist together.
Finally, he ties the last two ends above her fist together.
The six ends below Wandalina's fist are still hanging untied.

Now Queen katalana comes forward again. Although she was watching Gerik,
she has no idea which string end below Wandalina's fist belongs to which end above.
The Queen does the final step and randomly picks two of the ends below
and ties them together, then two more, and finally the last two.
So Wandalina now has six strings in her first with three knots above and three knots below.

Whether Wandalina and Gerik will be allowed to marry right away depends on what happens
when Wandalina opens her first. If the six strings form one large loop, then they will.
Otherwise, they will be required to wait and repeat the ritual in six months.

Questions:

A. When Wandalina opens her first and looks at the strings,
what combination of different size loops might there be?

There are 225 possible outcomes, but only 4 configurations:
. . Three loops of two strings each.
. . Two loops, one with two strings, one with four strings.
. . Two loops, both with three strings.
. . One big loop, using all six strings.

Quote:

B. What is the probability that the strings will form one big loop?
Number the strings from 1 to 6.

Gerik can tie the upper ends in 15 different ways.

. . $\begin{array}{c}(12, 34, 56) \\ (12,35,46) \\ (12,36,45) \end{array} \quad\begin{array}{c} (13,24,56) \\ (13,25,46) \\ (13,26,45) \end{array}$ . . $\begin{array}{c}(14,23,56) \\ (14,25,36) \\ (14,26,35) \end{array}\quad \begin{array}{c}(15,23,46) \\ (15,24,36) \\ (15,26,34) \end{array}$ . . $\begin{array}{c}(16,23,45) \\ (16,24,35) \\ (16,25,34) \end{array}$

The Queen has the same 15 choices for tying the lower ends.

Hence, there are: . $15 \times 15 \:=\:225$ possible outcomes.

For any one of Gerik's choices, the Queen has 8 choices for forming one big loop.

For example, if Gerik chooses this combination $(12,34,56)$
. . the Queen can form one big loop with one of these eight choices:

. . . . $\begin{array}{ccc}(13,25,46) &&(15,23,46) \\
(13,26,45) && (16,24,36) \\
(14,25,36) && (16,24,35) \\
(14,26,35) && (16,23,45) \end{array}$

I believe the probability of one big loop is $\frac{8}{15}$

And that is all the work I'm willing to do for this problem . . .
.
• October 13th 2009, 01:46 PM
Sarah-
Thanks for your suggestions! I'm still pretty lost in this one, but I'm trying to figure it out. If anyone else wants to chime in and give me assistance, it's greatly appreciated.