Micromass' big probability challenge

[micromass]

Probability theory is very nice. It contains many questions which are very easy to state, but not so easily solved. Let's see if you can solve these questions.

• For an answer to count, not only the answer must be given but also a detailed explanation.
• Any use of outside sources is allowed, but do not look up the question directly. For example, it is ok to go check probability books, but it is not allowed to google the exact question.
• If you previously encountered this statement and remember the solution, then you cannot participate in this particular statement.
• All mathematical methods are allowed.
• Please reference every source you use.

Here you go:

1. SOLVED BY BiGyElLoWhAt, PeroK The universe is about $432,043,200,000,000,000$ seconds old. Assume that from the very start of the universe, a solitary monk was flipping a coin over and over again. He flips a coin every second. A run of order $n$ are $n$ consecutive heads in a row, or tails in a row. On average, what is the largest run that the monk has had?
2. SOLVED BY ProfuselyQuarky Every cereal box contains one toy. There are in total $20$ different toys. How much boxes of cereal do you expect to buy - on average - in order to collect all the toys.
3. REMOVED
4. SOLVED BY PeroK A closet contains $100$ pairs of shoes. I choose at random $8$ shoes. What is the probability that I will have selected exactly one complete pair?
5. SOLVED BY mfb In a town of $n+1$ inhabitants. A person tells a rumor to two distinct numbers, the "first generation". These repeat the performance and generally, each person tells a rumor to two people at random without regard to the past development. Find the probability that the generation $r$ will not contain the person who started the rumor. Find the median of this distribution assuming $n$ large.
6. SOLVED BY andrewkirk Find the probability that in a random arrangment of $52$ cards no two aces are adjacent.
7. SOLVED BY Fightfish, mfb In a ballot, candidate Trump scores $304$ votes and Clinton scores $216$ votes. When the votes are counted, find the probability that throughout the counting, there are always more votes for Trump than Clinton.
8. SOLVED BY thephystudent I have a stick of $1$ meter. I break this stick in $3$ pieces in such a way that every point of the stick has as much chance of being a break point. Find the probability that I can combine the three pieces into an obtuse triangle.
9. SOLVED BY Charles Link, PeroK In a room of $1000$ people. How many people do you expect there to be so that nobody else in the room shares their birthday?
10. REMOVED

Thank you all for participating! I hope many of you have fun with this! Don't hesitate to post any feedback in the thread!

More information:

1. gato-docs.its.txstate.edu/mathworks/DistributionOfLongestRun.pdf
2. http://www.math.uah.edu/stat/urn/Coupon.html
   
3. REMOVED
4. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis" Exercise 26
5. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis"
6. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis" Exercise 42
   Solution: Look at $48$ non-aces. These determine $49$ gaps. Choose $4$ gaps.
7. Feller "An introduction to probability theory and its applications Vol1" Chapter III "Fluctuations in coin tossing and random walks"
8. http://www.math.uah.edu/stat/buffon/Triangles.html
9. Invented on my own
10. REMOVED

 

[BiGyElLoWhAt]

So here's my thinking (not a huge probability guy, so I might be missing something, but I think this makes sense):

We want to find the number of consecutive heads/tails such that the probability is 1 in 4.320432x10^17. The probability of flipping 1 heads is 1/2, the probability of two is (1/2)^2, and the probability of n heads is (1/2)^n. So we have (1/2)^n = 1/4.320432x10^(17)
or
$2^n = 4.320432E17$
$\text{log}_2(2^n) = \text{log}_2(4.320432E17)$ which gives us about $n=58.584$ which honestly seems a little low to me. But that's my guess, anyways.

 

[PeroK]

For number 4:

To have exactly one complete pair, you need one pair and six odd shoes. The pair is equally likely to be any two shoes, from the 1st and 2nd to the 7th and 8th. There are, therefore, $\binom{8}{2} = 28$ equally likely ways to get a pair.

The probability that the 1st and 2nd shoes are the pair is:

$p_{12} = \frac{100}{100} \times \frac{1}{99} \times \frac{98}{98} \times \frac{96}{97} \times \frac{94}{96} \times \frac{92}{95} \times \frac{90}{94} \times \frac{88}{93}$

Hence, the total probablity that there is exactly one pair is:

$p = 28 \times p_{12} = 0.24$

 

[micromass]

For number 4:

To have exactly one complete pair, you need one pair and six odd shoes. The pair is equally likely to be any two shoes, from the 1st and 2nd to the 7th and 8th. There are, therefore, $\binom{8}{2} = 28$ equally likely ways to get a pair.

