Help with simple probabilities

DeliciousTacosDeliciousTacos Registered User regular
edited May 2010 in Help / Advice Forum
Okay, so, I took a few statistics classes back in college but I can never remember a lot of the basics of probabilities. Here's my question:

Say there are two raffles, each of which already have 100 entries with one entry being drawn at random. If I can put two entries down, do I have a better chance of winning anything at all if I put both entries in one raffle or one entry in each raffle? (So it's either a 2/102 chance or two 1/101 chances). How do you calculate that? How about with me putting in more entries, say 20?

DeliciousTacos on

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  • DaenrisDaenris Registered User regular
    edited April 2010
    Since it's separate, independent raffles, the probability of winning in raffle A or B is just the sum of the probabilities of winning in raffle A and the probability of winning in raffle B. In this case putting a ticket in each will result in a (1/101) + (1/101) probability of winning something (.0198). This is very close to your 2/102 odds (.0196), just slightly in favor of the 1 ticket in each separate raffle. If you go to 20 tickets, putting all 20 in one yields a .167 probability, while putting 10 in each yields .182.

    Daenris on
  • KevdogKevdog Registered User regular
    edited May 2010
    Daenris wrote: »
    Since it's separate, independent raffles, the probability of winning in raffle A or B is just the sum of the probabilities of winning in raffle A and the probability of winning in raffle B. In this case putting a ticket in each will result in a (1/101) + (1/101) probability of winning something (.0198).

    Probabilities don't work that way. Suppose there were 101 raffles. By that logic, putting 1 ticket into each of them would give you probability 101 * (1/101) = 101/101 = 1 of winning. In other words, you'd be guaranteed to win, which clearly isn't true.

    OP, the two rules you're looking for are that, for independent events A and B:
    1. Probability(A and B) = Probability(A) * Probability(B)
    2. Probability(A or B) = 1 - Probability(not A and not B)

    So in this case, I would look at the chance of winning neither raffle, then subtract that from 1, and that's the chance of winning at least one raffle.

    Putting 1 ticket in each:
    Chance of winning neither = 100/101 * 100/101 = ~0.9803
    Chance of winning at least one = 1 - 0.9803 = 0.0197

    Putting 2 tickets in one:
    Chance of winning neither = 100/102 * 100/100 = ~0.9804
    Chance of winning at least one = 1 - 0.9804 = 0.0198

    Kevdog on
  • GdiguyGdiguy San Diego, CARegistered User regular
    edited May 2010
    Kevdog wrote: »
    Daenris wrote: »
    Since it's separate, independent raffles, the probability of winning in raffle A or B is just the sum of the probabilities of winning in raffle A and the probability of winning in raffle B. In this case putting a ticket in each will result in a (1/101) + (1/101) probability of winning something (.0198).

    Probabilities don't work that way. Suppose there were 101 raffles. By that logic, putting 1 ticket into each of them would give you probability 101 * (1/101) = 101/101 = 1 of winning. In other words, you'd be guaranteed to win, which clearly isn't true.

    Yeah, you're confusing Expectation (which is what you calculate - if you put one ticket in each, you EXPECT to win 1/101 + 1/101 times) with probability (as Kevdog accurately describes)

    The difference, as his example shows, is that the probability of winning at least one raffle can never be greater than 1 - even if you entered 1 million raffles, each with a 1/100 chance of winning, your probability of winning at least one raffle will max out at 1. Your expected number of wins, however, is 1 million * 1/100 = 10000. (As this is a binomial trial, expectation = n * p for n independent trials with p probability of success in each)

    Gdiguy on
  • DaenrisDaenris Registered User regular
    edited May 2010
    Gdiguy wrote: »
    Kevdog wrote: »
    Daenris wrote: »
    Since it's separate, independent raffles, the probability of winning in raffle A or B is just the sum of the probabilities of winning in raffle A and the probability of winning in raffle B. In this case putting a ticket in each will result in a (1/101) + (1/101) probability of winning something (.0198).

    Probabilities don't work that way. Suppose there were 101 raffles. By that logic, putting 1 ticket into each of them would give you probability 101 * (1/101) = 101/101 = 1 of winning. In other words, you'd be guaranteed to win, which clearly isn't true.

    Yeah, you're confusing Expectation (which is what you calculate - if you put one ticket in each, you EXPECT to win 1/101 + 1/101 times) with probability (as Kevdog accurately describes)

    The difference, as his example shows, is that the probability of winning at least one raffle can never be greater than 1 - even if you entered 1 million raffles, each with a 1/100 chance of winning, your probability of winning at least one raffle will max out at 1. Your expected number of wins, however, is 1 million * 1/100 = 10000. (As this is a binomial trial, expectation = n * p for n independent trials with p probability of success in each)

    Actually I wasn't confusing it with expectation... I just forgot to include a key piece of information in my haste. Since the events are not mutually exclusive (so you can win in both raffles) the probability of winning either is:
    P(A or B) = P(A) + P(B) - P(A and B), or
    1/101 + 1/101 - (1/101 * 1/101)

    Which works out the same as what Kevdog used, just framed differently.

    In the case of mutually exclusive events (B can't happen if A happens) this simplifies to just P(A) + P(B), which unfortunately is what I initially used.

    Daenris on
  • WylderneedshelpWylderneedshelp Registered User
    edited May 2010
    One ticket in each raffle

    Probability of NOT winning raffle A = 99/100 = 0.99
    Probability of NOT winning raffle B = 99/100 = 0.99

    Probability of NOT winning BOTH A & B = 0.99 * 0.99 = 0.9801

    Probability of NOT (NOT winning BOTH A & B) = Probability of winning A or B or Both = 0.0199

    = 1.99%


    Two tickets in the same raffle

    Probability of BOTH tickets NOT winning = 99/100 * 98/99 = 98/100 = 0.98

    Probability of NOT (BOTH tickets NOT winning) = 0.02

    = 2%


    You are marginally better with both tickets in the same raffle.



    Laymans Explanation

    You buy one ticket. Since this ticket is being drawn from either raffle (of 100 tickets) it doesn't matter which raffle it's from, the probability of it winning is 1%. There are two possibilities, it wins or it doesn't. If it wins, the argument is moot, so lets investigate the possibility if your first ticket doesn't win.


    If your first ticket doesn't win, you can buy a ticket now from either raffle.

    The raffle you just bought a ticket from has 99 tickets remaining. The "other" raffle has 100 tickets remaining.

    By definition, finding a winning ticket out of 99 possible tickets is more likely than finding a winning ticket out of 100 possible tickets. Therefore, you buy both tickets in the same raffle.



    Extreme example proving the theory

    Buy 100 tickets.

    If they are all in the same raffle, you just won. Guaranteed.

    If half is from each raffle, you could still lose.

    Chances of wining AT LEAST ONCE is better if all tickets are from the same raffle.



    So, is it worth it??

    There is a significantly small change (0.01%) that buying the tickets in different raffles will win TWICE. You are forfeiting this in exchange for a greater chance of winning a minimum of one prize, by buying them in the same raffle. On average, if you did this a LOT of times, you would end up with the same average number of prizes per ticket (0.01) regardless of which lotteries they are in. What action you take will depend entirely on whether you only care about getting AT LEAST ONE prize, or best potential maximum number of prizes.

    Wylderneedshelp on
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