More Challenging Problems

http://bsuvc.bsu.edu/~d004ucslabs/probability.html


Since the answer to one of the problems is not given, and the answer to another is completely wrong, my answers will appear in a post below.

1. The probability of not getting heads in 2 tosses is 3/4. To get two heads in a row on exactly the 10th toss, we have to NOT get two in a row in 4*2 flips, and then get two heads on throws 9 and 10. The probability of this is

(3/4)^4(1/2)(1/2) = 7.9%.


2. This result is counterintuitive. The posted solution is fine except the line:


Thus, the probability of both coins coming up with the given results is 2/27.


Should read:


Thus, the probability of both coins coming up with the given results is (1/2)(4/9) = 2/9.


3. (My solution) The posted solution would be correct only if the names were replaced in the hat after each draw. That isn't the way this kind of thing works, or some people might not get any presents, and some might get several. The correct solution is to note that if f(n) is the number of ways n people can draw names without drawing their own, then f(n) satisifes:


f(2) = 1

f(n) = (n-1)f(n-1)


Thus, f(n) = (n-1)!


So the probability is 9!/10! = 1/10.


4. The posted solution is clever. Mine is to note that you can only see two sides of something if one side slopes away from you less than 90 degrees. Since the angles of a pentagon are 108 degrees, the sides "stick out" 18%. We can see 3 sides as long as we are not more than 18 degrees off center to any one side in either direction for a total of 36 degrees per side. Since there are 5 sides, the probability is 36*5/360 = .5.


5. (My solution) The tallest person will always be visible. The second tallest person will be visible if and only if he is in front of the tallest person, which has probability 1/2. The third tallest person will be visible if and only if he is ahead of the tallest and second tallest people, which has probability 1/4, etc. Now notice that since one person OR MORE will be visible all the time, two people or more will be visible half the time, and three people or more will be visible 1/8 of the time, that means that exactly 1 person will be visible 1/2 the time, exactly 2 people visible 1/4 of the time, etc. So the expected value of the number of visible people is:


1/2 + 2/4 + 3/8 + ... (10 terms)


We know this sums to approximately 2 people visible on average.


5. Perhaps this is more intuitive. 50 errors were found by both proofreaders, and this was 50/70 of the errors found by the second reader alone. So if P(A) and P(B) are the fractions of the total errors found by readers A and B, then

P(A)P(B) = (50/70)P(B) so P(A) = 50/70. Since A found 50/70 of the errors and he found 60 errors, there must be (70/50)(60) = 84 errors total, so 4 errors were undetected.

Ignore my explanation for 1, and look the one on the other website. You can't just consider the individual pairs since you could get a head at the end of one pair and at the beginning of the next.

Great Problems BruceZ. I have some comments below.


Question 1 :

I agree that the posted solution to #1 is correct, but it is not very enlightening. It would be very hard to extend the given solution to n=10000. I will post my solution to problem #1 separately.


Question 3 :

I agree that the posted solution would be correct if the names were replaced in the hat after each draw.


I disagree with your solution to #3. You claim that


f(2) = 1

f(n) = (n-1)f(n-1)


where f(n) is the number of ways n people can draw names without drawing their own. Obviously f(2) = 1. It is easy to write out the 6 permutations to check that f(3) = 2 = 2*1 which agrees with your formula. When I wrote out the 24 permuatations for f(4), I got f(4) = 9, which does not obey your formula.


Question 5 : I like your reasoning on this one. I got slightly different numbers. You state that the 3rd tallest person would be visible 1/4 of the time. I think that he would be visible 1/3 of the time. There seem to be 6 equally likely possibilities to me: 123, 132, 213, 231, 312, 321. In 1/3 of these possibilities the 1 can be seen. Similarly, I think the nth person can be seen with probability 1/n, so then, using your reasoning, the average number of people seen in an N person line would be


1+ 1/2 + 1/3 + 1/4 + ... + 1/N

I fixed #5 above before I saw you answer. This is correct.

Problem # 1 :

Suppose you repeatedly toss a fair coin until you get two heads in a row. What is the probability that you stop on the 10th toss?


I am now getting a solution to problem #1 that differs from the posted answers.


Method 1 : Write a program that goes through all 1024 HT combinations. I did this. 880 of the combinations had a HH in them. Of those 880, only 34 had HH for the last two tosses and no previous HH. So the probability that HH occurs after the 10th toss, but not previously is 34/1024 = 3.3%


Method 2 : "Markov chain"

Assume that after each toss, there are 3 possible states


S1) Last Toss was Tails.

S2) Last Toss was Heads, but two consecutive heads have not been tossed.

S3) Two consecutive heads have been tossed.


The trasitions probabilities are


50% for S1->S1, S1-> S2, S2->S1, S2->S3,

100% for S3->S3,

0 otherwise.


To compute the probabilities the first n tosses, you can use the matrix


[0.5 0.5 0]^n

[0.5 0.0 0]

[0.0 0.5 1]


times the vector [ 1 0 0].


Computing this matrix for n = 9 and n = 10 gives


n = 9 S1 = 55/512 S2 = 17/256, S3 =423/512

n = 10 S1 = 89/1024 S2 = 55/1024, S3 =55/64.


So the probability of an HH occuring amoung the first 9 tosses is 432/512. The probability of an HH occuring among the first 10 tosses is 55/64. So the probability that the first HH will occur after 10 tosses is


55/64 - 423/512 = 17/512 = 3.3%

All 1024 combinations is not an appropriate sample space because sometimes we stop before we get to 10 flips, and the sequences have varying probabilities depending on how long they are. Both of your methods suffer from this problem.


What the posted solution is saying is that since after each number of flips only one sequence terminates, then if f(n) is the number of outcomes after n flips, then


f(2) = 4

f(n) = 2[f(n-1)-1]


f(10) is the size of the sample space.

we can still make all ten tosses, its just the ones where we should have stopped dont "win"

This is not as mathematically interesting, but I always feel a good old sledgehammer can show that for small numbers anyone can do maths....


We know that the series must finish THH and any series of ten tosses will finish this way 1/8 of the time.


Now we only have to look at seven tosses, out comes the hammer.

There are 128 possible tosses:


0 heads: 1 combination

1 heads: 7 combinations

2 heads: 21 combinations

3 heads: 35 combinations

4 heads: 35 combinations

5 heads: 21 combinations

6 heads: 7 combinations

7 heads: 1 combination


Picture the individual tosses as a series of boxes labelled:

1 2 3 4 5 6 7


for 0 and 1 heads, we know there are never two heads together.

for two heads, there are 6 ways of arranging so that two are together. (12,23,34,45,56,67)

for three heads there are 30 ways, but five are duplicated as triples, so 25 ways. (1,2*5,2,3*5 etc)

for four heads there is only one way not to have them together (1,3,5,7)

for five, six and seven, obviously none


so there are 0+0+6+25+34+21+7+1 ways = 94 combinations from 128 where there are heads together.

Therefore only 34/128 of the combinations have no heads together for seven tosses.

and we said at the start that this only works once in eight, so we get the figure 34/1024 as already shown.


I hope this inspires some hope in those with a passing interest in probability and only a light foundation in maths (I have massive interest in probability and a light foundation in maths [img]/images/smile.gif[/img] )