A local radio station is running a contest where they are selecting 107 people to participate in a drawing to win $1mm. In order to win the grand prize, you have to pick 2 matching grand prize certifictes from a pool of 30 moneybags.
In the radio spots promoting the contest, they give the odds of winning as 1:435 with no additional clarifications.
However, if you go to the website, the official rules say that only ONE of the 107 people will be selected to draw from the 30 moneybags, so your true odds are first 1:107 to even be the person allowed to draw for the $1mm, and THEN 1:435 for winning the cash.
Since I never fully grasped how to calculate odds, what are the true odds for winning this contest if you’re one of the 107 callers to get into the pool of contestants?
Is it shady for them to be advertising the odds of winning as 1:435 when your chances are really much, much slimmer?
You can just multiply the probabilities, if I’ve understood everything right.
Technically, 1:435 odds would mean a win probability of 1/436, but it might not really mean that because people can be careless when expressing these things — especially radio marketing sorts of people. (Moreover, if they’re picking one person from a pool of 107, then the odds of being picked are actually 1:106 (the probability is 1/107)).
But all that’s a minor quibble. The probability that a member of the pool will win a prize is (1/436) x (1/107) = 1/46652, assuming my interpretation above.
And yes, I’d agree they’re being shady — sleazy even — by claiming the odds are much better than they really are.
You also didn’t calculate the odds of being one of the 107 selected out of all the people who enter the contest. That reduces the oddsy by quite a bit more.
If winning means to correctly pick two of the bags out of 30, then the probability of winning is 1/435, which really makes the odds 1-to-434 of winning. The actual odds to the radio audience is significantly slimmer. First, you have to be one of the 107, and that could be tough odds already when you factor in the size of the listening audience. It’s not clear how this 107 will be selected; it does not appear to be random, so maybe the chances are not that slim. So let’'s suppose that you have a 10% chance of being in the pool of 107. The selection from the 107 appears to be random, so (with the 10% assumption) the probability of winning is (1/10)(1/107)(1/435) = 0.00000214846, or roughly the same as the probability of tossing a fair coin and, on consecutive tosses, get almost 19 consecutive heads.
I was going to note that, but the OP’s question was specifically, “what are the true odds for winning this contest if you’re one of the 107 callers to get into the pool of contestants?” So the condition of being in the pool has already been met.
The odds of getting to be one of the 107 contestants is basically 107 out of maybe 500,000 people in the listening audience (or out of 8 billion people on the planet, if you want to get technical. ), so it didn’t seem worth considering. The contestant selection process is happening over several weeks, where if you hear a certain cash register sound on the radio, you call in and caller #7 is added to the list of 107 finalists.
My issue with the way the promos are being run is that they explain the above, then say (without being clear that it will only be one of the 107 who gets a chance to draw the 2 out of 30 bags for the $1mm) that you need to get 2 matching bags out of 30 to win the big prize. The disclaimers at the end say ONLY that the odds are 1 in 435, without addressing the odds of getting to be the one person who draws at all.
I’m surprised their lawyers let them get away with it. I guess no one really cares that much…except me, because I’m anal that way.
), so it didn’t seem worth considering. The contestant selection process is happening over several weeks, where if you hear a certain cash register sound on the radio, you call in and caller