The probability that the 1st and 2nd shoes are the pair is:

$p_{12} = \frac{100}{100} \times \frac{1}{99} \times \frac{98}{98} \times \frac{96}{97} \times \frac{94}{96} \times \frac{92}{95} \times \frac{90}{94} \times \frac{88}{93}$

Hence, the total probablity that there is exactly one pair is:

$p = 28 \times p_{12} = 0.24$

Are you sure you're not overcounting? For example, if the shoes have colors, are you sure you're distinguishing between (blue, blue, red, purple, green, yellow) and (blue, blue, red, purple, yellow, green) ? In either case, I'm unsure how you found your $p_{12}$ probability. And are you distinguishing between left and right shoes? The first two shoes the same and left shoes is different from the first two shoes the same and right shoes.

 

[micromass]

So here's my thinking (not a huge probability guy, so I might be missing something, but I think this makes sense):

We want to find the number of consecutive heads/tails such that the probability is 1 in 4.320432x10^17. The probability of flipping 1 heads is 1/2, the probability of two is (1/2)^2, and the probability of n heads is (1/2)^n. So we have (1/2)^n = 1/4.320432x10^(17)
or
$2^n = 4.320432E17$
$\text{log}_2(2^n) = \text{log}_2(4.320432E17)$ which gives us about $n=58.584$ which honestly seems a little low to me. But that's my guess, anyways.

That's a nice approximation. Unless anybody can provide a more accurate result, I'll accept your solution.

 

[PeroK]

Are you sure you're not overcounting? For example, if the shoes have colors, are you sure you're distinguishing between (blue, blue, red, purple, green, yellow) and (blue, blue, red, purple, yellow, green) ? In either case, I'm unsure how you found your $p_{12}$ probability. And are you distinguishing between left and right shoes? The first two shoes the same and left shoes is different from the first two shoes the same and right shoes.

Sorry, I did 100 shoes (50 pairs), not 100 pairs! Out of time today to fix things now.

 

[CrackerMcGinger]

Just a guess by doing division, would number 4 be 0.16 or 1.6% chance of there being a whole pair? I'm probably wrong but you never know unless you try.

 

[CrackerMcGinger]

I don't understand what question 9 is asking, can someone elaborate?

 

[micromass]

I don't understand what question 9 is asking, can someone elaborate?

For example in a room with $5$ people, you have the following birthdays (June 13, June 28, May 1, June 13, May 1). In this case, only the second person has a birthday that isn't shared with anybody else. So in this particular situation, the answer is $1$. I'm looking for the average of all these situations.

 

[micromass]

Just a guess by doing division, would number 4 be 0.16 or 1.6% chance of there being a whole pair? I'm probably wrong but you never know unless you try.

That's not the correct answer.

 

[CrackerMcGinger]

That's not the correct answer.

I figured as much.

 

[micromass]

I figured as much.

You were close though!

 

[CrackerMcGinger]

In question number three are the balls numbered in order or are they each given a random number? If the former is the case then my answer for that question would be 213 since the last ball (the highest number of all five balls) was 213. If the latter is the case then my answer is five, since we know their to be at least five balls since you only picked five from the box.

 

[CrackerMcGinger]

You were close though!

Wait... how close, because my first answer was 0.24 or 2.4%.

 

[ShayanJ]

That's a nice approximation. Unless anybody can provide a more accurate result, I'll accept your solution.

I haven't thought about it myself and just tried to understand BiGy's reasoning. In the process, I realized it should be 1 in $\frac{4.320432 \times 10^{17}}{n}$, which makes the answer equal to almost 53.

 

[micromass]

In question number five are the balls numbered in order or are they each given a random number? If the former is the case then my answer for that question would be 213 since the last ball (the highest number of all five balls) was 213. If the latter is the case then my answer is five, since we know their to be at least five balls since you only picked five from the box.

The balls are numbered in order. So if there are $n$ balls, then there are numbered with $\{1,...,n\}$.

Can you give a reasoning for why you think $213$ is the best answer?

 

[micromass]

Wait... how close, because my first answer was 0.24 or 2.4%.

Incorrect. In either case, you should give a reasoning.

 

[CrackerMcGinger]

The balls are numbered in order. So if there are $n$ balls, then there are numbered with $\{1,...,n\}$.

Can you give a reasoning for why you think $213$ is the best answer?

Because the highest number ball from the box you picked from was labeled as ball number 213, which means their are at least a total of 213 balls in the box.

 

[Pro7]

For the cereals question I suppose , as the probability of getting a type of toy is 1/20, we have to buy 400 cereal boxes.I'm not so sure but a try anyways.

 

[micromass]

Because the highest number ball from the box you picked from was labeled as ball number 213, which means their are at least a total of 213 balls in the box.

Indeed, there are at least $213$. But what makes $213$ the optimal number of balls in the box? Sure, saying $213$ is the safest options, it's definitely a minimum number of balls. But what is most likely?

