Mixing
Explain the Birthday Paradox
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I recently read about the Birthday Paradox which states that in a group of 23 people, there's a probability of 50% that 2 people share their birthday, probability wise.
I calculated and don't think it's possible that it's true in any case (unless my math is wrong). So, can anyone please tell me how to prove or disprove it mathematically ?
probability paradoxes birthday
edited Nov 17 '17 at 20:46
Henry
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
$endgroup$
|
show 4 more comments
$begingroup$
I recently read about the Birthday Paradox which states that in a group of 23 people, there's a probability of 50% that 2 people share their birthday, probability wise.
I calculated and don't think it's possible that it's true in any case (unless my math is wrong). So, can anyone please tell me how to prove or disprove it mathematically ?
probability paradoxes birthday
edited Nov 17 '17 at 20:46
Henry
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
$endgroup$
5
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
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It has already been answered here: math.stackexchange.com/questions/25876/…
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– NicoDean
Jun 18 '14 at 13:13
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@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22
|
show 4 more comments
$begingroup$
I recently read about the Birthday Paradox which states that in a group of 23 people, there's a probability of 50% that 2 people share their birthday, probability wise.
I calculated and don't think it's possible that it's true in any case (unless my math is wrong). So, can anyone please tell me how to prove or disprove it mathematically ?
probability paradoxes birthday
edited Nov 17 '17 at 20:46
Henry
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
$endgroup$
I recently read about the Birthday Paradox which states that in a group of 23 people, there's a probability of 50% that 2 people share their birthday, probability wise.
I calculated and don't think it's possible that it's true in any case (unless my math is wrong). So, can anyone please tell me how to prove or disprove it mathematically ?
probability paradoxes birthday
probability paradoxes birthday
edited Nov 17 '17 at 20:46
Henry
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
edited Nov 17 '17 at 20:46
Henry
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
edited Nov 17 '17 at 20:46
Henry
101k481168
edited Nov 17 '17 at 20:46
Henry
101k481168
edited Nov 17 '17 at 20:46
Henry
101k481168
101k481168
asked Jun 18 '14 at 13:10
NibNib
1449
asked Jun 18 '14 at 13:10
NibNib
1449
asked Jun 18 '14 at 13:10
NibNib
1449
1449
5
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
$begingroup$
It has already been answered here: math.stackexchange.com/questions/25876/…
$endgroup$
– NicoDean
Jun 18 '14 at 13:13
$begingroup$
@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22
|
show 4 more comments
5
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
$begingroup$
It has already been answered here: math.stackexchange.com/questions/25876/…
$endgroup$
– NicoDean
Jun 18 '14 at 13:13
$begingroup$
@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22
5
5
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
$begingroup$
It has already been answered here: math.stackexchange.com/questions/25876/…
$endgroup$
– NicoDean
Jun 18 '14 at 13:13
$begingroup$
It has already been answered here: math.stackexchange.com/questions/25876/…
$endgroup$
– NicoDean
Jun 18 '14 at 13:13
$begingroup$
@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Okay, here are my calculations.
Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.
So the 1st person will easily choose any day according to his choice.
This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.
Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.
So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.
I hope this helps.
For more discussions you can refer here.
answered Jun 18 '14 at 13:24
purupuru
92468
$endgroup$
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
|
show 4 more comments
$begingroup$
Simple example with three balls: red, green and blue.
When we form a collection of two balls, we have
$$
3^2
$$
possibilities.
But some do not contain the same color - and that is given by
$$
3 times 2
$$
So the number of collection such that two balls have the same color is given by
$$
3^2 - 3 times 2 = 3
$$
So the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{3^2 - 4 times 3}{3^2} = frac{3}{9} = frac{1}{3}
$$
We can do the same for 4 balls and a collection of 2 balls.
the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{4^2 - 4 times 3}{4^2}
$$
The basic formula is then given by
$$
frac{F^n - F times (F-1) times (F-2) times cdot (F-n)}{F^n}
$$
where
$$
F
$$
is the 'freedom' - the number of different colors for the balls, and
$$
n
$$
is the number of balls in the collection.
Using some math we can write
$$
1 - frac{F!}{F^n big(F-nbig)!}
$$
Note that when $n>F$ we have
$$
k!
$$
for a negative number.
