jackpot:


You have 100 students only in a class. What is the expected number of students in the class that do not have birthday mates?

Unfortunately, this is not the question. I still maintain that you're confusing those 2 things. Question requires an unshared birthday and not individuals with an unshared birthday. That's one part you need to iron out to start with.

naturalwaves:

You've not answered my question. From the 15 people with shared birthdays, how many birthdays are shared
15 people can share just 1 birthday.

The question how many birth dates are shared can't be answers with does information, it is a lot of possibilities

jackpot:
You have 100 students only in a class. What is the expected number of students in the class that do not have birthday mates?

Your question doesn't make a shred of sense, or it is incomplete. If I have 100 students in a class, I need to know their birthdays before I can know the expected number of students that have no birthday mates because there are 365¹⁰⁰ possible ways a group of 100 students can be assigned birthdays.

Let's pick four possibilities with a class of 100:
1) If no one has a birthday mate, the expected number of students with no birthday mates is 100.
2) If only two students share the birthday and others don't have birthday mates, the expected number of students with no birthday mate is 98.
3) if three students share the same birthdays, two students share another birthday, and the rest have no birthday mates, the expected number of students with no birthday mates is 95.
4) If all students share the same birthday, the expected number of students with no birthday mates is 0.

So you see, you need to know the birthdays of each student to know the expected number of students with no birthday mates.

Edit: I could list many more possibilities. In fact, I can list a possibility to arrive at any expected number (from 0 to 100) of students with no birthday mates.

naturalwaves:

That means that 15 players share birthdays but only 12 birthdays are shared.

no, 6+1=7. I get your point about shared birthdays but that isn't the angle of the question.

But I mean how many students in a group of 100 do we expect to not share their birthdays with others in the group? In other words, how many do not have birthday mates?

Like in the data, 85 didn't have birthday mates.

Mrshape:

The question how many birth dates are shared can't be answers with does information, it is a lot of possibilities

I mean from the figure she supplied about the 15 people. She has answered that and I saw that it is 13 birthdays that were shared amongst the 15 people. I asked because I wanted to make her realize that the numbers cannot be equal.

jackpot:
no, 6+1=7. I get your point about shared birthdays but that isn't the angle of the question.

But I mean how many students in a group of 100 do we expect to not share their birthdays with others in the group? In other words, how many do not have birthday mates?

Like in the data, 85 didn't have birthday mates.

I am coming. Let me check what you posted again but you got the basic drift anyway. shared birthdays may be 7 but 15 players may have a shared birthday.

jackpot:

Among the 15 players, 6 pairs of players shared 6 different birthdays while 3 players shared the same birthday.

So, 6×2+3=15.

The rest 85 players do not have birthday mates.

See what you wrote now Jackpot
You initially said 15 have shared birthdays.
Now, 6 pairs =12
Hiw did they share different birthdays again. I think you didn't understand my question.

I actually meant that, for the number that have their birthdays shared, what is the distribution like? E. G if 15 people share birthdays, 4 can be April 30 and the remaining 11 can be May 1. That's the sort of info I requested.

Mrshape:

The question how many birth dates are shared can't be answers with does information, it is a lot of possibilities

That is the reason why I used expected number.

It is expected that the possibilities should cluster around a particular value (the answer in this case)

It's similar to asking "expected number of times 6 will appear in 60 tosses of a fair die". Of course, there are different possibilities but the answer is 10.

@Jackpot
In my last reply to you, I didn't mention that I haven't been following this thread thoroughly. Is the rephrased question you sent to me complete? Perhaps you left out the fact that certain number of people shared birthdays in a certain way. Perhaps, your rephrased question is indeed complete, and therefore irrational.

Here is my last reply:
https://www.nairaland.com/1147658/nairaland-mathematics-clinic/270#89041840

naturalwaves:


See what you wrote now Jackpot
You initially said 15 have shared birthdays.
Now, 6 pairs =12
Hiw did they share different birthdays again. I think you didn't understand my question.

I actually meant that, for the number that have their birthdays shared, what is the distribution like? E. G if 15 people share birthdays, 4 can be April 30 and the remaining 11 can be May 1. That's the sort of info I requested.

