🔒 Locked
Probability question.
Advertisement
Original Post
Alright. So suppose Sprite tells me that I have a 1 in 3 chance of getting a free iTunes download if I buy a bottle of their soda. If I buy 150 bottles, what is the probability that I will get 50 free downloads? 100%? I know it's not 100% in practice, but it that even a mathematically correct statement?
Quote:
Original post by uncle_rico
Alright.
So suppose Sprite tells me that I have a 1 in 3 chance of getting a free iTunes download if I buy a bottle of their soda.
If I buy 150 bottles, what is the probability that I will get 50 free downloads? 100%?
No, the probability is not 100% -- if it were 100%, then you would be guaranteed 50 bottles if you bought 150 bottles. Clearly that is not the case. The expected value of the number of free downloads you'll get from 150 bottles is 50, however.
So what's the real probability? Well, this is a series of Bernoulli trials. Each time you open a bottle, it's either a success (you win) or a failure (you don't), and each bottle is independent of all the others (opening one bottle doesn't affect the probability you'll succeed after opening a different bottle). To find the number of successes in n trials, we use a binomial distribution. We seek the probability P of obtaining n successes out of N trials, where a success occurs with probability p and a failure occurs with probability q = 1 - p. This is given by:
C(N,n) * [pn] * [qN - n]
--> [N! / (n! * (N - n)!)] * [pn] * [qN - n]
--> [150! / (50! * 100!)] * [(1/3)50] * [(2/3)100]
--> [(1 * ... * 150) / ((1 * ... * 50) * (1 * ... * 100))] * [(1/3)50] * [(2/3)100]
--> [(101 * ... * 150) / (1 * ... * 50)] * [(1/3)50] * [(2/3)100]
--> [(101 * ... * 150) / 50!] * [(1/3)50] * [(2/3)100]
If someone else with a calculator wants to compute the exact numerical value, be my guest. =)
edit: Never mind. Thanks, Google! The answer is about 6.8%. Thus, the probability that you'll get exactly 50 wins is 6.8%.
[Edited by - kSquared on March 16, 2005 5:45:40 PM]
I want it on the record that I've drank damn near 150 Pepsi products, and only won once. Somebody out there must be making bank for these odds to play out [frown]. Whoever it is would probably get the music off a P2P network anyhow. Lucky bastard.
CM
CM
It occurred to me that the value of "exactly 50 successes" is probably not what you really wanted. After all, you want to win 50 downloads -- you would surely be just as happy getting 51 downloads or more, no? In this case, we use the cumulative distribution function (cdf). If we let SUM{i = k, n, fi} represent a standard summation, then the cdf for the binomial distribution is given by
SUM{i = 0, k, C(N,i) * [pi] * [qN - i]}
where N is the number of trials. This expression represents the probability of obtaining k or fewer successes in N trials, so it will be a number between 0 and 1. We're actually interested in probability that you get *more* than k successes, however. Since the events E1 = "k or fewer successes in N trials" and E2 = "at least k + 1 successes in N trials" are mutually exclusive, you can evaluate the probability of one using the other: 1 - P(E1) = P(E2). As we showed above, there is a closed-form expression for the cdf of a binomial distribution, but it would require computing many terms by hand (fifty terms, in this case).
So we'll let the computer do the dirty work. You can use this page to see for yourself that the probability of getting 49 successes or fewer is approximately 0.4697. Subtracting this from 1 gives the value of approximately 0.5303. Thus, given a random sample of 150 Sprite bottles, your chance of finding 50 or more winners is 53.03%.
Happy drinking! ;)
SUM{i = 0, k, C(N,i) * [pi] * [qN - i]}
where N is the number of trials. This expression represents the probability of obtaining k or fewer successes in N trials, so it will be a number between 0 and 1. We're actually interested in probability that you get *more* than k successes, however. Since the events E1 = "k or fewer successes in N trials" and E2 = "at least k + 1 successes in N trials" are mutually exclusive, you can evaluate the probability of one using the other: 1 - P(E1) = P(E2). As we showed above, there is a closed-form expression for the cdf of a binomial distribution, but it would require computing many terms by hand (fifty terms, in this case).
So we'll let the computer do the dirty work. You can use this page to see for yourself that the probability of getting 49 successes or fewer is approximately 0.4697. Subtracting this from 1 gives the value of approximately 0.5303. Thus, given a random sample of 150 Sprite bottles, your chance of finding 50 or more winners is 53.03%.
Happy drinking! ;)
Wow, Ksquared, I never knew someone could care so much about something that matters so little.
Quote:
Original post by lethalhamster
Wow, Ksquared, I never knew someone could care so much about something that matters so little.
