The Birthday Problem
Part 1: How many total strangers must you lure into a room in order to have a 50/50 chance that at least two of them have the same birthday (month and day)?
Quote: PerditionDo we account for the those zany Leap Year Kids?
Fair question.
Let's ignore 'em. That way we only have an even 365 days with which to work.
N=23=0.507297Quote: MathExtremistProduct(1-((i-1)/365)) for i=1..N becomes less than 50% at N=23.
What's the second part?
the 2nd part
I say it could be the "almost-birthday problem"
in a randomly assembled group of N people at least two people will have birthdays within R
days of each other
how about 50% for R=1
This is one my new questions at casino night party events
goes over well
as does
where's the beef?
my second guess to second part
how many persons are needed to have a group of persons in which all 365
possible birthdays (excluding February 29) are represented with a probability
of at least 50%.
good for large casino night party events
for 38#s in Roulette that would be 152 spins at 0.501599
added:
part 2 not as exciting IMO
here goes
square root of 2*n*LN(1/(1-p))
this approximates
n: 2,598,960
p: 0.5
1,898.136874
so 1899 hands (a bit more than 3 hours of you play)
using ME formula and Excel
0.999001: 5991
0.750153: 2685
0.500217: 1899
0.250265: 1224
0.100128: 741
some R code to check
> n <- 1899
> 1-prod( ((2598960-(n-1)):2598960)/rep(2598960,n) )
[1] 0.500217
Quote: MathExtremistProduct(1-((i-1)/365)) for i=1..N becomes less than 50% at N=23. What's the second part?
OK, now you are actually scaring me. It appears that you have really "matched my meat".
I'm starting to wonder if I have the horsepower to even slow you down, let alone stump you.
Having made proper mincemeat of Part I, allow me to present you with Part II (slightly ahead of schedule).
I will attempt to be precise, and I formally invite you to chastise me if I'm not.
"I play Jacks or Better video poker which uses a standard 52 card deck (no jokers). I average about 600 hands per hour, and often get an eerie sense of deja vous that I've been dealt the identical initial set of five cards previously that morning. I wondered how likely that really was, or whether I was simply becoming "punchy" from too much play. Therefore, I put it to you (and all others who dare) the following quest: Disregarding permutations, how many hands must I play to have a 50/50 chance of being dealt two combinatorically identical initial hands?
Quote: dblanch256OK, now you are actually scaring me. It appears that you have really "matched my meat".
I'm not sure what that means, and I'm also not sure I want to... :)
Quote:"I play Jacks or Better video poker which uses a standard 52 card deck (no jokers). I average about 600 hands per hour, and often get an eerie sense of deja vous that I've been dealt the identical initial set of five cards previously that morning. I wondered how likely that really was, or whether I was simply becoming "punchy" from too much play. Therefore, I put it to you (and all others who dare) the following quest: Disregarding permutations, how many hands must I play to have a 50/50 chance of being dealt two combinatorically identical initial hands?
In the same vein of "first understand the problem," do you mean
A) combinatorially identical with respect to a standard 52-card deck
or do you mean
B) combinatorially identical with respect to a Jacks or Better video poker paytable?
As an example, Ah 9h 3h 4d 6c has a particular distribution of expectations for each of the 32 possible draw strategies, and one of those represents the optimal play. Ad 9d 3d 4h 6s has exactly the same distribution and optimal strategy. Are those combinatorially identical as you're using it? Edit: both hands, from a poker standpoint, are "Ace 9 3 suited with 4, 6 offsuit."
If A, then substitute 2,598,960 for 365 in the formula above, then solve for N. If B, substitute 134,459 for 365 and solve for N.
See:
http://www.math.utah.edu/~ethier/sample.pdf
https://wizardofodds.com/games/video-poker/methodology/
Classical textbook exercise. I do it every year with my students. D.Blanch is probably a first-year student discovering all these simple problems and feeling elated and superior. Sorry boy, we went there before you...Quote: MathExtremistProduct(1-((i-1)/365)) for i=1..N becomes less than 50% at N=23. What's the second part?
One year, I proposed the bet to the class and was surprised by their unanimous readiness to bet that there IS a "birthday coupling". Until one added: "We have twins in the class".
To be sure: births are not equiprobably distributed along the year, so the traditional answer is incorrect (even without taking into account 29 Feb and twins). Simple reasoning shows that the probability of similar birthdays must be higher than the hypergeometric answer. Hence, the number of people necessary to bring the probability to 50% might well be less than 23. This depends on the birth distribution in the mother population. In my country, I remember computing it was slightly above 21 (hence 22).