Okay, here's the info. I had to look for the sheet of paper that I used


2 shared 13th January
2 shared 5th February
2 shared 9th August
2 shared 31st August
3 shared 2nd October
2 shared 16th October
2 shared 16th December
Total of 15.


The rest 85 didn't have birthday mates.


I hope all is clear now?

Thanks Mr. Shape for talking about the birthday paradox. I had to quickly read on it on wiki and then I was lucky to get this video. Please, let us take time to watch it. The dude in the video even used 100 people to illustrate from time stamp 5:33 Please, try and watch.
CC: Martinez39s, jackpot.


https://www.youtube.com/watch?v=ofTb57aZHZs

Ps: The dude also got 0.00003 as his answer.

Martinez39s:


Your question doesn't make a shred of sense, or it is incomplete. If I have 100 students in a class, I need to know their birthdays before I can know the expected number of students that have no birthday mates because there are 365¹⁰⁰ possible ways a group of 100 students can be assigned birthdays.

Let's pick four possibilities with a class of 100:
1) If no one has a birthday mate, the expected number of students with no birthday mates is 100.
2) If only two students share the birthday and others don't have birthday mates, the expected number of students with no birthday mate is 98.
3) if three students share the same birthdays, two students share another birthday, and the rest have no birthday mates, the expected number of students with no birthday mates is 95.
4) If all students share the same birthday, the expected number of students with no birthday mates is 0.

So you see, you need to know the birthdays of each student to know the expected number of students with no birthday mates.

Edit: I could list many more possibilities. In fact, I can list a possibility to arrive at any expected number (from 0 to 100) of students with no birthday mates.

that is why we are assuming a random group of 100 nah.

By the way, in a random experiment of the question, the possiblity of the cases you mentioned happening is almost zero.

See, I believe that we are learning.

Download that PDF and run the experiment with the last 100 players. Tell me your result.

jackpot:


Okay, here's the info. I had to look for the sheet of paper that I used


2 shared 13th January
2 shared 5th February
2 shared 9th August
2 shared 31st August
3 shared 2nd October
2 shared 16th October
2 shared 16th December
Total of 15.


The rest 85 didn't have birthday mates.


I hope all is clear now?

Thanks for the clarification.
This is my points exactly.
Based on this example, 15 people have shared birthdays but the number of shared birthdays is just 7. That's the difference I was talking about.

The question requires.... Number of unshared birthdays from the group and not number of students with an unshared birthday.

naturalwaves:
Thanks Mr. Shape for talking about the birthday paradox. I had to quickly read on it on wiki and then I was lucky to get this video. Please, let us take time to watch it. The dude in the video even used 100 people to illustrate from time stamp 5:33 Please, try and watch.
CC: Martinez39s, jackpot.


https://www.youtube.com/watch?v=ofTb57aZHZs

my question is different from the birthday paradox. Birthday paradox has it's origin in the probability of at least a member of a group of size k (an integer>2) having a birthday mate.

But birthday paradox helps a bit to understand the question.

If the probability that a member of a group having a birthday mate in a group of 100 is 99.99997%, one can infer that on average those that have birthday mate(s) in a random group of 100 are a significant number x.

I'm looking for (100-x) being number of those that don't have birthday mates.

3 in ten million that's the answer

jackpot:
my question is different from the birthday paradox. Birthday paradox has it's origin in the probability of at least a member of a group of size k (an integer>2) having a birthday mate.

But birthday paradox helps a bit to understand the question.

If the probability that a member of a group having a birthday mate in a group of 100 is 99.99997%, one can infer that on average those that have birthday mate(s) in a random group of 100 are a significant number x.

I'm looking for (100-x) being number of those that don't have birthday mates.

3 out of 10000000 people

naturalwaves:


Thanks for the clarification.
This is my points exactly.
Based on this example, 15 people have shared birthdays but the number of shared birthdays is just 7. That's the difference I was talking about.

The question requires.... Number of unshared birthdays from the group and not number of students with an unshared birthday.

That was not the question she asked before she said number of people who do not share birth date

jackpot:
my question is different from the birthday paradox. Birthday paradox has it's origin in the probability of at least a member of a group of size k (an integer>2) having a birthday mate.

But birthday paradox helps a bit to understand the question.

If the probability that a member of a group having a birthday mate in a group of 100 is 99.99997%, one can infer that on average those that have birthday mate(s) in a random group of 100 are a significant number x.