The state of Nevada's entire economy is based on the assumption that that knowledge is worth quite a bit. That, and the assumption that nobody bothers to care about it.
CM
Quote:
Original post by Conner McCloud Quote:
Original post by lethalhamster
Wow, Ksquared, I never knew someone could care so much about something that matters so little.
The state of Nevada's entire economy is based on the assumption that that knowledge is worth quite a bit. That, and the assumption that nobody bothers to care about it.
CM
Truer words were never spoken. While we're on the subject, the best bet at most casinos (without card counting) is at the craps table -- take full odds behind the pass line. The house edge is less than four-tenths of a percent.
Quote:
Original post by kSquared
Truer words were never spoken. While we're on the subject, the best bet at most casinos (without card counting) is at the craps table. The house edge is less than four-tenths of a percent.
I knew that,
Quote:
-- take full odds behind the pass line
but I don't know what this means.
Quote:
Original post by Zahlman Quote:
Original post by kSquared
Truer words were never spoken. While we're on the subject, the best bet at most casinos (without card counting) is at the craps table. The house edge is less than four-tenths of a percent.
I knew that,Quote:
-- take full odds behind the pass line
but I don't know what this means.
When you start each "round" of craps, the shooter (person rolling the dice) makes an initial roll called the come-out roll. If the come-out roll is 4, 5, 6, 8, 9, or 10, that number becomes the "point" number. When you bet on the pass line, you are betting that the point number will be rolled before a 7. The house edge on a pass line bet is 1.4%.
But once a point is made on the come-out roll, most casinos allow pass line bettors to "take odds" -- you get to place a multiple of your original bet if you so choose. This additional bet pays the true statistical odds, 2-to-1 if 4 or 10 is the point, 3-to-2 if 5 or 9 is the point, and 6-to-5 if 6 or 8 is the point. Because it pays true odds, the house has no edge on behind-the-line odds bets. Thus, by taking the maximum odds (which vary by casino) you can lower the house edge for any given bet down substantially.
In the long run, you still come out worse though. I've never gambled any money in my life. If I ever do, I'll go in with a substantial sum of (disposable) money, wage it all on a single roll, and walk away with whatever the result is.
Quote:
Original post by Nemesis2k2
In the long run, you still come out worse though. I've never gambled any money in my life. If I ever do, I'll go in with a substantial sum of (disposable) money, wage it all on a single roll, and walk away with whatever the result is.
If you're smart, it's not gambling, it's just another way of spending your money. Every once in a while, I can go to a casino with some friends, spend four or five hours, and only "lose" twenty dollars. It's roughly the same as watching a movie, but you don't have to pay for beverages. And sometimes, you get lucky and don't feel bad buying yourself something nice, because you've already budgeted that money for entertainment.
CM
Quote:
Original post by Nemesis2k2
In the long run, you still come out worse though. I've never gambled any money in my life. If I ever do, I'll go in with a substantial sum of (disposable) money, wage it all on a single roll, and walk away with whatever the result is.
If the excitement, comps, and free drinks are worth less to you than the house advantage on your disposable income, you shouldn't play. But otherwise, there's really no reason not to at least give it a shot. The $30 you're spending at the movies for popcorn and seats is better spent at a casino, where you at least have a chance of winning additional money (unless you're really into watching Gigli over and over). Play to be entertained, not to win; and set a daily limit for yourself. You can't lose with that strategy.
Quote:
Original post by kSquared
The $30 you're spending at the movies for popcorn and seats is better spent at a casino, where you at least have a chance of winning additional money (unless you're really into watching Gigli over and over).
The money is better spent at a casino...unless you are watching Gigli? I'm speechless.
CM
Quote:
Original post by Conner McCloud Quote:
Original post by kSquared
The $30 you're spending at the movies for popcorn and seats is better spent at a casino, where you at least have a chance of winning additional money (unless you're really into watching Gigli over and over).
The money is better spent at a casino...unless you are watching Gigli? I'm speechless.
CM
De gustibus non est disputandum, as they say. Gigli didn't have zero sales, so that means at least one person found it interesting.
Quote:
Original post by kSquared Quote:
Original post by Conner McCloud Quote:
Original post by kSquared
The $30 you're spending at the movies for popcorn and seats is better spent at a casino, where you at least have a chance of winning additional money (unless you're really into watching Gigli over and over).
The money is better spent at a casino...unless you are watching Gigli? I'm speechless.
CM
De gustibus non est disputandum, as they say. Gigli didn't have zero sales, so that means at least one person found it interesting.
They were all tricked into thinking it would be good. The man tricked them out of their money. The man!
Topic Locked
This topic has been locked by a moderator. New replies are not allowed.
Sign in to reply to this topic.
Advertisement