Quote: kubikulannClassical textbook exercise. I do it every year with my students. D.Blanch is probably a first-year student discovering all these simple problems and feeling elated and superior. Sorry boy, we went there before you...
One year, I proposed the bet to the class and was surprised by their unanimous readiness to bet that there IS a "birthday coupling". Until one added: "We have twins in the class".
To be sure: births are not equiprobably distributed along the year, so the traditional answer is incorrect (even without taking into account 29 Feb and twins). Simple reasoning shows that the probability of similar birthdays must be higher than the hypergeometric answer. Hence, the number of people necessary to bring the probability to 50% might well be less than 23. This depends on the birth distribution in the mother population. In my country, I remember computing it was slightly above 21 (hence 22).
I apologize for offering such simple problems. Suggest you take a look at Wizard Bait (two topics ago) and see how simple you find it to be. ;)
Quote: kubikulannClassical textbook exercise. I do it every year with my students. D.Blanch is probably a first-year student discovering all these simple problems and feeling elated and superior. Sorry boy, we went there before you...
One year, I proposed the bet to the class and was surprised by their unanimous readiness to bet that there IS a "birthday coupling". Until one added: "We have twins in the class".
To be sure: births are not equiprobably distributed along the year, so the traditional answer is incorrect (even without taking into account 29 Feb and twins). Simple reasoning shows that the probability of similar birthdays must be higher than the hypergeometric answer. Hence, the number of people necessary to bring the probability to 50% might well be less than 23. This depends on the birth distribution in the mother population. In my country, I remember computing it was slightly above 21 (hence 22).
I apologize for offering such simple problems. Suggest you take a look at Wizard Bait (two topics ago) and see how simple you find it to be. ;)
Or, if that one is beneath you also, try this one:
I know an infinite number of wizards and told them to prepare for the following contest and see if they can derive a method that guarantees a winning probability of at least 90 percent.
Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?
[Should be a piece of cake for someone who has "seen it all before" (!)
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"Sex is like a snotty French math teacher ... mostly chafing and hot air." -- Jean Claude Van Dave
Quote: dblanch256I apologize for offering such simple problems. Suggest you take a look at Wizard Bait (two topics ago) and see how simple you find it to be. ;)
Or, if that one is beneath you also, try this one:
I know an infinite number of wizards and told them to prepare for the following contest and see if they can derive a method that guarantees a winning probability of at least 90 percent.
Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?
[Should be a piece of cake for someone who has "seen it all before" (!)
=======================================================================================
"Sex is like a snotty French math teacher ... mostly chafing and hot air." -- Jean Claude Van Dave
I don't think it's a question of "beneath" anyone, but you are playing in some rarified air here, and what's evident is that you haven't read back in the forum enough to know your audience is at least your equal. I'm quite certain you're welcome to pose your questions, but yes, they are simplistic to many of the guys here. I'm also certain that if you're looking for fellow math playmates, you will be welcome to settle in and challenge people for fun, but it's likely you will have to work some to be the top dog you seem to be portraying yourself to be (referring collectively to your comments in your several threads ). Enjoy; I came here to solve a particular strategy problem; I stuck around because of the incredible concentration of very smart people. Hopefully that's what you're looking for as well.
Quote: beachbumbabsQuote: dblanch256I apologize for offering such simple problems. Suggest you take a look at Wizard Bait (two topics ago) and see how simple you find it to be. ;)
Or, if that one is beneath you also, try this one:
I know an infinite number of wizards and told them to prepare for the following contest and see if they can derive a method that guarantees a winning probability of at least 90 percent.
Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?
[Should be a piece of cake for someone who has "seen it all before" (!)
=======================================================================================
"Sex is like a snotty French math teacher ... mostly chafing and hot air." -- Jean Claude Van Dave
I don't think it's a question of "beneath" anyone, but you are playing in some rarified air here, and what's evident is that you haven't read back in the forum enough to know your audience is at least your equal. I'm quite certain you're welcome to pose your questions, but yes, they are simplistic to many of the guys here. I'm also certain that if you're looking for fellow math playmates, you will be welcome to settle in and challenge people for fun, but it's likely you will have to work some to be the top dog you seem to be portraying yourself to be (referring collectively to your comments in your several threads ). Enjoy; I came here to solve a particular strategy problem; I stuck around because of the incredible concentration of very smart people. Hopefully that's what you're looking for as well.