 

[micromass]

I haven't thought about it myself and just tried to understand BiGy's reasoning. In the process, I realized it should be 1 in $\frac{4.320432 \times 10^{17}}{n}$, which makes the answer equal to almost 53.

Can you tell us why?

 

[ProfuselyQuarky]

For number 2, I'm not sure, but wouldn't that only require a Monte Carlo simulation? Too bad I can't use my polyhedral dice set on this ...

So, there are $20$ toys. After buying one box, the probability of getting a different toy would be $\frac{19}{20}$ and so on. The the number of boxes purchased for one different prize would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {19}{20})^1(\frac {1}{20})^{k-1}$, for two prizes it would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {18}{20})^1(\frac {2}{20})^{k-1}$, for three prizes it would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {17}{20})^1(\frac {3}{20})^{k-1}$, and so forth. The $k$ would represent how many purchases it takes from $1$ to $\infty$. This pattern continues until the purchases needed to acquire every toy is $\displaystyle\sum_{n=1}^{\infty}k(n)(1-n)^{k-1}$. The $n$ stands for probability.This can be simplified:

$\displaystyle\sum_{n=1}^{\infty}k(n)(1-n)^{k-1}=(n)\frac{1}{n}(\frac {1}{1-(1-n)})=(n)\frac {1}{n}(\frac {1}{n})=(n)\frac {1}{n^2}=\frac {1}{n}$

Thus, to acquire the number of boxes of cereal to get every toy:

$\frac {20}{20}+\frac {20}{19}+\frac {20}{18}+\frac {20}{17}+\frac {20}{16}+\frac {20}{15}+\frac {20}{14}+\frac {20}{13}+\frac {20}{12}+\frac {20}{11}+\frac {20}{10}+\frac {20}{9}+\frac {20}{8}+\frac {20}{7}+\frac {20}{6}+\frac {20}{5}+\frac {20}{4}+\frac {20}{3}+\frac {20}{2}+\frac {20}{1}=71.9548\approx 72$

My guess is that you have to buy 72 boxes of cereal to acquire every toy?

 

[CrackerMcGinger]

Indeed, there are at least $213$. But what makes $213$ the optimal number of balls in the box? Sure, saying $213$ is the safest options, it's definitely a minimum number of balls. But what is most likely?

Since you give me 5 numbers, each of which represent the ball you picked at random, and the highest of all five numbers being 213, I personally can't find a higher number of balls that we know are their, and since both the GCF and HCF are 1 I will stay at my answer of 213.

 

[micromass]

For number 2, I'm not sure, but wouldn't that only require a Monte Carlo simulation? To bad I can't use my special polyhedral dice on this ...

So, there are $20$ toys. After buying one box, the probability of getting a different toy would be $\frac{19}{20}$ and so on. The the number of boxes purchased for one different prize would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {19}{20})^1(\frac {1}{20})^{k-1}$, for two prizes it would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {18}{20})^1(\frac {2}{20})^{k-1}$, for three prizes it would be $\displaystyle\sum_{n=1}^{\infty} k(\frac {17}{20})^1(\frac {3}{20})^{k-1}$, and so forth. The $k$ would represent how many purchases it takes from $1$ to $\infty$. This pattern continues until the purchases needed to acquire every toy is $\displaystyle\sum_{n=1}^{\infty}k(n)(1-n)^{k-1}$. The $n$ stands for probability.This can be simplified:

$\displaystyle\sum_{n=1}^{\infty}k(n)(1-n)^{k-1}=(n)\frac{1}{n}(\frac {1}{1-(1-n)})=(n)\frac {1}{n}(\frac {1}{n})=(n)\frac {1}{n^2}=\frac {1}{n}$

Thus, to acquire the number of boxes of cereal to get every toy:

$\frac {20}{20}+\frac {20}{19}+\frac {20}{18}+\frac {20}{17}+\frac {20}{16}+\frac {20}{15}+\frac {20}{14}+\frac {20}{13}+\frac {20}{12}+\frac {20}{11}+\frac {20}{10}+\frac {20}{9}+\frac {20}{8}+\frac {20}{7}+\frac {20}{6}+\frac {20}{5}+\frac {20}{4}+\frac {20}{3}+\frac {20}{2}+\frac {20}{1}=71.9548\approx 72$

My guess is that you have to buy 72 boxes of cereal to acquire every toy?

That is correct!

 

[micromass]

Since you give me 5 numbers, each of which represent the ball you picked at random, and the highest of all five numbers being 213, I personally can't find a higher number of balls that we know are their, and since both the GCF and HCF are 1 I will stay at my answer of 213.

Do you think the probability is high for you to pick exactly the highest number? Because that's what happens in your reasoning.