But as
$$
big(n-1big)! = frac{n!}{n}
$$
we see that
$$
big(-1big)! = frac{0!}{0} rightarrow infty
$$
So in case $n > F$ the change becomes $1$
Instead of color - we can consider birthdays, so $F=365$ and we get
$$
1 - frac{365!}{365^n big(365-nbig)!}
$$
The case $n=23$ gives
$$
1 - frac{365!}{365^23 big(365-23big)!} = 50.7%
$$
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
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add a comment |
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/r/eli5 explains more simply. I've rewritten 3 comments that stand alone and can be read separately.
Explanation 1 with Arithmetic
I misunderstood the birthday problem the first time, as I'd read about it and think: "If I find 22 (so a group of 23, not 70) other people, there is a 50% chance that one of them will have the same birthday as me."
However, the probability isn't that any particular person will have a match, but that at least one pair will have a match. It's much easier to understand the problem when you realize that there are $dfrac{23 times 22}{2} = 253$ unique pairs in the group.
Now reword the conclusion as "Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday."
Explanation 2 by visualizing a spinning prize wheel
Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29.
Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $~1/6$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $~1/6$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $(5/6)^{10}$ which is about a $~16%$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $~84%$ chance of landing on at least 1 previously landed-on slot.
In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group.
Explanation 3 by visualizing the Pigeonhole Principle
Once you understand what the "problem" itself is, you can easily visualize why the probability of a match is high, by using the Pigeonhole Principle.
Assume you have 365 different holes in a wall, and you randomly put pigeons into holes without looking if there's already a pigeon in there or not. As you add more pigeons, what increases steadily is the odds of you accidentally putting a pigeon into a hole that already has a pigeon.
The same principle can be applied to the birthday problem; treat each hole as a different day of the year and each pigeon representing a person. When a pigeon is placed into hole that already has a pigeon, this sharing of the hole means that someone in the room shares the same birthday as someone else.
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
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add a comment |
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3 Answers
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3 Answers
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active
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active
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active
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$begingroup$
Okay, here are my calculations.
Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.
So the 1st person will easily choose any day according to his choice.
This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.
Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.
So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.
I hope this helps.
For more discussions you can refer here.
answered Jun 18 '14 at 13:24
purupuru
92468
$endgroup$
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
|
show 4 more comments
$begingroup$
Okay, here are my calculations.
Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.
So the 1st person will easily choose any day according to his choice.
This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.
Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.
So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.
I hope this helps.
For more discussions you can refer here.
answered Jun 18 '14 at 13:24
purupuru
92468
$endgroup$
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
|
show 4 more comments
$begingroup$
Okay, here are my calculations.
Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.
So the 1st person will easily choose any day according to his choice.
This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.
Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.
So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.
I hope this helps.
For more discussions you can refer here.
answered Jun 18 '14 at 13:24
purupuru
92468
$endgroup$
Okay, here are my calculations.
Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.
So the 1st person will easily choose any day according to his choice.
This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.
Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.
So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.
I hope this helps.
For more discussions you can refer here.
answered Jun 18 '14 at 13:24
purupuru
92468
edited Jun 18 '14 at 13:30
answered Jun 18 '14 at 13:24
purupuru
92468
answered Jun 18 '14 at 13:24
purupuru
92468
answered Jun 18 '14 at 13:24
purupuru
92468
92468
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
|
show 4 more comments
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
Allright, @puru, but how do you know that the series 1.(364/365).(363/365)....(343/365) is approximately 50% ???
$endgroup$
– Nib
Jun 18 '14 at 13:25
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
It's basically $(365)!/(353!)$*$1/(365)^{23}$
$endgroup$
– puru
Jun 18 '14 at 13:27
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@Nib I'd just use a calculator/computer to show that.
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:28
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
$begingroup$
@ThomasAndrews It might overflow, however, if we go by my method, it's better to write a code and calculate!
$endgroup$
– puru
Jun 18 '14 at 13:29
1
1
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
$begingroup$
@JoeyBF It is, but you need to know how to write the correct formula. For example, the formula you linked to was for 24 people...
$endgroup$
– Thomas Andrews
Jun 18 '14 at 13:31
|
show 4 more comments
$begingroup$
Simple example with three balls: red, green and blue.
When we form a collection of two balls, we have
$$
3^2
$$
possibilities.