I'm looking for (100-x) being number of those that don't have birthday mates.

I think I get what you are trying to say now

Modified :
....... but you must realize that what you're asking is impossible to determine without towing the probability lane. The best we can do is to calculate the probability of an unshared birthday and then tweak it. Take for instance if the probability of an unshared birthday is 1/5. It means that for every 5 people, 1 won't share birthday and since we are looking at a total of 100 students, you will get 20 students in all as what you desire.

Unfortunately, in this case the probability is marginal and the number you require is about 3 in a 10 million samples or something like that.

Cc: jackpot, Mr. Shape

naturalwaves:


Thanks for the clarification.
This is my points exactly.
Based on this example, 15 people have shared birthdays but the number of shared birthdays is just 7. That's the difference I was talking about.

Okay.

naturalwaves:

The question requires.... Number of unshared birthdays from the group and not number of students with an unshared birthday.

my question remains number of students that do not share their birthdays with any other group members.

However, what you asked is another different legit question.

Maybe you should pose this question formally? I'll be happy to attempt it.

Can you run an experiment with the last 100 players in the PDF file and give your findings regarding:

1. Number of unshared birthdays?
2. Number that do not have birthday mates?

Mrshape:

3 out of 10000000 people

when a random experiment gave me 85 without birthday mates (and 15 with birthday mates) out of 100

If anything, the two are not close. Random experiments often give clue to magnitude of the answer.

Try and run the experiment with a section of the players data. It helps to appreciate results when they come.

But you may use a consecutive 100 players for the experiment to have semblance of randomness

jackpot:
when a random experiment gave me 85 without birthday mates (and 15 with birthday mates) out of 100

If anything, the two are not close. Random experiments often give clue to magnitude of the answer.

Try and run the experiment with a section of the players data. It helps to appreciate results when they come.

But you may use a consecutive 100 players for the experiment to have semblance of randomness

But random experiment can also give 100 with birth day mates and zero without birthday mate

jackpot:
when a random experiment gave me 85 without birthday mates (and 15 with birthday mates) out of 100

If anything, the two are not close. Random experiments often give clue to magnitude of the answer.

Try and run the experiment with a section of the players data. It helps to appreciate results when they come.

But you may use a consecutive 100 players for the experiment to have semblance of randomness

A random sample can also give you 0 people without birthday mates in a size of 100.

jackpot:
that is why we are assuming a random group of 100 nah.

By the way, in a random experiment of the question, the possiblity of the cases you mentioned happening is almost zero.

See, I believe that we are learning.

Download that PDF and run the experiment with the last 100 players. Tell me your result.

My reasoning is valid for a randomly selected group of 100 students. In fact, throughout this thread, based on my presumption of what your question could possibly mean (as you stated it at first), I have been working with the idea that the group of 100 students were randomly selected.

Let me explain something. When you randomly select a group of 100 students, you have randomly selected a group of 100 birthdays. Each selected birthday has been selected from 365 possible birthdays (dates in a year), and repetition is allowed as you randomly select other birthdays in the group of 100 birthdays. Using the selection of humans, the emboldened statement is tantamount to saying each person from the randomly selected group of hundred has a birthday that is one of the 365 dates (birthdays) in a year, and a person in the group could have a birthday mate in the group (remember that repetition of birthdays is allowed).

When you want to randomly selecting a group of 100 people (birthdays), here are some possible logical questions you can ask if no further information is given or no possible consideration is made (assuming a year contains 365 days):
(1) What is the probability that no two people share the same birthday?

Ans: ³⁶⁵P₁₀₀ ÷ 365¹⁰⁰ = 365! ÷ (265! × 365¹⁰⁰) = 0.00000030725

(2) What is the probability that at least two people share the same a birthday

Ans: 1 – (³⁶⁵P₁₀₀ ÷ 365¹⁰⁰) = 1 – [365! ÷ (265! × 365¹⁰⁰)] = 0.99999969275

Naturalwaves got this, but he applied it to your question which doesn't have anything to do with probability.

(3) What is the number of ways of selecting a group of hundred people such that no two shares the same birthday.

Ans: ³⁶⁵P₁₀₀ = 365!/265!

