# Calling probability gurus - help me figure odds on this radio station contest

A local radio station is announcing a contest with a grand prize of \$25,000. It’s a multistage contest, with weekly winners qualifying for a chance at the big bucks.

At the end of the summer, the qualifiers gather on a certain date at a certain local park. It’s a must-be-present-to-win deal. All the qualifiers who are there get their names put in a drawing. One name is drawn.

Here comes the tricky part. The “winner” gets his or her shot at a box with 40 envelopes.

Five of the envelopes are marked \$25,000.

Seventeen of the envelopes have a merchandise prize worth approx \$900.

Eighteen of the envelopes have another merchandise prize worth approx \$750.

Winner gets to draw envelopes until he or she gets five that match. They win that prize.

If it’s the \$25,000, they get their check after “verification,” which takes 2-3 weeks.

If one of the two merchandise prizes shows up first , another name is drawn from those present, and that person gets the other merchandise prize.

So only the first person chosen has a chance at the 25 grand.

It seems as if the fewest envelopes that person could pick would be five (the first five were all the same), and the most a person could pick would be 13 (four of one kind, four of another kind and five of the third kind).

And that’s where my probability skills fall off the table.

I suspect that the odds are very long indeed, because the payoff is being handled through a third-party company, perhaps akin to those which sell a package offering a cash prize or a new car to someone who hits a hole-in-one at a charity golf tournament.

What’s the Straight Dope on this contest?

(5 / 30 ) * ( 4 / 29) * ( 3 / 28 ) * ( 2 / 27 ) * ( 1 / 26 )

I believe that’s your odds of winning the \$25K.

Oops … I have had a long day. The first denominator should be 40, then 39, etc.

I’m afraid my skills are also pretty limited, but I think I was able to calculate a couple of admittedly highly improbable scenarios.

The first is of drawing a merchandise prize w/o any \$25,000 picks, and the merch picks being evenly distributed across those two categories, so that you end up having to draw the maximum nine envelopes for this “merch-only” scenario. This is, if I reckon roughly:

35/40 x 34/39 x 33/38 x 32/37 x 31/36 x 30/35 x 29/34 x 28/33 x 27/32, which works out to a deceptively small-seeming odds of 29.51089% of drawing the booby prize in this way, without ever seeing a single big-prize envelope. The computations are fairly precise (to four decimals throughout), but the reasoning is looser, because I didn’t bother to differentiate more precisely between the 17-envelope (call them “A”) and 18-envelope ("B) categories. In the real world, even if your first nine envelopes were all mechandise picks, it’s unlikely that it would be in the order of A-B-A-B-A-B…

Conversely, the odds of drawing the big prize straight out, without seeing a merchandise envelope, is 5/40 x 4/39 x 3/38 x 2/37 x 1/36, or an infinitisemal .00015% chance.

The overwhelming odds is that your envelope draws would fall somewhere in between, with more randomly-distributed picks. The tricky part comes in trying to calculate those more complicated permutations, and I’m not going to attempt it. But these two extremes give you a very rough idea of the general odds involved.

I suppose the question of whether the contest is worth playing depends on the driving distance to the park site, the value of your time, the cost of a gallon of gas, the tax bite afterwards should you win something (odds determined by how many weeks the contest runs, as well as the factors determining your odds of being a given week’s winner), and even the cost of entering the contest (calls or postcards or whatever).

It may even work out that your odds of winning the big prize are no greater as the big-draw winner than the odds of going home with a merchandise prize as the optional lucky weekly winner, but figuring that out would involve still more work!

You may well enjoy better returns on a higher-odds but low-yield state lottery ticket. Or a throw of the roulette wheel in a casino…

This is the sort of problem where the hypergeometric distribution is what you want. In that case, the answer is [sub]5[/sub]C[sub]5[/sub][sub]35[/sub]C[sub]0[/sub]/[sub]40[/sub]C[sub]5[/sub].

That answer gives you the same value as ccwaterback’s corrected product, so it’s looking good.

But that’s the odds of choosing the five prize envelopes on the first five picks–as I understand it, you could pick 2 of them, and four each of the others, and then 3, and still win the grand prize.

So, the odds are better than that. Probably not good, but better than that.

AH HA!!! Yes, this is true. My assumption was that as soon as you picked a “non-\$25K” envelope you were out of the competition.

So, the odds would be a quite a bit more complex to calculate.

Not doing the calculations, but here is one way to look at it: 5/40 or 1/8 of your population is “good” and you need to choose that unique 1/8 of the population before you choose 1/8 of the population consisting of matching “bad” members. And the “bad” members account for 7/8 of your population.

Some code:

``````
public class TestDrawingChance {
public static void main(String[] args) {
int p25000 = 0;
int p900 = 0;
int p750 = 0;
int numTrials = 10000000;
for (int i=0; i<numTrials; i++) {
int env750 = 18;
int env900 = 17;
int env25000 = 5;
while (true) {
int which = (int)Math.floor((env750+env900+env25000)*Math.random());
if (which < env25000)
env25000--;
else if (which < env25000+env900)
env900--;
else
env750--;
if (env25000 == 0) {
p25000++;
break;
} else if (env900 == 12) {
p900++;
break;
} else if (env750 == 13) {
p750++;
break;
}
}
}
System.out.println("After "+numTrials+" trials:");
System.out.println("25000: "+p25000);
System.out.println("900:   "+p900);
System.out.println("750:   "+p750);
}
}

``````

Gives results:
After 10000000 trials:
25000: 11690
900: 4590918
750: 5397392

So, a bit better than 0.1%, it looks like.

The odds of choosing 5 GP envelopes, 4 P1s, and 4 P2s, in the first 13 choices is C(5,5)*C(17,4)*C(18,4)/C(40,13), which usoft xcel says is .000605, or 1 out of 1652. So, the odds have to be better than that, of getting the grand prize.

Wow! Just as I suspected, you are all amazing.

Thanks to all.

Hometownboy (who won’t be bothering with this particular contest)

Man, they make it particularly difficult, don’t they? Aren’t they legally required to publish the odds? The huge difficulty in claculating them is that you don’t have a fixed number of drawings, but rather the total number of picks depends on what get picked. Now I am going to have to calculate this, thanks a lot.

I think the overall probability of winning is given by sum(sum(C(5, 5)*C(17, i)*C(18, j)/C(40, i + j + 5), 0 < i < 4), 0 < j < 4). Maple gives that as 431725/166847208, or about .0026.

No, that’s too high. I didn’t debug Sofis’s code, but it seems OK.

The problem with that, is it includes strings where you choose the five GP envelopes first, and then all possible combinations of other envelopes after. Try using C(5,4) (and C(40, i + j + 4) ) in your calculation and then multiply by 1/(40-4-i-j) to choose the fifth GP envelope last.