Real world math question

[Angus McPresley]

A friend of mine on a mailing list came across this math question. I gave him an answer, but I want to see if you all agree with me before I post it.

He was at a Chinese restaurant with his wife and two friends. At the end of the meal they all got fortune cookies, the kind with lottery numbers on the back. These are the kind with five numbers, a colon, and the a final bonus number, e.g. 34, 23, 3, 5, 14: 21.

What happened is that each of the four ended up with their birth date (the day of the month) as the last (bonus) number. He was wondering, what (exactly) are the odds of such an occurrence?

For your answer, assume the following (some of which may be superfluous):

1. Each number in the list ranges from 1 to 48.
2. There are never repeats of numbers. Do the earlier numbers matter?
3. He didn't say which day of the month anyone was born. Does that matter?

I feel confident the answer I gave him was correct, but I'd love to get confirmation from some real math types.

 

[RenaissanceBiker]

Well, you can calculate the odds of it happening one time easy enough. But if you go to chinese restaurants enough times, and open a few fortune cookies each time, it is bound to happen. It may have happened to me once or twice but I don't pay attention to those numbers.

 

[andyandy]

He was at a Chinese restaurant with his wife and two friends. At the end of the meal they all got fortune cookies, the kind with lottery numbers on the back. These are the kind with five numbers, a colon, and the a final bonus number, e.g. 34, 23, 3, 5, 14: 21.

What happened is that each of the four ended up with their birth date (the day of the month) as the last (bonus) number. He was wondering, what (exactly) are the odds of such an occurrence?

.

i don't understand this sentence

"What happened is that each of the four ended up with their birth date (the day of the month) as the last (bonus) number."

did the four different people at the restaurant all have the same birthday (day of the month) and that day was the same as the one fortune cookie they opened?
Or did they have 4 separate cookies and each of those had a bonus number the same as their birthday (day of the month)?

 

[Crossbow]

Off hand, I would say the odds would be
1:48⁴or put another way
1:5,308,416.

By the way, I am not terribly well versed in probability so if someone can double-check my math I would appreciate it as well.

 

[andyandy]

if it's the second, assuming that there's a even distribution of numbers from 1-49, then the odds of any one person having their birthday match is just 1/49....and provided a large enough number of these cookies are produced, then the choices are effectively independent - so for all 4 people the odds would just be (1/49)^4

ETA

oh 48 rather than 49......well, same process

 

[cyborg]

Um, surely the fact that only 31 of the 49 numbers CAN be valid birthdates and then only for certain months of the year complicates things somewhat?

 

[andyandy]

Um, surely the fact that only 31 of the 49 numbers CAN be valid birthdates and then only for certain months of the year complicates things somewhat?

there's still a 1/48 (or 1/49 ) chance that your birthdate will match the bonus ball regardless.

 

[RenaissanceBiker]

andyandy for the win.

 

[cyborg]

Yes, I'm looking at that backwards. Best not to attempt maths whilst hungover I guess.

 

[Gwyn ap Nudd]

Each walks into the restaurant with a number (the birthday) that he is hoping to match. All of those numbers (1-31) can be matched if the cookies' numbers go from 1-48.

The probability of any one matching his number is 1/48. The odds are 1:47

Since each match is independant of the others, you multiply the probabilities to get the event that four match (1/48⁴) So the odds are 1: (48⁴-1) = 1: (5308416 - 1) = 1:5308415

But then, you added the condition that there are no repeats of numbers. That complicates the problem in two ways. First, if any two (or more) have the same birthday, then it is not possible for all four to match, and second, once the first person opens and reveals his cookie, the second person has only 47 numbers to worry about, etc.

The probability that two people have different birthdays is 30/31. Given this, the probability that a third does not match either of the first two is 29/31. Given that, the probability that the fourth does not match any of the other three is 28/30. So the probability that the four don't match is (28*29*30)/31³.

Similarly, the probability of the first matching his cookie is 1/48, the second has a 1/47 probability, etc.

The final probability is (28*29*30)/31³*48*47*46*45 or 24360/139121586720 = 406/2318693112. The odds are 406:2318692706

 

[andyandy]

But then, you added the condition that there are no repeats of numbers. That complicates the problem in two ways. First, if any two (or more) have the same birthday, then it is not possible for all four to match, and second, once the first person opens and reveals his cookie, the second person has only 47 numbers to worry about, etc.

yes, i'm not sure why the "no repeats" was introduced to model this situation.....i took it as meaning that the numbers before and including the bonus ball weren't repeated within that sequence....in which case it wouldn't effect the bonus ball for this model - it would still be a number between 1-48 and therefore there would still be a 1/48 chance of a birthday matching it.....

 

[Jekyll]

andyandy for the win.

Yes, but he answered the wrong question .
Calculating the probability of one particular hit doesn't tell us very much, you have to ask what is the probability of any hit occurring.
So you need to consider all the probabilities of it happening in the first column or the birthdates being shifted round the table one to the left or shifted chronologically etc. .

