Thursday, January 11, 2007

Problems of the Day 1-12-07

Hope you enjoyed the 'inserting the digit 9' problem. Algebraically, 9 is indeed special:
(100t + 90 + u) - (10t + u) = 90t + 90 = 90(t + 1), where t = tens' digit and u = units' digit of the original number.

Some probability today...

For Grade 7 and above:
This just happened. A bio teacher asked the math teachers for a mathematical explanation of a standard Punnett square problem. The question asks for the probability of a Brown-eyed offspring given a Brown-eyed (dominant) male parent and a Blue-eyed (recessive) female parent. Everyone knew the answer was 75% but a clear explanation was desired.

Here are the Punnett diagrams for the two equally likely possibilities.
B = Brown; b = blue; B is dominant so any BB or Bb combination produces a Brown-eyed child. To refresh your student's memory, recall that the paternal genotype is in the left column and the maternal genotype is across the top.

Case I (BB male parent, bb female parent)
Note: Spacing may be off - the b's at the top of each table should be indented to appear as headings in the 2nd and 3rd columns. I didn't have time to enter the html code for Tables.

---b---b
B Bb Bb

B Bb Bb

Case II (Bb male, bb female)

---b---b
B Bb Bb

b bb bb


It's easy to see that in 6 of the 8 cases (75%), the offspring will have brown eyes. (fairly standard stuff you can find in a bio text, but it's always fun to play with this). The science teacher recognized that the probability of a Brown-eyed offspring was 100% in Case I and 50% in Case II, but didn't want to simply tell them to average these. He wanted a more precise method that would work in other problems.
See if your students can explain why it's 75%!

Comments: Here's the explanation given by one of our outstanding math teachers, who thoroughly understands the probability concepts underlying the Punnett square. She said it is like an average but the key is that the 2 cases are equally likely each with a 50% chance of occurring. Thus, (0.5)(1) + (0.5)(0.5) = 0.75 or 75%. Voila! Her explanation allows students to handle more complicated genetic problems with 3 or more cases. Hey, this stuff isn't calculus but math-science articulation is always healthy and doesn't happen as often as it should, right?
--------------------------------------------------------------------------------------------------------------
Now for the challenge problem for your hs students who want to be or believe they are probability experts. Remember, most of us feel insecure about these problems just as our students often feel about math in general.

You have 8 playing cards: 2 jacks, 2 queens, 2 kings and 2 aces. If 4 of these are dealt to you at random, what is the probability you will have at least one matching pair?

Answer (no solution yet): Approx. 77.1% or 27/35 [corrected from earlier posting - thanks to Eric!]
Send me your comments.

7 comments:

Mrs. Temple said...

I assume you are dealing with non-replacement? Since you want at-least 1 pair, you could use the complement rule, find the probability of no pairs and subtract from 1

Eric Jablow said...

Like many other combinatorics problems, it will be easier to calculate the opposite probability, the probability that there is no pair. Let's ask the following questions.

How many ways can one deal 4 cards out of 8, order mattering?

8·7·6·5=1680.

There are 8 choices for the first card, 7 for the second, 6 for the third, and 5 for the fourth.

How many ways can one deal 4 cards, with no pairs occuring?

8·6·4·2=384. There are 8 choices for the first card. After that, there are only 6 cards left that won't pair it. If the first is an Ace, for example, the second card can't be the other Ace. Then, there will be only 4 possibilities for the third card. Finally, there will be only 2 for the last.

So the chance of no pair occurring is 384/1680, and the chance of at least one pair occurring is 1 −384/1680 = 1296/1680.

A more sophisticated solution, if your students know their binomial coefficients, is this.

How many ways can one deal 4 cards from 8 (order not mattering)?

8 choose 4=70. Sorry, I can't type the standard notation in HTML.

How many hands of one Ace, one King, one Queen, and one Jack?

2·2·2·2=16. There are two choices for the Ace, 2 for the King, 2 for the Queen, and 2 for the Jack.

So, the answer is 1−16/70=54/70.