But some do not contain the same color - and that is given by
$$
3 times 2
$$
So the number of collection such that two balls have the same color is given by
$$
3^2 - 3 times 2 = 3
$$
So the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{3^2 - 4 times 3}{3^2} = frac{3}{9} = frac{1}{3}
$$
We can do the same for 4 balls and a collection of 2 balls.
the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{4^2 - 4 times 3}{4^2}
$$
The basic formula is then given by
$$
frac{F^n - F times (F-1) times (F-2) times cdot (F-n)}{F^n}
$$
where
$$
F
$$
is the 'freedom' - the number of different colors for the balls, and
$$
n
$$
is the number of balls in the collection.
Using some math we can write
$$
1 - frac{F!}{F^n big(F-nbig)!}
$$
Note that when $n>F$ we have
$$
k!
$$
for a negative number.
But as
$$
big(n-1big)! = frac{n!}{n}
$$
we see that
$$
big(-1big)! = frac{0!}{0} rightarrow infty
$$
So in case $n > F$ the change becomes $1$
Instead of color - we can consider birthdays, so $F=365$ and we get
$$
1 - frac{365!}{365^n big(365-nbig)!}
$$
The case $n=23$ gives
$$
1 - frac{365!}{365^23 big(365-23big)!} = 50.7%
$$
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
$endgroup$
add a comment |
$begingroup$
Simple example with three balls: red, green and blue.
When we form a collection of two balls, we have
$$
3^2
$$
possibilities.
But some do not contain the same color - and that is given by
$$
3 times 2
$$
So the number of collection such that two balls have the same color is given by
$$
3^2 - 3 times 2 = 3
$$
So the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{3^2 - 4 times 3}{3^2} = frac{3}{9} = frac{1}{3}
$$
We can do the same for 4 balls and a collection of 2 balls.
the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{4^2 - 4 times 3}{4^2}
$$
The basic formula is then given by
$$
frac{F^n - F times (F-1) times (F-2) times cdot (F-n)}{F^n}
$$
where
$$
F
$$
is the 'freedom' - the number of different colors for the balls, and
$$
n
$$
is the number of balls in the collection.
Using some math we can write
$$
1 - frac{F!}{F^n big(F-nbig)!}
$$
Note that when $n>F$ we have
$$
k!
$$
for a negative number.
But as
$$
big(n-1big)! = frac{n!}{n}
$$
we see that
$$
big(-1big)! = frac{0!}{0} rightarrow infty
$$
So in case $n > F$ the change becomes $1$
Instead of color - we can consider birthdays, so $F=365$ and we get
$$
1 - frac{365!}{365^n big(365-nbig)!}
$$
The case $n=23$ gives
$$
1 - frac{365!}{365^23 big(365-23big)!} = 50.7%
$$
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
$endgroup$
add a comment |
$begingroup$
Simple example with three balls: red, green and blue.
When we form a collection of two balls, we have
$$
3^2
$$
possibilities.
But some do not contain the same color - and that is given by
$$
3 times 2
$$
So the number of collection such that two balls have the same color is given by
$$
3^2 - 3 times 2 = 3
$$
So the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{3^2 - 4 times 3}{3^2} = frac{3}{9} = frac{1}{3}
$$
We can do the same for 4 balls and a collection of 2 balls.
the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{4^2 - 4 times 3}{4^2}
$$
The basic formula is then given by
$$
frac{F^n - F times (F-1) times (F-2) times cdot (F-n)}{F^n}
$$
where
$$
F
$$
is the 'freedom' - the number of different colors for the balls, and
$$
n
$$
is the number of balls in the collection.
Using some math we can write
$$
1 - frac{F!}{F^n big(F-nbig)!}
$$
Note that when $n>F$ we have
$$
k!
$$
for a negative number.
But as
$$
big(n-1big)! = frac{n!}{n}
$$
we see that
$$
big(-1big)! = frac{0!}{0} rightarrow infty
$$
So in case $n > F$ the change becomes $1$
Instead of color - we can consider birthdays, so $F=365$ and we get
$$
1 - frac{365!}{365^n big(365-nbig)!}
$$
The case $n=23$ gives
$$
1 - frac{365!}{365^23 big(365-23big)!} = 50.7%
$$
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
$endgroup$
Simple example with three balls: red, green and blue.
When we form a collection of two balls, we have
$$
3^2
$$
possibilities.