Yes there is clearly rarefied air here, and that's exciting for me as well. If anyone feels I've "talked down" to them, or have otherwise maligned them, I apologize. But I didn't come here to be insulted either. Apparently posting "traditional problems" is offensive to some members. I didn't realize that before, but that's easy enough to correct. I guess if all else fails, there's always the BLOCK option (sigh).
I may be a bit high-spirited, but part of that is, like you said, the giddiness of finding this new group. If you have reviewed my comments, you already know that I have made a special effort to give credit (and help) wherever I can. The few times I've lashed out [e.g. "no soup for you"] were aimed at people whose replies had nothing to do with the topic (e.g. "what color is the missile?"), were neither clever nor funny, and should have been posted to a "who gives a shit" blog somewhere else. I've been trying to follow a thread and some dude who's "bored out of his gourd" keeps posting "That's what she said ..." or some equally jejune clutter, between every other legitimate reply. I don't flag them because, at heart, they are only dumb-ass annoyances.
That said, you make some good points and I'm glad you were "man enough" to offer them. Thank you for taking the time to post.
And the distribution of births is, as I recall, not equivalent between different countries and cultures. The traditional notion of a June bride, coupled with the traditional notion of an immediately consummated marriage, ought to lead to a larger than average probability of a springtime birth. But only in the northern hemisphere where that tradition holds: in the southern hemisphere, June is in the middle of winter. I wonder if Mike has any global statistics on this from his time with the government.Quote: kubikulannTo be sure: births are not equiprobably distributed along the year, so the traditional answer is incorrect (even without taking into account 29 Feb and twins).
Quote: 7crapsN=23=0.507297
the 2nd part
I say it could be the "almost-birthday problem"
in a randomly assembled group of N people at least two people will have birthdays within R
days of each other
how about 50% for R=1
This is one my new questions at casino night party events
goes over well
as does
where's the beef?
my second guess to second part
how many persons are needed to have a group of persons in which all 365
possible birthdays (excluding February 29) are represented with a probability
of at least 50%.
good for large casino night party events
for 38#s in Roulette that would be 152 spins at 0.501599
added:
part 2 not as exciting IMO
here goeswikipedia has the formulas (there are a few ways)
square root of 2*n*LN(1/(1-p))
this approximates
n: 2,598,960
p: 0.5
1,898.136874
so 1899 hands (a bit more than 3 hours of you play)
using ME formula and Excel
0.999001: 5991
0.750153: 2685
0.500217: 1899
0.250265: 1224
0.100128: 741
some R code to check
> n <- 1899
> 1-prod( ((2598960-(n-1)):2598960)/rep(2598960,n) )
[1] 0.500217
Yes, 1898 is what I got also. Well played! [Clearly there is some rarefied air in this forum.]
Since I've had some snotty complaints for posting "easy" problems, I'll try to ramp up the challenges, and hopefully the interest level as well.
FWIW, I appreciate your willingness to confirm this number because I wasn't completely sure my own calculations were correct. Like I said, after about three hours of play I was unable to recall the complete set of previous hands I'd had, so I wanted to determine if it was even plausible. It appears that it is.
Quote:I don't think it's a question of "beneath" anyone, but you are playing in some rarefied air here, and what's evident is that you haven't read back in the forum enough to know your audience is at least your equal. I'm quite certain you're welcome to pose your questions, but yes, they are simplistic to many of the guys here. I'm also certain that if you're looking for fellow math playmates, you will be welcome to settle in and challenge people for fun, but it's likely you will have to work some to be the top dog you seem to be portraying yourself to be (referring collectively to your comments in your several threads ). Enjoy; I came here to solve a particular strategy problem; I stuck around because of the incredible concentration of very smart people. Hopefully that's what you're looking for as well.
Babs --
Again, you make several good points. Just so you know, I have no illusions about being top dog here. I've already met at least two people who can run rings around me, and that's both humbling and rewarding.
What set me off was this gratuitous remark:
Classical textbook exercise. I do it every year with my students. D.Blanch is probably a first-year student discovering all these simple problems and feeling elated and superior. Sorry boy, we went there before you...
Does that sound reasonable to you?
