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Real world math question

Sorry if this has been dealt with (haven't read the whole thing), but the odds of 1 in 5-whatever million are only relevant if it was actually specified before-hand that the bonus number would specifically relate to each person's day of birth. I doubt that was the case because surely then the bonus number would only be in the range 1..31.

It's like throwing a dozen dice then saying hey, what are the odds on getting this particular combination? The odds are irrelevant unless the anticipated match was pre-specified.
 
lenny said:
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?)
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I imagine it depends on the printing technology they use. If they were using a computer printer, then yes, it would be very easy to use random or psuedo-random numbers. If they are using some other technology, then they may need a different whatchamacallit for each set of numbers. In that case not all numbers would be equally representative.

I bring up this issue not to criticize previous answers (I agree with the math presented) but rather because the title included the phrase "real world." In the real world numbers aren't necessarily as easy to track as they are in theory.
 
ChristineR said:
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.
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I think you are underestimating human laziness. My guess is that they don't really care and they use a simple random number generator.

So, let's assume the true odds are 1 in 5,308,416. Now how many times in a single year does a group of four people go to a chinese restaurant somewhere in the USA? I think it is at least an order of magnitude higher. If that is the case then this probably happens several times a year but it is usually not noticed.
 
I have been with a group where everyone got the same fortune, including the same lucky numbers. I assume that this was an error in the distribution process, but it does tell us something about the printing technology used in that case at least.

My guess would be that whoever writes the little bon mots just types in a string of numbers after them. However, I think it very likely that someone, somewhere, has used a random number generator, so I would think (1/42)^4 would be a reasonable upper limit.
 
The issue of it being the bonus number effects the chances. There have been numbers drawn already and they cannot be repeated.

Put it this way: are the chances of the first number listed being the birth date for the four people the same as the bonus number matching the birth date?

There is a constraint in place for the generation of a random bonus number. That constraint changes the odds. Almost 10% of the time, it will be impossible that the bonus number matches the day because that number has already been drawn. It's not a simple exercise.

I'm convinced that the answer given is wrong and I'll see if the correct answer is offered.

BTW, I am assuming entirely random number generation along with no biaas in printing/distribution.
 
NiallM, as I said before, it makes no difference whether the bonus number is drawn first or last, unless the bonus number is in some way special. In Michigan they have a pick 6 of 48 lotto, and then the bonus ball is drawn from a separate tub of seven red balls. Obviously then the bonus number is different.

However, if the bonus number is an just an extra ball drawn last from the tub of 48 numbers, then its chances are still just 1 in 48.

Let me do the math for two numbers drawn from 48 balls 1-48. What is the chance that 1 will be drawn first? 1 in 48.

What about the second ball? There are two cases--1 is drawn first, and one is not drawn first.

In case 1, the chance of 1 being drawn second is zero.
In case 2, the chance of 1 being drawn second is 1 in 47.

Net results of 1 being drawn second is

(1/48 * 0) + (47/48 * 1/47) = 0 + 1/48 = 1/48. Same as 1 being drawn first.

You can also do this with three or more balls, but your cases start to get pretty ugly. The point is that the numbers will work out to 1/48.
 
NiallM: But all numbers are equally likely to have appeared (or not) earlier in the sequence. No numbers are particularly excluded from that final position. All numbers remain equally likely in that last position.

[Christine beat me to it]
 
I'll leave the math to others but this reminds me of James Randi's piece on the 1/10/2007 episode of the podcast Skeptics Guide to the Universe. He talks about 2 cars involved in an accident that have consecutive VIN numbers. His point was that there are many possible coincidences that would register equally as amazing. His examples were consecutive license plates or driver's license numbers.
So the question becomes not what are the chances of this coincidence but what are the chances of any outcome that would be seen as unusual. In the case at hand I'm sure they would have felt the same if the first number in the sequence had been their birth dates instead of the bonus number. How about if they were in the second through fifth positions? You could probably think of dozens of combinations that would be amazing coincidences. So the odds that any one of them would come up are not as long as you might think.

No idea how to calculate any of that. Just thought I would throw it out as food for thought.
 
Yes, I know that you're both right, but it's my intuition that is telling me that it is wrong.

That is until I imagine 48,000,000 draws and ask myself how often 48 will be the bonus number in the draw. My guess is somewhere in teh region of 1,000,000.

A simulation quickly written up in c shows stunning adherence to this conclusion.

Intuition is an enemy in maths - in particular when you haven't studied stats in over 25 years.
 
baron said:
Sorry if this has been dealt with (haven't read the whole thing), but the odds of 1 in 5-whatever million are only relevant if it was actually specified before-hand that the bonus number would specifically relate to each person's day of birth. I doubt that was the case because surely then the bonus number would only be in the range 1..31.