 

[andyandy]

Yes, but he answered the wrong question .
Calculating the probability of one particular hit doesn't tell us very much, you have to ask what is the probability of any hit occurring.
So you need to consider all the probabilities of it happening in the first column or the birthdates being shifted round the table one to the left or shifted chronologically etc. .

yes - but that wasn't the question asked.......which was specifically about the bonus ball....

why answer a harder question if an easier one will suffice?

but it's true - just because the odds of this happening are small, it doesn't mean this is particularly impressive - why choose the bonus ball? Why choose birthdays? Why that particular day at the restaurant? This is just post-hoc analysis....

 

[RenaissanceBiker]

What happened is that each of the four ended up with their birth date (the day of the month) as the last (bonus) number. He was wondering, what (exactly) are the odds of such an occurrence?

He asked about the bonus number. The others don't matter. I thought the part about repeats of numbers referred to the numbers in individual cookies, not birthdays. I assumed that it would be possible for two of them to have the same birthday.

 

[Ladewig]

I think the assumption that all numbers from 1-48 occur with equal probability on fortune cookie fortunes might not be correct. I am not convinced that the printers really use a random number function when generating the numbers.

 

[Jekyll]

yes - but that wasn't the question asked.......which was specifically about the bonus ball....

why answer a harder question if an easier one will suffice?

Mainly because it needs beating into people that just because the chances of something happening are vanishingly small, it doesn't mean that we should be surprised that it did happen.

Having said that, I didn't try to answer the harder version either .

 

[Angus McPresley]

Here was the answer I gave him; I'm happy that I came up with the same answer as many here... It was he that initially expressed a concern about the earlier numbers in the list, so I included that in my statement of the problem to you all, but you all seem to have basically come to the same conclusion about them, that they didn't matter...

I love math problems that pop up in real life, so I'll give this a
shot. Be warned, I've had four glasses of red.

We can't be fully sure how they pick the numbers, but I think it's
fairly safe to say that the earlier numbers don't come into play at
all -- any number is equally likely to appear as the last number, so
the last number is essentially random.

The next problem is figuring out the chances that just one of your
birth dates -- a number that's 1 in 28, 30, or 31 -- matches a number
that's 1 in 48. I think that's second nature to a statistics guru to
figure out, but I had to think long and hard. I figure the odds are
just 1 in 48. I think the rule must be, for any 1 in n to match
another 1 in m, the odds are 1 in max(m,n). I'm fairly sure that's
correct. So that's the odds of just one person matching their birth date.

So now, the odds of four of you matching is 1 in 48 times itself four
times, or --

1 in 5,308,416

That's so unlikely that I wonder if my math is wrong. Or else it was
just a wonderful coincidence!

Of course, 1000 things (conservatively) happen to you a day, at which
rate something that unlikely would happen to you once every 14 years
or so...

 

[Art Vandelay]

"I think the rule must be, for any 1 in n to match
another 1 in m, the odds are 1 in max(m,n). "

Another way of looking at it is that it is o/m*n, where o=overlap. If they both start at one, then o=min(n,m). But this formula generalizes to other cases. For instance, what if the numbers ranged from 10 to 60? Then it would be 22/51*31 (pretending that there are 31 dates in each month).

 

[lenny]

I think the assumption that all numbers from 1-48 occur with equal probability on fortune cookie fortunes might not be correct. I am not convinced that the printers really use a random number function when generating the numbers.

i expect that "they" do use a random number generator on a digital computer; which means of course the numbers are not random.

and i expect that they may print the same fortune and number in many different cookies. (any data here?)

neither fact changes the odds much, as long as you only count one dinner.(and, as Jekyll and andyandy noted, the probability was arguably one when the calculation was done; analysis post-hoc is fundamentally misleading.)

 

[ChristineR]

I would also say that it's 1 : (48)^4, assuming all the assumptions are correct.

The "no repeats" rule is not really relevant here. First of all, you can repeat. Each person could have a 1 on their slip for example. Second, there is no special reason why two of the diners could not have been born on the first and both had a 1 as their bonus numbers.

The right way to look at is to treat each slip as a completely independent entity. Then assuming that all the slips were made according to the same rules, the answer must be (something) to the fourth power.

If you look at each slip as a separate entity, then simply imagine that the bonus ball is drawn first. The odds of a match are 1:48. The rest of the numbers aren't really relevant. If you want to imagine that the bonus ball is drawn last you can do some somewhat complicated math and find the chances are still 1:48. Basically, if the target ball is not drawn in the first six picks it becomes more likely that it will be drawn in the latter picks and the chances that it cannot be drawn last (because it was already drawn) cancels out the increased chance that it will be drawn last.

Now I suspect that these numbers are not random, nor even sufficiently psuedorandom as they would be if computer generated. First of all, lotto players love dates, and as such the numbers 1-31 are overrepresented when humans pick lotto numbers. I find it interesting that the example given was 34, 23, 3, 5, 14: 21, which does seem to overrepresent the range 1-31. I see no reason why a human writing these numbers would not over represent dates.

Secondly, not all lottos are pick five of 1-48 with a 1-48 bonus number. It seems to me that the cookie makers might be trying to cover eventualities and not generating random numbers at all--that is, a human is in fact writing them. That's my guess.

So overall, I would suspect that the numbers are 1-42, not 1-48, and that 1-31 will be overrepresented. It's still a pretty nice coincidence though.