An A&Me quiz: What are the odds?

Posted Dec 28, 2012 at 9:49 am in Threads > Entertainment

I love the end of year contests. There’s lots of activity on the site, the prizes are awesome and, even on a day when there are over 1000 entries, I feel like I have a reasonable chance to win. Even though I didn’t qualify for day 3, I really like the way the grand prize was handled. I respect the reward for loyalty.

So here’s my quiz: Assume there were 1500 entries on day 1; 300 on day 2; and 25 on day 3. One contestant will be chosen as a winner on each day, and each of those contestants will have a 1 in 3 chance of claiming the prize. If you qualified to enter all 3 days, what is the probability you will win? (Assume, also, that 3 different winners were drawn on the 3 days of the contest.)

I know – you hated your statistics class. But sometimes it can have practical applications.

  • decker

    When none of the prices had been given out, I calculated 1:40 chance but this was before the 3 day grand prize give away and the announced winners for the first couple days. I also had to assume that the entrant would be able to enter all contests which was not the case for me. After winners are announced the probability decreases. There is a formula for calculating probability based on multiple events (in our case chances to win).

  • Puzzled by Choice

    Math is my Kryptonite, so I’ll just hang out for the answer or a near guess by those smarter than me.

  • raichleb

    I’d like to wait at least another day before posting the probabilities, just to see if the thread generates any more interest. One of the things I found interesting is the advantage that accrues with being able to enter all 3 days. Specifically, a person who can enter all 3 days is 66 times more likely to win than a person who enters only on day one, and 11 times more likely to win than a person who enters on days 1 & 2.

    BTW, I’m guessing that a regular poster to this site, like yourself, isn’t half as bad as math as you think you are.

  • amIT29

    i think the subject would be probability and statistics per say… assuming fair play

    I think the odds goes like this
    day 1 odds = 1/1500
    day 2 odds = 1/299 (omit 1 person who already won day 1)
    day 3 odds = 1/23 (omit 1st 2 days winner)

    so overall probability to be picked as winner on any of the 3 days is sum of every day odds = is 0.0475 (1/21 approx)

    hence final winning chance will be 1/21 * 1/3 = 1/63 (approximately) is the odds of bagging the prize.. (correct me if iam wrong i havent touched probabilistic theories in 10 years)

  • raichleb

    Exactly! (Either that or we’re both wrong.) And I gotta add – that’s pretty impressive for someone who hasn’t looked at probability for 10 years.

    I figured it just a little differently, because the rules didn’t say a person could not be chosen on multiple days. But the probability of that occurring is extremely small, so it only changes the total probability by about 3/10,000.

    Here are the approximate probabilities for each day, assuming, as amlT29 did, a unique winner for each day:

    If you only entered day one you have about a 0.02% chance of winning the grand prize.
    If you entered day one and day two, your chance of winning increased to about 0.13%
    And if you entered all 3 days, your chance increased to about 1.58%.

    • amIT29

      thanks mate glad i didn’t goofed it up :P , but hey “because the rules didn’t say a person could not be chosen on multiple days” i thought line in the thread starter
      ” (Assume, also, that 3 different winners were drawn on the 3 days of the contest.)” meant exactly that :)

      • raichleb

        You’re right. That’s pretty bad when I lay down my own rules and then ignore them.