If the question has order disregarded, your answer is ³⁶⁵C₁₀₀ = 365!/(100! × 265!)

(4) What is the number ways of selecting a group of hundred people such that at least two have the same birthday

365¹⁰⁰ – ³⁶⁵P₁₀₀ = 365¹⁰⁰ – (365!/265!)

This question of yours doesn't make sense:

You have 100 students only in a class. What is the expected number of students in the class that do not have birthday mates?

Either you mistakenly left out something or you didn't state the question properly. If this question is complete, and there is nothing more to it, then it's illogical even if you have a unique group of 100 students or a randomly selected group of 100 students.


MrShape, naturalwaves,

naturalwaves:

A random sample can also give you 0 people without birthday mates in a size of 100.

That's like 1 in 100 cubic trillion trillion chance.

That's why I am saying you should run the experiment. Let's see possibility of getting zero.

I bet that any random sample of 100 that you chose at random will give you about 15-30 persons with birthday mates. That's also surprising considering the fact that there are 365 days in our year.

jackpot:
That's like 1 in 100 cubic trillion trillion chance.

That's why I am saying you should run the experiment. Let's see possibility of getting zero.

I bet that any random sample of 100 that you chose at random will give you about 15-30 persons with birthday mates. That's also surprising considering the fact that there are 365 days in our year.

The chances of winning a mega lottery in the United States of America is approximately 0 yet people still win. It would therefore be false to say that ANY person who plays the lottery will fail. The fact that a large group of people played the lottery 200 times without success isn't proof that no one can win the lottery. The odds of an event might be very large, but as long as the probability of that event isn't exactly 0 then that event can happen.

naturalwaves:

I think I get what you are trying to say now

Modified :
....... but you must realize that what you're asking is impossible to determine without towing the probability lane. The best we can do is to calculate the probability of an unshared birthda.y and then tweak it. Take for instance if the probability of an unshared birthday is 1/5. It means that for every 5 people, 1 won't share birthday and since we are looking at a total of 100 students, you will get 20 students in all as what you desire.

Unfortunately, in this case the probability is marginal and the number you require is about 3 in a 10 million samples or something like that.

Cc: jackpot, Mr. Shape

It's actually very possible to determine by going experimentally.

The hard way is to run the experiment about ten thousand times using a PC and find the average.

The easy way should be probability distribution functions and expected values from statistics. I strongly feel so.

Martinez39s:
The chances of winning a mega lottery in the United States of America is approximately 0 yet people still win. It would therefore be false to say that ANY person who plays the lottery will fail. The odds of an evening might be very large, but as long as the probability of that event isn't exactly 0 then that event can happen.

i know, but try first

I bet you might even get up to 23 players with birthday mates if you run the experiment.

jackpot:
i know, but try first

I bet you might even get up to 23 players with birthday mates if you run the experiment.

Sure! Even though any outcome is possible, I will way most likely get at least two people with the same birthday. This is because the probability of having at least two people with the same birthday is 0.99999969275 (99.999969275% chance). The probability of not having anyone with a birthday mate is 0.00000030725 (0.000030725% chance). So you see that event of having a selection in which no one has a birthday mate is a rare one just like winning a mega lottery. I calculated these probabilities in a post of mine to you.

MrShape, naturalwaves, Jackpot

Martinez39s:
My reasoning is valid for a randomly selected group of 100 students. In fact, throughout this thread, based on my presumption of what your question could possibly mean (as you stated it at first), I have been working with the idea that the group of 100 students were randomly selected.

Let me explain something. When you randomly select a group of 100 students, you have randomly selected a group of 100 birthdays. Each selected birthday has been selected from 365 possible birthdays (dates in a year), and repetition is allowed as you randomly select other birthdays in the group of 100 birthdays. Using the selection of humans, the emboldened statement is tantamount to saying each person from the randomly selected group of hundred has a birthday that is one of the 365 dates (birthdays) in a year, and a person in the group could have a birthday mate in the group (remember that repetition of birthdays is allowed).

When you want to randomly selecting a group of 100 people (birthdays), here are some possible logical questions you can ask if no further information is given or no possible consideration is made (assuming a year contains 365 days):
(1) What is the probability that no two people share the same birthday?

