Working out overall odds

Hi,

I'm wondering how you calculate the overall chances of winning something with multiple attempts.

For example if a scratch card has 0.5% chance of winning and I bought 30 of them what are my overall chances of finding a winner?

I hope that makes sense and thanks.

Quote: isidi

Hi,

I hope that makes sense and thanks.

It does, kind of.
I have never ever bought a scratch card. so I use will use a slightly different example and then give my thought on the card example

Say a box contains 200 small plastic balls, all same size and weight but all are white in color except one and that is red in color.
The game is you pay $1 to attempt to draw one ball from the box (it has a cover so you can not see inside but can reach into it to remove a ball)
and you win if it is the red ball and are paid $200 for your win that included your $1 you wagered. Draw a white ball, you lose you $1 and return the ball to the box. (sample with replacement)
sounds fair to me and maybe hard to do on just one attempt.

so you have $30 in your pocket and decide to try this 30 times. Maybe you can win 2 times and really have some spending money.

your chances of winning at least one time (p) is equal to
1 - the probability of losing all 30 times. (or 100% - the probability as a percentage of losing all 30 times)

your chance of losing = 199/200 = q
so q^30 = 0.860384 or
q^30*100 = 86.0384% = the chance of losing every time you play.
p = the probability of winning = 1 - 0.860384 = 0.1396158 (about 1 in 7 chance)
the chance of winning in one attempt = 1/200 so we have increased our chances of winning at least one time by playing more times. That has to be case.

Now if you want to know the chances of winning 1 time or exactly 2 times, that is a different question and we can use the binomial probability distribution to answer questions about the number of successes in N trials.
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat


with your scratch card if the chance of winning on each card = 1/200 then my math would also apply.

But say they print 2000 cards and only 10 are winners (still 1 in 200 ratio)
it may be possible that you do not know how many tickets are left and how many are still winners. yuck!
maybe there are no winners left!

or you may know the distribution of the number of cards left and the number of winners left, this seems fair to me,
and now you buy just 1 ticket and it is not a winner.
there are 1999 cards left and 10 are winners. 0.50025% is the chance of winning the next card.
the probability of a win has just slightly increased. maybe buy another!

this is a way different calculation, and has to do with something called the hypergeometric distribution.
http://stattrek.com/online-calculator/hypergeometric.aspx

and it is easy to do using a calculator for this or learn the math to do this yourself
1 - no wins probability = 1 - 0.859434972 = 0.140565028

a slightly better deal than my earlier game I showed.

so if you really want to know about your specific scratch card, you should find out exactly how it wins and loses.
any questions?
Sally

At 0.5% your EXPECTATION is 1 winner for every 200 tickets. With 30 tickets, your EXPECTATION is 30/200 or 0.15 winners. This is not the same as "you have a 15% chance to win". This isn't what you asked, but perhaps this helps you with whatever it is you're trying to do or figure out.

Quote: RS

At 0.5% your EXPECTATION is 1 winner for every 200 tickets. With 30 tickets, your EXPECTATION is 30/200 or 0.15 winners. This is not the same as "you have a 15% chance to win".

yes and a nice point to make too.
0.15 average # of winning tickets sure looks like a 0.15 winning probability
and for very small values of P the expected number of successes over N trials
is almost equal to the probability of at least one success in N trials

how about more cards or tickets
say 300 tickets (the office workers pooled their monies)
so we could expect 300/200 or 1.5 winners on average.
Sweet!
How to spend our winnings!

But (199/200)^300 = the probability of no winners in 300 tickets or cards = 0.222292
and only a 0.7777 (1-0.222292) probability of at least 1 winning card.

Sally

Quote: isidi

For example if a scratch card has 0.5% chance of winning and I bought 30 of them what are my overall chances of finding a winner?

If there is a 0.5% chance of a winning ticket, there is a 99.5% chance of a losing ticket.
With this information, you can calculate the chance of finding all 30 tickets as losing tickets, which is P("30 losses") = 0.995^30 = 86%.

Since you either have 30 losing tickets or you have a winner among your 30 tickets (notice that both results are mutual exclusive), the change of finding a winner (a single winner or more than one) is 1 - 86% = 14%.

Thanks folks, you have been really helpful and informative.