Dave Marain said...

Thank you eric for catching my careless error!
No excuses here but I was actually giving the answer to the problem of 8 pairs of cards (16 in all) instead of 4; I had decided to simplify the problem but I forgot to revise the answer!
Your methods were outstanding and clear. I actually would liken this problem to the 'Birthday Problem', the one in which if you have 23 people randomly selected, the probability is better than 50% that at least 2 would have the same birthday (like Jan 12th). Thus, I approached it like this:
(8/8)(6/7)(4/6)(2/5) = 16/70 = 8/35. Therefore the desired probability is 1 - 8/35 = 27/35. This is virtually the same as your first approach.
Again, thanks for being kind!
Also, thanks to Mrs. Temple for pointing out that dealing 4 cards from a deck implies non-replacement.
Dave M

Mrs. Temple said...

Dave - I was also reminded of the birthday problem, now if only my students would recall our discussion of it!

Jonathan said...

Well, I came late, but there is still more to be done.

On notation: C(4,2) is perfectly acceptable, as is 4C2 with the numbers being subscripted and (42) with the 4 aligned vertically over the 2, much as a fraction in parentheses, but without the fraction bar.

Conditional Probability solution (almost identical to Eric)

1 - (8/8)*(6/7)*(4/6)*(2/5)
1 - 8/35 = 27/35

Combinatorial solution (also counting the hands without pairs, and subtracting):

[C(8,4) - C(2,1)*C(2,1)*C(2,1)*C(2,1)] / C(8,4)

(numerator is all hands minus those made up of 1 out of 2 J's, 1 out of 2 Q's, etc, all over total number of hands)

(70 - 16)/70 = 27/35


P(2 Pair)
Combinatorial approach first:
C(2,2)*C(2,2)*C(4,2)/C(8,4) =

(the C(4,2) is for the different pair pairs we could have, ex J's and Q's or J's and A's)

1*1*6/70 = 3/35

Conditional Probability
(8/8)*(1/7)*(6/6)*(1/5) (for the case where it comes up AABB) * 3 (for the ABBA and ABAB cases) = 144/1680 = 3/35


Probability(exactly one pair)
subtract! 1 - 27/35 - 3/35 = 5/35

Eric Jablow said...

The birthday problem is not simply trivia about birthdays; it has some important consequences you should tell your students about.

One of the most important probles in applied mathematics is that of factorizing large numbers. It's pretty easy to tell whether a number is prime or not. Techniques using the Little Fermat theorem and generalizations of it can determine primality in nearly polynomial time. Agrawal, Kayal, and Saxena found a polynomial-time method in 2002; that isn't practical yet.

On the other hand, prime factoring is difficult. One technique was first discovered by Fermat.
Suppose you want to factor n. Try to find two numbers a and b whose squares are equivalent modulo n. Then, n | (a−b) (a+b), and there's a good chance that n has a prime factor dividing a−b. Just choose candidates for a and b at random less than n. Numbers a little larger than the square root of n work best.

How many candidates do you need to try? Assuming that squares modulo n are as random as birthdays, it's the same answer—about 1.2 times the square root of n. This is called the birthday attack.

The same techniques are used to crack hashing functions like MD5; these are used for digital signatures. You can look these ideas up on MathWorld.

It's not all fun and games. Your students really need to know that mathematics has consequences.

Dave Marain said...

Thank you Eric and Jonathan for your thoughtful and rigorous replies. Eric, I'm not sure where you got the idea that I approach the Birthday Problem frivolously! I use it to engage student interest but develop the concepts carefully. I am indebted to you for showing us the mathematical implications of this to number theory and computer science. I will definitely share this with our students. Out of curiosity, is combinatorics one of your areas of interest? I sense strong computing background as well. I'm very lucky that you and jonathan are reading this and keeping me honest. I now feel the need to edit my work much more carefully or at least when I am semi-conscious! As you can see from my latest post, I am shifting my focus back to national curriculum issues, but I will have new problems this weekend.