so what r the chances of gettn in the 1st 2 rows

24

Comments

  • cbbjr
    cbbjr Posts: 241
    Solat13 wrote:
    Actually they won't. ;) Each show is an independent entity unto its own. Sort of how your odds of spinning one particular number on a roulette wheel are always the same.

    But your odds do very slightly decrease and increase due to the number of seats allocated to each venue through presales.

    But the idea that going to more shows increases your odds is not true.

    But while the odds for each individual show don't change, other than the number of seats allocated to each venue as you mention, doesn't the probability that you will be selected at least once increase? Isn't the probability of two independent events calculated as P(a) + P(b)? So if there is on average, based on number of seats in a row and ticket pairs per venue, a 9% chance you are in the first two rows for any given show, if you go to three shows isn't the probability of sitting in the first two rows once .09+.09+.09 = 27%?

    It has been many years since my last statistics class so I could be completely wrong, but I thought while the odds that you get in the first two rows for any given show are relatively constant, the odds that you do it once increases with the number of shows you attend.

    Either way, this is the best thread I have read in a while.
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  • nfanel
    nfanel Posts: 2,558
    the odds increase.

    example:
    The probability of two mutually exclusive events (events where the outcome of the first does not affect the outcome of the second) occurring two times in a row is calculated by multiplying the probability of the first happening by the probability of the second. Take a coin toss for example. The probability of flipping a coin and getting heads once is 1/2. The probability of flipping a coin the second time and getting heads is also 1/2. Now the probability of getting heads twice in a row is 1/2 x 1/2 or 1/4. This statistical law can be used in the same way to calculate the probability or odds of flipping a coin multiple times.

    Example
    Q. What are the odds of flipping a coin multiple times and getting heads 5 times in a row?

    A. We use the probability of getting heads once which is 1/2 and multiply it 5 times to get the solution.

    1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32 or 3.125%


    so the odds of not ever getting first or second row actually goes down with each show you add since you're multiplying by the odds. in this case you're looking for the odds of not getting first or second row at multiple shows.

    looks like that villanova education actually came in handier. ;)
  • pjsteelerfan
    pjsteelerfan Maryland Posts: 9,905
    Even still, I will take my 10% now chance over my old 0% chance any day.
    ...got a mind full of questions and a teacher in my soul...
  • cbbjr
    cbbjr Posts: 241
    nfanel wrote:
    the odds increase.

    example:
    The probability of two mutually exclusive events (events where the outcome of the first does not affect the outcome of the second) occurring two times in a row is calculated by multiplying the probability of the first happening by the probability of the second. Take a coin toss for example. The probability of flipping a coin and getting heads once is 1/2. The probability of flipping a coin the second time and getting heads is also 1/2. Now the probability of getting heads twice in a row is 1/2 x 1/2 or 1/4. This statistical law can be used in the same way to calculate the probability or odds of flipping a coin multiple times.

    Example
    Q. What are the odds of flipping a coin multiple times and getting heads 5 times in a row?

    A. We use the probability of getting heads once which is 1/2 and multiply it 5 times to get the solution.

    1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32 or 3.125%


    so the odds of not ever getting first or second row actually goes down with each show you add since you're multiplying by the odds. in this case you're looking for the odds of not getting first or second row at multiple shows.

    looks like that villanova education actually came in handier. ;)

    When calculating independent probability, don't you add, not multiply? Don't blame me if I am wrong, I am a product of public schools.
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  • nfanel
    nfanel Posts: 2,558
    cbbjr wrote:
    When calculating independent probability, don't you add, not multiply? Don't blame me if I am wrong, I am a product of public schools.
    nope, you multiply:
    To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.

    Multiplication Rule 1: When two events, A and B, are independent, the probability of both occurring is:
    P(A and B) = P(A) · P(B)

    Experiment 1: A dresser drawer contains one pair of socks of each of the following colors: blue, brown, red, white and black. Each pair is folded together in matching pairs. You reach into the sock drawer and choose a pair of socks without looking. The first pair you pull out is red -the wrong color. You replace this pair and choose another pair. What is the probability that you will choose the red pair of socks twice?
    Probabilities: P(red) = 1/5
    P(red and red) = P(red) · P(red) = 1/5 · 1/5 = 1/25
  • AmentsChick
    AmentsChick Posts: 6,969
    This thread gives me a migraine. What do you expect, though, I'm an SDSU dropout. :(
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  • chromiam
    chromiam Posts: 4,114
    cbbjr wrote:
    When calculating independent probability, don't you add, not multiply? Don't blame me if I am wrong, I am a product of public schools.

    Even if you add, the probability decreases....


    nfanel- Yeah yeah Nova edumacation ;)
    Aments- I dropped out of Penn too :D
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  • cbbjr
    cbbjr Posts: 241
    nfanel wrote:
    nope, you multiply:
    To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.