But some do not contain the same color - and that is given by
$$
3 times 2
$$
So the number of collection such that two balls have the same color is given by
$$
3^2 - 3 times 2 = 3
$$
So the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{3^2 - 4 times 3}{3^2} = frac{3}{9} = frac{1}{3}
$$
We can do the same for 4 balls and a collection of 2 balls.
the change of finding two balls with the same color in a collection of 2 balls is given by
$$
frac{4^2 - 4 times 3}{4^2}
$$
The basic formula is then given by
$$
frac{F^n - F times (F-1) times (F-2) times cdot (F-n)}{F^n}
$$
where
$$
F
$$
is the 'freedom' - the number of different colors for the balls, and
$$
n
$$
is the number of balls in the collection.
Using some math we can write
$$
1 - frac{F!}{F^n big(F-nbig)!}
$$
Note that when $n>F$ we have
$$
k!
$$
for a negative number.
But as
$$
big(n-1big)! = frac{n!}{n}
$$
we see that
$$
big(-1big)! = frac{0!}{0} rightarrow infty
$$
So in case $n > F$ the change becomes $1$
Instead of color - we can consider birthdays, so $F=365$ and we get
$$
1 - frac{365!}{365^n big(365-nbig)!}
$$
The case $n=23$ gives
$$
1 - frac{365!}{365^23 big(365-23big)!} = 50.7%
$$
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
answered Jun 18 '14 at 13:51
johannesvalksjohannesvalks
5,4391117
5,4391117
add a comment |
add a comment |
$begingroup$
/r/eli5 explains more simply. I've rewritten 3 comments that stand alone and can be read separately.
Explanation 1 with Arithmetic
I misunderstood the birthday problem the first time, as I'd read about it and think: "If I find 22 (so a group of 23, not 70) other people, there is a 50% chance that one of them will have the same birthday as me."
However, the probability isn't that any particular person will have a match, but that at least one pair will have a match. It's much easier to understand the problem when you realize that there are $dfrac{23 times 22}{2} = 253$ unique pairs in the group.
Now reword the conclusion as "Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday."
Explanation 2 by visualizing a spinning prize wheel
Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29.
Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $~1/6$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $~1/6$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $(5/6)^{10}$ which is about a $~16%$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $~84%$ chance of landing on at least 1 previously landed-on slot.
In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group.
Explanation 3 by visualizing the Pigeonhole Principle
Once you understand what the "problem" itself is, you can easily visualize why the probability of a match is high, by using the Pigeonhole Principle.
Assume you have 365 different holes in a wall, and you randomly put pigeons into holes without looking if there's already a pigeon in there or not. As you add more pigeons, what increases steadily is the odds of you accidentally putting a pigeon into a hole that already has a pigeon.
The same principle can be applied to the birthday problem; treat each hole as a different day of the year and each pigeon representing a person. When a pigeon is placed into hole that already has a pigeon, this sharing of the hole means that someone in the room shares the same birthday as someone else.
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
$endgroup$
add a comment |
$begingroup$
/r/eli5 explains more simply. I've rewritten 3 comments that stand alone and can be read separately.
Explanation 1 with Arithmetic
I misunderstood the birthday problem the first time, as I'd read about it and think: "If I find 22 (so a group of 23, not 70) other people, there is a 50% chance that one of them will have the same birthday as me."
However, the probability isn't that any particular person will have a match, but that at least one pair will have a match. It's much easier to understand the problem when you realize that there are $dfrac{23 times 22}{2} = 253$ unique pairs in the group.
Now reword the conclusion as "Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday."
Explanation 2 by visualizing a spinning prize wheel
Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29.
Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $~1/6$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $~1/6$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $(5/6)^{10}$ which is about a $~16%$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $~84%$ chance of landing on at least 1 previously landed-on slot.
In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group.
Explanation 3 by visualizing the Pigeonhole Principle
Once you understand what the "problem" itself is, you can easily visualize why the probability of a match is high, by using the Pigeonhole Principle.
Assume you have 365 different holes in a wall, and you randomly put pigeons into holes without looking if there's already a pigeon in there or not. As you add more pigeons, what increases steadily is the odds of you accidentally putting a pigeon into a hole that already has a pigeon.
The same principle can be applied to the birthday problem; treat each hole as a different day of the year and each pigeon representing a person. When a pigeon is placed into hole that already has a pigeon, this sharing of the hole means that someone in the room shares the same birthday as someone else.
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
$endgroup$
add a comment |
$begingroup$
/r/eli5 explains more simply. I've rewritten 3 comments that stand alone and can be read separately.
