I was just stumped on a lottery themed question that seems like it should be easy...:

In RI they do second chance drawings where you send in spent scratchers and they draw from all of the ones sent in at random for small prizes. So conceptualizing as a ball problem:

1,000,000 balls in a jar (total tickets). 1,679 are red (mine), 998,321 are blue (not mine) If 1,112 balls are drawn without replacement, what is the overall probability that one red ball will be selected?

What formula/setup would you use to solve this?

  • 1
    $\begingroup$ Exactly one red ball, or at least one red ball? $\endgroup$ – one_observation Jan 11 '16 at 18:24
  • $\begingroup$ At least one red ball $\endgroup$ – Mike M Jan 11 '16 at 18:53

Edit: Sorry, dumb mistake! The answer can be gotten exactly because the probability of getting zero reds is analytically tractable, even with replacement. My bad. I'll change that. In the meantime though, I think the approximation is still decent.

To flesh out MikeP's answer, let's imagine that you mean at least one red ball, because I assume that's equivalent to "I get some money," which I assume is what you really want to know.

IN THE GENERAL CASE, this, and maybe this is where you're getting stuck, is just really hard to know exactly. That's because (again in the general case) to get a precise answer, you need to work through a tree of conditional probabilities that's 1,112 layers deep. What if this happens on the first draw -- okay well given that, what if this happens on the next draw ... there may be efficient approximations (I googled a little bit, but no luck) but I don't know them, and calculating the whole tree (in a brute force way) just isn't tractable. If it were, you would have the probability for every possible drawing sequence, and then add together the probabilities of the cases you wanted, i.e. those with at least one ball. And again, I may be missing something more sophisticated.

If we're approximating anyway though, here's a way to force the problem to be simple (which is a fundamental and trainable math skill, not just a way to cheat). Your numbers are so different in magnitude that I would just assume they're drawn WITH replacement. Most of the time, you're going to draw a blue ball (999 times out of a thousand), and that unreplaced blue ball isn't going to make a dent on the remaining ones. And even with a thousand draws, in expectation, you're only going to draw one red ball (because you get it ~1 draw in a thousand), so the population of red balls isn't going to be affected much either. In which case you've just got

$ P(N_{red} \geq 1) = 1 - P(N_{red} = 0) = 1 - P(x = blue)^{draws} = 1 - 0.998321^{1112} = 85\%$

which again heuristically, doesn't seem too far off -- ~1000 shots at drawing a ~1/1000 ball, you expect to get about 1, i.e. you're pretty likely to get at least one, but aren't likely to get many more, and there's still a fair chance of getting none.

However here, you're in luck, because now that you see you only need to calculate the probability of zero reds, that's easy again - even without replacement.


The odds for exactly one ball are the odds of 1,111 non-red times the odds of a red times the number of ways that can happen.

So, for the 1111 non-red, you multiply: $$\frac{998,321}{1,000,000} * \frac{998,320}{999,999} * \frac{998,319}{999,998} * ... * \frac{998,321-1,110}{1,000,000-1,100}$$ for the one red ball: $$\frac{1,679}{1,000,000-1,111}$$ since the red ball doesn't have to be last, you could mix the denominators any which way just so long as you get all the factors of: $$ 1,000,000! / (1,000,000-1,112)!$$ And the number of ways this can happen is 1,112 (1st through last)

  • $\begingroup$ Ahh, I see your comment that it was at least. In this case, I would use Sophologist's approximation. The exact solution would get kind of ugly. I calculated it for exactly 1 (29%) and 2 (27%), but I bet these would converge to a sum of about 85% as he found. $\endgroup$ – MikeP Jan 11 '16 at 19:11
  • $\begingroup$ Oops, made the same mistake - simple for the at least one case: Take step 1 one step further to 998,321-1,111 over 1,000,000-1,111 and subtract that from 1. For at least one you get 0.8458 $\endgroup$ – MikeP Jan 11 '16 at 21:01
  • $\begingroup$ Cool to see with and without replacement turning out similar, thanks for the answer! $\endgroup$ – Mike M Jan 11 '16 at 22:22

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