 

[CrackerMcGinger]

I applaud Miss ProfuselyQuarky.

 

[ShayanJ]

Can you tell us why?

It seems to me that BiGy's reasoning is, that the longer the run, the lower the probability for it to happen, so the longest run should happen only once. That's why he says the probability is 1 in the number of flips. But it can't be 1 in the number of flips because each run is a sequence of flips. So we should group the number of flips into sequences of flips and consider the longest run to be one of those sequences in the group and so the probability for it should be 1 in the number of sequences in the group. The only condition on the groupings is that they should contain the longest run so there is a huge number of possibilities for the groupings. But the simplest is grouping the flips into sequences with the same number of flips as the longest run.

 

[CrackerMcGinger]

Do you think the probability is high for you to pick exactly the highest number? Because that's what happens in your reasoning.

Their is a decent probability that there is a higher number of balls than my estimate, I just don't have the knowledge to figure it out. If the same experiment were repeated several time then I would be able to form a hypothesis on the highest number of balls, for example; If you picked five balls five times and if only one you received the 213 ball and you not one ball was a higher number than that ball I would say that 213 would be the highest number of balls, but if you did the same thing and a higher ball came up on one of the picks it would immediately disprove my answer.

 

[micromass]

It seems to me that BiGy's reasoning is, that the longer the run, the lower the probability for it to happen, so the longest run should happen only once. That's why he says the probability is 1 in the number of flips. But it can't be 1 in the number of flips because each run is a sequence of flips. So we should group the number of flips into sequences of flips and consider the longest run to be one of those sequences in the group and so the probability for it should be 1 in the number of sequences in the group. The only condition on the groupings is that they should contain the longest run so there is a huge number of possibilities for the groupings. But the simplest is grouping the flips into sequences with the same number of flips as the longest run.

I understand what you're saying, but I'm not sure how to formalize this... Not that BiGy's answer is all that formal. In either case, BiGy's answer is closer to yours, I'm afraid.

 

[ShayanJ]

I understand what you're saying, but I'm not sure how to formalize this... Not that BiGy's answer is all that formal. In either case, BiGy's answer is closer to yours, I'm afraid.

I think the final answer is the average over all the possible groupings which surely changes the number but I don't know in which way and I don't think its easy to find out!

 

[ProfuselyQuarky]

I have a stick of $1$ meter. I break this stick in $3$ pieces in such a way that every point of the stick has as much chance of being a break point. Find the probability that I can combine the three pieces into an obtuse triangle.

I am utterly dumbfounded by this one because you specified that the triangle must be obtuse.

The pieces of the $1$ meter stick (after the stick is broken in two places) has the side lengths $a$, $b$, and $(1-a-b)$. If the three parts are greater than half the meter stick, then you can't make a triangle. That leaves that no piece can be greater than half the meter stick, which I believe implies that there is a 25% chance that the meter stick can even make any sort of triangle. For instance, if you break the stick into equal thirds, you can get and equilateral triangle, but that's not obtuse!

Can't wait to see the answer to this one.

 

[micromass]

I am utterly dumbfounded by this one because you specified that the triangle must be obtuse.

The pieces of the $1$ meter stick (after the stick is broken in two places) has the side lengths $a$, $b$, and $(1-a-b)$. If the three parts are greater than half the meter stick, than you can't make a triangle. That leaves that no piece can be greater than half the meter stick, which I believe implies that there is a 25% chance that the meter stick can even make any sort of triangle. For instance, if you break the stick into equal thirds, you can get and equilateral triangle, but that's not obtuse!

Can't wait to see the answer to this one.

How can three pieces be larger than half the meter stick? Won't you need 1.5 meters for that?

 

[micromass]

I have removed two questions since they are about statistics and not about probability. Don't worry, I will post them in a statistics challenge soon!

 

[ProfuselyQuarky]

How can three pieces be larger than half the meter stick? Won't you need 1.5 meters for that?

I meant a piece being greater than HALF of the stick. Sorry.

 

[CrackerMcGinger]

I am utterly dumbfounded by this one because you specified that the triangle must be obtuse.

The pieces of the $1$ meter stick (after the stick is broken in two places) has the side lengths $a$, $b$, and $(1-a-b)$. If the three parts are greater than half the meter stick, than you can't make a triangle. That leaves that no piece can be greater than half the meter stick, which I believe implies that there is a 25% chance that the meter stick can even make any sort of triangle. For instance, if you break the stick into equal thirds, you can get and equilateral triangle, but that's not obtuse!

Can't wait to see the answer to this one.

Well, what is the probability of a stick being broken into three unequal parts, one shorter, one longer, and another shorter again? If it has a probability that each point of the stick can be broken, then what's the probability that the stick can be broken into 3 equal parts?