    Multiplication Rule 1: When two events, A and B, are independent, the probability of both occurring is:
    P(A and B) = P(A) · P(B)

    Experiment 1: A dresser drawer contains one pair of socks of each of the following colors: blue, brown, red, white and black. Each pair is folded together in matching pairs. You reach into the sock drawer and choose a pair of socks without looking. The first pair you pull out is red -the wrong color. You replace this pair and choose another pair. What is the probability that you will choose the red pair of socks twice?
    Probabilities: P(red) = 1/5
    P(red and red) = P(red) · P(red) = 1/5 · 1/5 = 1/25

    But isn't the question what is the probability that you will get in the first two rows ONCE? Your formula works for getting in the first two rows each time.
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  • Solat13
    Solat13 Posts: 6,996
    Here's the thing. The odds will actually be greater for MSG and any other arena shows than they will be for Camden and Mansfield and other amphitheatre.

    The rows in arenas for the most part are the same length as any other row. The rows in amphitheaters start with the smallest rows (Rows 1 and 2) and get wider and wider.

    Unless you know the actual number of seats in each row like we do in MSG, this question is just impossible to answer. In Camden Row 1 is like half the size of Row 3. I sat in Row 2 in Camden for night one of 2006 on Mike's side and there was nothing but stage in front of me. (About 5-10 feet to my right were the three guys who Ed brought on stage that night that were released by the Innocence Project .)
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  • PureandEasy
    PureandEasy Posts: 5,818
    You people are spending WAY too much time thinking about this. Don't get yourself so worked up thinking you might be in the first two rows, then be disappointed when you're not.

    It's all PJ, it's all good.

    Hope to see a lot of you at the Camden shows, especially my blood brother and sister

    you know who you are :D
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  • nfanel
    nfanel Posts: 2,558
    chromiam wrote:
    Even if you add, the probability decreases....


    nfanel- Yeah yeah Nova edumacation ;)
    Aments- I dropped out of Penn too :D
    and actually i dropped out of a statistics masters program.... so maybe i'm wrong. guess i should've stuck it out. :p this is starting to hurt my head, too. :D
  • nfanel
    nfanel Posts: 2,558
    PureandEasy wrote:
    You people are spending WAY too much time thinking about this. Don't get yourself so worked up thinking you might be in the first two rows, then be disappointed when you're not.

    It's all PJ, it's all good.

    Hope to see a lot of you at the Camden shows, especially my blood brother and sister

    you know who you are :D
    yes, we are. and i am done thinking for the day.
    i will see you there! :)
  • whistle
    whistle Posts: 268
    I still think that no matter how many shows you go to your odds will stay the same. The probability of you getting the upgrade increases the more you go. But you need to take into account whether or not 10C keeps track of the 'winners' if they do not then in theory the same ppl could get the upgrade twice in a row. The chances of this are slim... (look at the coin flipping ex. above) but still possible.

    Im only going to 1 show this tour so i have a 10% chance, so the probability would not really matter. But for someone going to 10 shows they 'should' get it once, but since the events are independent only your probability increases
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  • lukin137
    lukin137 Posts: 19
    well if it is truly a ten percent chance..for each show..(and that it is what i was trying to figure out in the first place) then like someone commented earlier..that is much better than my previous 0% so lets all hope 4 the best people..n i cant friggin wait to find out.i'll be like a kid on christmas mornin openin that envelope if i see the bran new bike ...... i wont be able to contain myself.......and if it is clothes..dissappointed.yes.but i'll realize soon that the clothes aint too bad either
  • ECM
    ECM Posts: 1,687
    I give up..... you girls have fun with it
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  • whistle
    whistle Posts: 268
    Evil Closet Monkey wrote:
    I give up..... you girls have fun with it

    hahah didnt you start all this mess??? :p
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  • cbbjr
    cbbjr Posts: 241
    chromiam wrote:
    Even if you add, the probability decreases....

    I wasn't a math major, but how do you figure? If there is a 9% chance and you add .9+.9+.9 it increases. If you go to three shows, the odds that you will be in the first two rows for any given show remain the same, the odds that you will be in the first two rows for one of the shows increase, and the odds that you will be in the first two rows for all shows decrease.
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  • ECM
    ECM Posts: 1,687
    whistle wrote:
    hahah didnt you start all this mess??? :p

    I was kind of enough to make a guestimation for the guy who started the thread....

    I should have used the word "probability" over "odds"....
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  • nfanel
    nfanel Posts: 2,558
    lukin137 wrote:
    well if it is truly a ten percent chance..for each show..(and that it is what i was trying to figure out in the first place) then like someone commented earlier..that is much better than my previous 0% so lets all hope 4 the best people..n i cant friggin wait to find out.i'll be like a kid on christmas mornin openin that envelope if i see the bran new bike ...... i wont be able to contain myself.......and if it is clothes..dissappointed.yes.but i'll realize soon that the clothes aint too bad either
    lol. great analogy.
  • whistle
    whistle Posts: 268
    Evil Closet Monkey wrote:

    I should have used the word "probability" over "odds"....

    Yeah i think that is what has ppl confused... Odds are constant, probability changes in the case...
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