Explanation 1 with Arithmetic
I misunderstood the birthday problem the first time, as I'd read about it and think: "If I find 22 (so a group of 23, not 70) other people, there is a 50% chance that one of them will have the same birthday as me."
However, the probability isn't that any particular person will have a match, but that at least one pair will have a match. It's much easier to understand the problem when you realize that there are $dfrac{23 times 22}{2} = 253$ unique pairs in the group.
Now reword the conclusion as "Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday."
Explanation 2 by visualizing a spinning prize wheel
Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29.
Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $~1/6$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $~1/6$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $(5/6)^{10}$ which is about a $~16%$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $~84%$ chance of landing on at least 1 previously landed-on slot.
In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group.
Explanation 3 by visualizing the Pigeonhole Principle
Once you understand what the "problem" itself is, you can easily visualize why the probability of a match is high, by using the Pigeonhole Principle.
Assume you have 365 different holes in a wall, and you randomly put pigeons into holes without looking if there's already a pigeon in there or not. As you add more pigeons, what increases steadily is the odds of you accidentally putting a pigeon into a hole that already has a pigeon.
The same principle can be applied to the birthday problem; treat each hole as a different day of the year and each pigeon representing a person. When a pigeon is placed into hole that already has a pigeon, this sharing of the hole means that someone in the room shares the same birthday as someone else.
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
$endgroup$
/r/eli5 explains more simply. I've rewritten 3 comments that stand alone and can be read separately.
Explanation 1 with Arithmetic
I misunderstood the birthday problem the first time, as I'd read about it and think: "If I find 22 (so a group of 23, not 70) other people, there is a 50% chance that one of them will have the same birthday as me."
However, the probability isn't that any particular person will have a match, but that at least one pair will have a match. It's much easier to understand the problem when you realize that there are $dfrac{23 times 22}{2} = 253$ unique pairs in the group.
Now reword the conclusion as "Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday."
Explanation 2 by visualizing a spinning prize wheel
Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29.
Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $~1/6$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $~1/6$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $(5/6)^{10}$ which is about a $~16%$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $~84%$ chance of landing on at least 1 previously landed-on slot.
In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group.
Explanation 3 by visualizing the Pigeonhole Principle
Once you understand what the "problem" itself is, you can easily visualize why the probability of a match is high, by using the Pigeonhole Principle.
Assume you have 365 different holes in a wall, and you randomly put pigeons into holes without looking if there's already a pigeon in there or not. As you add more pigeons, what increases steadily is the odds of you accidentally putting a pigeon into a hole that already has a pigeon.
The same principle can be applied to the birthday problem; treat each hole as a different day of the year and each pigeon representing a person. When a pigeon is placed into hole that already has a pigeon, this sharing of the hole means that someone in the room shares the same birthday as someone else.
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
edited Jan 22 at 4:58
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
answered Jan 22 at 4:48
Greek - Area 51 ProposalGreek - Area 51 Proposal
3,196769105
3,196769105
add a comment |
add a comment |
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5
$begingroup$
You could read the Wikipedia article en.wikipedia.org/wiki/Birthday_problem or show us your calculations
$endgroup$
– Henry
Jun 18 '14 at 13:12
$begingroup$
It has already been answered here: math.stackexchange.com/questions/25876/…
$endgroup$
– NicoDean
Jun 18 '14 at 13:13
$begingroup$
@henry, Ummm, I'm not sure if I can understand those solutions, but here are my calculations: We know that there are 253 possible combinations in the 23 ppl group, now, we also know that, each of them has birthday date A is 1/365, so, probability that two people share their birthday is 1/365^2, and so, any one pair sharing their birthday, the probability will be, 253/365^2, which of course, is very less, hence I believe this paradox is false.
$endgroup$
– Nib
Jun 18 '14 at 13:17
$begingroup$
Alo, doesn't the birthday paradox state that 2 ppl have the same birthday out of group of 23 ppl, whereas some of the commentators have pointed that the question's a duplicate of one which says 3 ppl have common birthdays, also, in that question, the number of ppl taken in room are 30, not 23, so, that means that my question and that question are quite different.
$endgroup$
– Nib
Jun 18 '14 at 13:19
$begingroup$
And also, @Henry, the wikipedia article is saying the same thing as I do till a certain point, after which, I can't understand at all what they're trying to tell me
$endgroup$
– Nib
Jun 18 '14 at 13:22