It's like throwing a dozen dice then saying hey, what are the odds on getting this particular combination? The odds are irrelevant unless the anticipated match was pre-specified.
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Exactly!

And of course the meaningless probability calculations continue.

(I would have thought this answer was obvious to skeptics)
 
Squishua said:
Exactly!

And of course the meaningless probability calculations continue.

(I would have thought this answer was obvious to skeptics)
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Sometimes the prediction of an event is implicit rather than explicit. (such as, for instance, a shared birthdate). Are you telling me that if you started work in an office with three other people and they all turned out to share your birthdate that you wouldn't ask yourself "what are the odds on that?"

In this case it is still appropriate to ask a "What are the odds?" question and still subject it to a probability calculation as if the event had been predicted.
 
NiallM said:
In this case it is still appropriate to ask a "What are the odds?" question and still subject it to a probability calculation as if the event had been predicted.
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It's "appropriate" for a bit of fun, yes. It's not, however, an appropriate method by which to mathematically gauge liklihood.

A few years ago I worked with two colleagues in a sectioned-off part of the office. We all played the National Lottery and one draw I was lucky enough to get 5 numbers (1 in 55,000 odds). The next month, my friend on the left got 5 numbers too. Three months later, my colleague on my right got 5 numbers.

Now any way you look at it, that's unlikely. But what are the odds? Are they really 1 in 166,375,000,000,000? Of course not, that would practically prove the intervention of a cosmic superpower. The actual fact is that the odds aren't relevant, because nobody specifically predicted we three people would win specific amounts on specific days. There are so many combinations that would prove equally unlikely that attempting to apply mathematics to it is just ridiculous.
 
NiallM said:
Put it this way: are the chances of the first number listed being the birth date for the four people the same as the bonus number matching the birth date?
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How could there be any difference? Suppose that five different people were to use five different methods for ensuring that they are all differemt. The first person chooses them in order. The second chooses them in reverse order, so that the bonus number is chosen first. The third person chooses them all at once, and if any number is chosen twice, he throws the results out, and tries again until they're all different. The fourth person assigns a number to all the cards of a deck, take four cards out the deck out, spreads six of them out on a table, and takes them to represent the numbers, left to right. The fifth person takes the same cards, but takes them right to left. Can any of these people have different probabilities of getting a certain bonus number?

There is a constraint in place for the generation of a random bonus number. That constraint changes the odds.
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Not on average.

Almost 10% of the time, it will be impossible that the bonus number matches the day because that number has already been drawn.
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But tyhe rest of the time, it will correspondingly more likely. If none of the other numbers match, then there will now be a 1/43, rather than 1/48, chance of the last number matching.

It's not a simple exercise.
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Actually, the math isn't very complicated.
The first number has a 1/48 chance of matching.
For the second number to match, we need the first not to match (p=47/48), then we need the second number to match (since we have only 47 numbers left, p=1/47). Multiply those together, and we get 1/48.
For the third number, we need the first two to not match (p= (47/48)*(46/47)) and the third to match (p=45/46). Again, the product is 1/48.
And so on.
 
baron said:
A few years ago I worked with two colleagues in a sectioned-off part of the office. We all played the National Lottery and one draw I was lucky enough to get 5 numbers (1 in 55,000 odds).
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lucky devil! How much did you win? :)
 
Art Vandelay said:
How could there be any difference? Suppose that five different people were to use five different methods for ensuring that they are all differemt. The first person chooses them in order. The second chooses them in reverse order, so that the bonus number is chosen first. The third person chooses them all at once, and if any number is chosen twice, he throws the results out, and tries again until they're all different. The fourth person assigns a number to all the cards of a deck, take four cards out the deck out, spreads six of them out on a table, and takes them to represent the numbers, left to right. The fifth person takes the same cards, but takes them right to left. Can any of these people have different probabilities of getting a certain bonus number?

Not on average.

But tyhe rest of the time, it will correspondingly more likely. If none of the other numbers match, then there will now be a 1/43, rather than 1/48, chance of the last number matching.

Actually, the math isn't very complicated.
The first number has a 1/48 chance of matching.
For the second number to match, we need the first not to match (p=47/48), then we need the second number to match (since we have only 47 numbers left, p=1/47). Multiply those together, and we get 1/48.
For the third number, we need the first two to not match (p= (47/48)*(46/47)) and the third to match (p=45/46). Again, the product is 1/48.
And so on.
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That's what I acknowledged. I wrote a simulation. A few lines into it I realised the compensatory forces at work in the whole equation. I finished off the thing and the results, of course, square up.
 
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