Let’s say I had a million cards and one of the cards has a picture of a dog on it and all of the others were blank. If I flipped over one card, my odds of flipping over the cardetails with the dog on it would be one in a million. What would my odds be if I had a million chances?
It depends. Do you separate out the cards that have already been flipped? Or do you put the flipped cards back in the deck and reshuffle each time?
You put the flip card back in the deck and reshuffle every time
63.21% It’s 1 - (999999/1000000)^1000000
In other words, what’s the chance of you NOT picking the dog card? It’s 999999/1000000. What’s the chances of you not picking the dog card a million times in a row? That’s the ^1000000 part. So the probability of you picking the dog card would be 1 - that number.
There’s also a quick estimate that achieves almost the same number (to six decimal places) expressed as 1-(1/e).
I don’t understand a word of what you just said, but that’s the answer and you are a genius!
Two more questions. Is that 63.21 even, or did you approximate? How many times would it take to get to 50 or more percent?
Or he/she took Calculus.
The limit of (1+1/n)[sup]n[/sup] is e. This is extremely well known. Change it to (1-1/n)[sup]n[/sup] and you get 1/e. Subtract that from 1 to get the answer. While a million isn’t infinity, it’s close enough for the convergence rate of this limit.
Knowing that a lot of probability stuff converges to either 1/e or 1-1/e is useful in many situations. E.g., in Computer Science it pops in hashing table stuff. Also in choosing a place to park your car.
And no, that’s not remotely “even”. But enough digits to get the general idea.
(1) In the limit, with a hyuuuge number of cards, a better approximation is
63.21205588285576784044762298385391325541888690%
For the smaller number, 1000000, use
63.2120742768354905714201874729630340981109564%
(2) From an urn of size n, the number of picks (with replacement) needed to have at least p chance of getting the unique joker is
ln(1-p) / ln((n-1)/n)
When n is very large (e.g. n = 1 million), this is approximated by
- n * ln(1-p)
For n = 50%, this approximation becomes
-1000000*ln(.5)
You can type that last line into Google and see, approx.
693147
Why do I get the feeling this is a homework question? :rolleyes:
The basic equation is:
y = 1 - (999,999/1,000,000)^x
In your original question, you were asking “What is y when x is 1,000,000?”
In your quoted question, you are asking “What is x when y = .5?” Solve that and you have your answer.
Notwithstanding the re-shuffling as described above … and thank you folks for the wonderful answers in the case you’ve presented … I just wanted to be the first to point out that flipping these million cards in order, the odds of flipping the dog card is 1 … [giggle] … I learned that playing blackjack
Do you know about basic probabilities? Let’s simplify this so we can see what’s going on. Let’s take the case of a 6-sided die. What is the chance I roll a 6 on one roll? It’s 0.166… or 1/6. Right? Easy enough.
Now how do we figure it out that in two rolls what the odds of getting at least one six are? The way we figure it out is to work in the other direction: that is, figure out what the chances are we don’t roll a six in each roll. The probability of not rolling a 6 in each roll is 5/6 or 0.8333… So there’s roughly an 83% chance I don’t roll a six in the first roll, and an 83% chance I don’t roll a six on the second roll. To figure the probabilty that in two rolls we don’t roll a 6, we multiply the probabilities (.83333… * .83333… = .694444, so the probability that we roll at least one six is 1 minus that, or .305555…)
Let’s check by enumerating all the possible combinations:
1-1,1-2,1-3,1-4,1-5,1-6
2-1,2-2,-2-3,2-4,2-5,2-6
3-1,3-2,3-3,3-4,3-5,3-6
4-1,4-2,4-3,4-4,4-5,4-6
5-1,5-2,5-3,5-4,5-5,5-6
6-1,6-2,6-3,6-4,6-5,6-6
If you count them up, you see there’s 36 possible outcomes, and 11 of them have at least one six in them. That is 11/36 or 0.305555… Let’s check with the number in the calculation we used above. Yep, they agree!
So, the general formula for an event X happening in Y trials is 1 minus the probability of the event X NOT happening after Y trials:
What’s the probability you don’t roll a 6? 5/6
How many times are you rolling the die? You have to multiply 5/6 by itself that many times to get the probability all your rolls do not contain a 6. What does multiplying something by itself X number of Y times translate to? Taking that probability to the power of Y.
So, to figure out what the probability of not getting a dog in your example, we have 999999/1000000. Since we’re doing it 1000000 times, we take the 999999/1000000 probability and multiply it by itself 1000000 times, or (999999/1000000)^1000000. This gives us an answer of about 0.3679. Now, this is the probability we don’t get a dog. So we have to subtract this answer from 1 to get the probability we do get at least one dog, so approximately 0.6321.
Wolfram Alpha is good for doing these large sorts of calculations. Here is the exact calculation and the full answer to however many digits you want of the probability.
Now, as mentioned above, there is a way of estimating it for this particularly type of case using the formula 1-(1/e), but I won’t get into that. That specifically applies to an event with 1/N times of happening in N number of trials. As N approaches infinity, the probability approaches 1-(1/e).
It’s not necessary to take Calculus. This is probability, something they teach in first-year Algebra. I for one learned it a second time in Statistics and a third time in a Political Science undergrad course.
You didn’t learn about the limit definition of e in any of those classes.
And that’s not needed for the question that was asked. The question is like day 1 or day 2 ,if the first class is mostly going over the syllabus, of an intro to probability.
Chessic was responding to ftg’s useful post, which explained how knowing a little about e can help with probability calculations. But you don’t generally learn that aspect until Calculus, because it relies on knowing about how to define e with limits.
Using calculus to explain or solve an elementary probability question is like taking a sledgehammer to crack a nut.
No one is “using calculus” to solve or explain the problem.
A person with knowledge of the calculus would likely understand the relationship between a limit (such as e) and a recursive probability calculation (like raising the same fraction to increasingly larger powers. Thus, offering to approximate the answer by referring to e becomes easy; you don’t need to know the calculus to use it.
For a different perspective on OP’s problem:
If you draw (with replacement) a million times from a million-sized deck, you expect to see the joker an average of 1 time exactly. You’ll see it zero times 36.79% of the time, one time 36.79%, two times 18.39%, three times 6.13% , and so on.
Note that these probabilities are in the approximate ratio {1; 1; 1/2; 1/6; …; 1/k!; …} The ratios become more exact as the deck size increases. This is the Poisson distribution.
Let’s check that the joker sightings do indeed average to 1. The expected sightings are
36.79% * (01 + 11 + 2/2 + 3/3! + 4/4! + 5/5! … ) =
36.79% * (0 + 1 + 1 + 1/2! + 1/3! + 1/4! + … )
And indeed the second factor is a standard definition of e (Euler’s number) and the first factor (approx. 36.79%) is 1/e.
Until recently, using calculus was the only way to solve this question. It’s elementary to say the answer is “1 - (999999/1000000)^1000000”, but there’s no way to calculate the value by hand without calculating the approximation using calculus. I don’t think even a scientific calculator would evaluate it correctly.
Of course advanced math software like Mathematica have arbitrary precision arithmetic and can calculate the value correctly. Nowadays you can just type it in Google and get the correct answer.