My son is in the 4th grade. They have four 4th grade classes in school. Every year classes assignment is done differently so that kids will have a chance to get to know everyone in their age group. There is one mom that always jokes with me that our kids are never in the same class and that perhaps there is some suspicious manipulation behind this "random" class assignment. What are the chances that our kids would never be in the same class during the whole 9 years they spend together in school?
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11 comments:
It's been quite a few years since my last probability class, but I think you want to do (.75)^9 = 7.5% chance of never being in the same class, assuming it's a random, independent selection each year.
Oh, can I submit a second answer? If you factor in the fact that the first few years have already passed and the two kids haven't yet been in the same class, then the probability for previous years is 100%. If the school starts in kindergarten and the mom's comment was spoken after the 4th grade assignments were announced, then that's 5 years out of the 9 where the probability was 100% because those events already occurred. So that leaves only 4 years left: (.75)^4=31.6%
I think Barry is absolutely right in both of his answers. Taking all 9 years and considering only the remaining 4 years.
The probability that two kids are not in the same class is 3/4. The probability of this happening 8 years in a row is (3/4) in the power of 8 = 7.5%
Small but feasible.
New puzzle tomorrow!
I think this is ALMOST, but not exactly right. Say there are 100 kids in 4th grade and each class has 25 (yes, we'd all like smaller classes, but let's go with this because the numbers are easy), then there are 99 kids other than my kid, and 24 slots for other kids in his classroom. So each year there is a 24/99 chance of a particular kid being in his class, or a 75/99 that he won't be. So after nine years, the chance would be (75/99)^9.
Yep, like I said, it's been a while since I studied this. I think Kim is absolutely right.
That is the beauty of math - one could convince others using data and logical reasoning. And Kim so thoroughly and gracefully showed us that we were wrong. She gets a special honorary point and now qualifies for a prize.
Kim, please email me your snail mail address to send the prize to maria at marialando dot com.
I think that ,the class assignment puzzle, hides a "trap" which I fell into and so did the other puzzle solvers. The trap is the assumption that each year's random class assignment is an event which is dependent on the previous year's class assignment.
If each year's class assignment is random (The kid is assigned to a class by a draw) then the probability to be in a class with a specific other kid will be 1:4 no matter how many years you attend school.In other words: It is not as if I was drawn to go to class number 1 last year reduces my chances to draw the same class (number 1) this year.
Barry was right, Kim's answer is wrong. I think that Kim's answer is right for any particular room, but there are four rooms. Imagine there are 4 kids, and 2 rooms. Using ABCD as the four kids, you can list all of the room assignments possible -- there are 12 total. Of these A and B are together 1/4 of the time. Also, A and C are 1/4, and A and D are 1/4. So using brute force shows Barry is right, not Kim.
This turned out to be slightly more complex but much more interesting puzzle than what I originally meant. Love it! Comments' format does not allow for a thorough explanation, but let’s try:
Kim was right. To make absolutely sure I am not misleading anyone, I consulted a friend of mine, Professor of Math in University of Chicago. He confirmed that Kim was right and praised her explanation.
Let's see what is different between Kim and Barry’s solutions:
Forget about the power of 9 that we all agree about.
Kim says we will get a 75/99 chance of kids not being in the same class. Barry says it is 75/100.
Lets write it with variables, so that we can substitute any number of kids and classes.
Assume that N is the total number of kids, C is the number of classes.
Kim’s solution generalization: ((C-1)/C x N) / (N-1)
Barry’s: ((C-1) / C * N ) / N = (C-1) / C
So, in Barry’s solution it does not matter how many kids, only number of classes matters.
When we have 8 kids and 4 classes, N=8, C=4:
Kim: (¾ x 8) / 7 = 6/7
Barry: ¾
Thinking empirically. Put first kid in any class. You have 7 kids left. There is only one more spot in the same class where first kid is. There is 1/7 chance that a specific kid out of this 7 will get this spot, and 6/7 chance that he won’t.
When we have 4 kids and 2 classes, N=4, C=2:
Kim: (½ x 4) / 3 = 2/3
Barry: ½
Now, as prluhmann suggested lets name kids A, B, C, D
Possible class assignments are:
AB CD
AC BD
AD BC
We see that A and B are together 1/3 of the cases and apart 2/3. Just as Kim’s approach says.
You all are amazingly smart!
Kim,
will you come with me next time I go to the race track?
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