Worse, this same douche bag went back to a totally different thread and balanced my initial five star rating with a zero. And yes, I intentionally offered him a problem I knew would be over his head. He needs to understand that I'm not some defenseless newbie who will tolerate a cowardly attack on his character. [I give as good as I get.]
Anyway, he seems to have crawled back into his classroom, at least for now. I say good riddance.
Quote: dblanch256
Worse, this same douche bag went back to a totally different thread...
Anyway, he seems to have crawled back into his classroom, at least for now. I say good riddance.
Personal insult, three-day Suspension.
(What is a "douche bag", by the way?)
I think I must plead guilty, about condescension. Actually, when talking scientific stuff, I stick to the facts and don't deem important to take care of feelings. My fault, I suppose.
On other stuff, I sometimes choose to be tactful, sometimes not when I wish to give a lesson.
I observe a behaviour, I draw factual conclusions and state them blankly. No insult intended, but it sure may not please those exposed.
Do you also have that saying: "Only truth hurts" ? If you see how young Blanch reacted, his feeling insulted, doesn't that prove the point? Maybe he is not a first-year student, but feeling it as an insult is not very respectful for all the first-year students. Or does he resent the age difference? (Cf. the term "snotty" and his hatred of teachers) Also, "elated" or "superior" don't mean insults in my book. Why his fury, if not because he has difficulty accepting that other people have learnt what he did before him?
Oh! and: I am not one to play with star ratings and such...
Quote: kubikulannThank you, Babs.
(What is a "douche bag", by the way?)
I think I must plead guilty, about condescension. Actually, when talking scientific stuff, I stick to the facts and don't deem important to take care of feelings. My fault, I suppose.
On other stuff, I sometimes choose to be tactful, sometimes not when I wish to give a lesson.
I observe a behaviour, I draw factual conclusions and state them blankly. No insult intended, but it sure may not please those exposed.
Do you also have that saying: "Only truth hurts" ? If you see how young Blanch reacted, his feeling insulted, doesn't that prove the point? Maybe he is not a first-year student, but feeling it as an insult is not very respectful for all the first-year students. Or does he resent the age difference? (Cf. the term "snotty" and his hatred of teachers) Also, "elated" or "superior" don't mean insults in my book. Why his fury, if not because he has difficulty accepting that other people have learnt what he did before him?
Oh! and: I am not one to play with star ratings and such...
A "douche bag" is American slang for a man who is an indiscriminate ass regardless of the subject or the situation. It is derived from the appearance of a clear plastic bag and long nozzle provided in a commercial product to wash a woman's inner genitalia, after the product's use. The words "Prick" or "Dick" would be a rough equivalent; I think I saw you used one of those in another post.
We also have the saying "the truth hurts". It is often used as a condescending remark as well, though generally factual in statement. You are nothing if not straighforward and blunt; something I've had to learn about you, myself.
I think I was in error saying you were from Denmark; it's Belgium, correct? Est-ce que vous etes Belge? Sorry about that.
In many everyday situations, I have to endure that facts or "truth" is considered tactless and/or condescending. After years of unsuccessfully trying to grasp how I should say things, I gave up, accepting that I am "not like the rest of them" (i.e. those charming tactful people) and living with it.
Like Epicurus said, stop suffering about things which you cannot change, and change those you can.
Quote: kubikulannYes, I'm Belgian.
In many everyday situations, I have to endure that facts or "truth" is considered tactless and/or condescending. After years of unsuccessfully trying to grasp how I should say things, I gave up, accepting that I am "not like the rest of them" (i.e. those charming tactful people) and living with it.
Like Epicurus said, stop suffering about things which you cannot change, and change those you can.
Well said.
Quote: beachbumbabsWell said.
At first I was going to send this message privately to Kubikulann, but I think he deserves better, so I posting it publicly instead.
Kubikulann, my recent incarceration gave me time to reflect on this forum, our altercation, and most painfully of all, my own actions. There is no excuse for the verbal barrage I subjected you to and I am deeply and sincerely sorry. All indications (including heartfelt endorsements from your many friends here) suggest that you are a long-standing and treasured resource in this forum, and that I'd be a fool to think otherwise. I agree with that.
I did you wrong. I both recognize and regret it. I hope you can find it within yourself to forgive me. Either way, I'm very sorry.
The internet age is dangerous precisely because it allows such quick retorts. I do not blame you at all, and made an effort on myself not to be incensed by words that hurt me.
I hope the proof is made that we are both above that. Cheers to a protracted mathematical collaboration and exhange!