(2) What is the probability that at least two people share the same a birthday Naturalwaves got this, but he applied it to your question with doesn't have anything to do with probability.

(3) What is the number of ways of selecting a group of hundred people such that no two shares the same birthday.

(4) What is the number ways of selecting a group of hundred people such that at least two have the same birthday

It's unfair to tag what we don't understand as illogical

If I see a question that is indeterminate, I will know.

For instance, do you know that some great guys think that that angle question I posted earlier was not complete? Yet, Mrshape and others dealt mercilessly with the question here.

Martinez39s:

This question of yours doesn't make sense: Either you mistakenly left out something or you didn't state the question properly. If this question is complete, and there is nothing more to it, then it's illogical even if you have a unique group of 100 students or a randomly selected group of 100 students.


MrShape, naturalwaves,

Do you know that questions like the following have an answer?

Let a, b, c be real numbers selected randomly from the interval (0, 1). What is the probability that the quadratic equation ax²+bx+c=0 has at least one real solution?

Just imagine where one will start. Mine is better that it can even be experimented

@Jackpot

Here is an edited post of mine explaining your observation on the experiment of randomly selecting 100 people (I wrote this when you asked me to try the experiment):

Martinez39s:
Sure! Even though any outcome is possible, I will way most likely get at least two people with the same birthday. This is because the probability of having at least two people with the same birthday is 0.99999969275 (99.999969275% chance). The probability of not having anyone with a birthday mate is 0.00000030725 (0.000030725% chance). So you see that the event of having a selection in which no one has a birthday mate is a very very very rare one just like winning a mega lottery. I calculated these probabilities in a post of mine to you.

MrShape, naturalwaves,

From these probabilities, you can see why you keep getting people with the same birthdays in your selection—the odds are greatly stacked in your favour (you have a 99.999969275% chance of getting at least two people with the same birthday). This is the same reason why people keep getting unsuccessful in their attempt to win the lottery— the odds are incredibly stacked against them. The popular Mega Millions multi-state lottery has odds of approximately 1 in 175,711,536. This is a 0.000000569% chance.

jackpot:

It's unfair to tag what we don't understand as illogical
If I see a question that is indeterminate, I will know.

Nah! *yimu* Your question wasn't properly phrased, and it didn't make sense the way you presented it. I now understand what you were trying to ask. You should have asked "if 100 people are selected randomly what's the LIKELY number (or Likely range of number) of people who have birthday mates? This question is incredibly difficult to solve, that's if it has an answer.

For instance, do you know that some great guys think that that angle question I posted earlier was not complete? Yet, Mrshape and others dealt mercilessly with the question here.

The angle question is complete; I never thought it wasn't. Any mathematician who thinks that question isn't complete isn't that great afterall.

Do you know that questions like the following have an answer?

Let a, b, c be real numbers selected randomly from the interval (0, 1). What is the probability that the quadratic equation ax²+bx+c=0 has at least one real solution?

Just imagine where one will start. Mine is better that it can even be experimented

Ahhh! I won't even bother with this question. I can't kill myself.


MrShape, naturalwaves

Martinez39s:
@Jackpot

Here is an edited post of mine explaining your observation on the experiment of randomly selecting 100 people (I wrote this when you asked me to try the experiment): From these probabilities, you can see why you keep getting people with the same birthdays in your selection—the odds are greatly stacked in your favour (you have a 99.999969275% chance of getting at least two people with the same birthday). This is the same reason why people keep getting unsuccessful in their attempt to win the lottery— the odds are incredibly stacked against them. The popular Mega Millions multi-state lottery has odds of approximately 1 in 175,711,536. This is a 0.000000569% chance.

Nah! *yimu* Your question wasn't properly phrased, and it didn't make sense the way you presented it. I now understand what you were trying to ask. You should have asked "if 100 people are selected randomly what's the LIKELY number (or Likely range of number) of people who have birthday mates? This question is incredibly difficult to solve, that's if it has an answer.

The angle question is complete; I never thought it wasn't. Any mathematician who thinks that question isn't complete isn't that great afterall.

Ahhh! I won't even bother with this question. I can't kill myself.


MrShape, naturalwaves

Lol. I will try that tomorrow I am tired too o.

Mrshape:

Lol. I will try that tomorrow I am tired too o.

I don't think I would ever try it.