Someone Else’s Homework: A Probability Question

My friend’s finished the last of the exams and been happy with the results. And I’m stuck thinking harder about a little thing that came across my Twitter feed last night. So let me share a different problem that we had discussed over the term.

It’s a probability question. Probability’s a great subject. So much of what people actually do involves estimating probabilities and making judgements based on them. In real life, yes, but also for fun. Like a lot of probability questions, this one is abstracted into a puzzle that’s nothing like anything anybody does for fun. But that makes it practical, anyway.

So. You have a bowl with fifteen balls inside. Five of the balls are labelled ‘1’. Five of the balls are labelled ‘2’. Five of the balls are labelled ‘3’. The balls are well-mixed, which is how mathematicians say that all of the balls are equally likely to be drawn out. Three balls are picked out, without being put back in. What’s the probability that the three balls have values which, together, add up to 6?

My friend’s instincts about this were right, knowing what things to calculate. There was part of actually doing one of these calculations that went wrong. And was complicated by my making a dumb mistake in my arithmetic. Fortunately my friend wasn’t shaken by my authority, and we got to what we’re pretty sure is the right answer.


Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

2 thoughts on “Someone Else’s Homework: A Probability Question”

  1. So! It’s the odds of drawing a [1,2,3] or a [2,2,2], as no other combinations yield a total of 6.

    For the first part, we can draw that combination out six ways: 1, 2, 3 or 1, 3, 2 or 2, 1, 3 or 2, 3, 1 or 3, 1, 2 or 3, 2, 1. Each of those should have the same probability: 5/15 * 5/14 * 5/13, thus 125/2730. Multiplied by 6, that’s 750/2730.

    The second part’s the easy one: 5/15 * 4/14 * 3/13, thus 60/2730.

    The odds totalled are then 810 / 2730, which reduces to 27 / 91 . Approximately a 29.67% chance!

    Now, let me rough-check this by simplifying to an approximate. If we approximate each time was a 1/3 chance of getting any number, we’d have 7 winning combinations out of 27 possibilities. That’s 25.93% . Six of the seven winning combinations increase the odds of drawing a ‘winning’ ball over that, due to a ‘losing’ ball’s removal. The seventh decreases the odds due to the removal of a ‘winning’ ball. So we can assume the odds will go up somewhat from 25.93%, which jibes well with 29.67%.

    Now for a followup that’s much more challenging: Let’s assume you have that same set of balls, but you’re drawing numbers out one at a time as long as your total is 5 or less. What are the odds then that your end total will be 6? And before you make the calculations, would you wager even money on a ‘hit’ or a ‘bust’?



    1. So for the problem I set you, yes, the 29.67% chance is just what I worked out. For the 1-2-3 case I noticed a way to figure out the probability that didn’t require thinking out how there are six different ways to draw a one, a two, and a three from this. It goes as so:

      The only way to draw a six on three balls with no repeated digits is if each ball is a different number to the balls drawn before. So your first draw you draw any ball; there’s a 15/15 chance you can do that. For the second draw there’s 10 balls of a different value among the 14 remaining. So the chance that you draw a different one is 10/14. For the third draw there’s 5 balls of a different value from the first two among the 13 balls remaining. So the chance you draw a different one here is 5/13. So the chance of a six by this route is 15/15 * 10/14 * 5/13, about 27.5%.

      Now on your problem … well, I’m thrown off by genre-awareness. It seems obvious that busting would happen; therefore, why ask the question? And with just one route to a perfect six having a chance of over 25 percent, it’s hard not to suspect something’s up. So I have to say I think a hit is more likely, but not as a result of deep qualitative thinking about the problem.

      The exact problem seems like work so I’ll rough out a simpler version. You can’t draw more than five balls without a decision, one way or the other. So there’s at most 3^5, that is 243, different draws. I’ll pretend for the sake of making this a rough calculation that all these draws are equally likely, even though (say) 1-1-1-1-1 is less probable than 1-1-2-2-3.

      If I haven’t missed any there’s six starting sequences that would end with drawing a six. That’s picking 1-1-1-1-2; 1-1-2-2-x, 1-1-1-3-x, 1-2-3-x-x, 2-2-2-x-x-, and 3-3-x-x-x. Here ‘x’ means drawing anything else at all, whether 1, 2, or 3.

      There’s five ways of drawing the 1-1-1-1-2 sequence: that way, 1-1-1-2-1, 1-1-2-1-1, 1-2-1-1-1, or 2-1-1-1-1. There’s five ways to start out with 1-1-2-2, and three ways to end that sequence. There’s four ways to start out with 1-1-1-3, and three ways to end that sequence. Six ways to start out 1-2-3, and nine ways to end that sequence. There’s one way to start out with 2-2-2, and nine ways to end that sequence. One way to start out with 3-3, and twenty-seven ways to end that sequence.

      So if I’m not missing things that’s a total of 5 + 15 + 12 + 54 + 9 + 27 configurations that start out by hitting six exactly. That’s 122 combinations out of 243 which is just barely over half the possible configurations.

      What holds me off from declaring that to be The Answer is that I’m aware what a cheat it is to pretend that all 243 configurations are equally likely. For the sake of a rough calculation it’s okay to pretend they are, but (for example) 1-2-2-1-2 is more likely than is 1-2-2-1-1. And 1-2-2-2-1 is as likely as 1-2-2-1-2, but it’s a bust instead. I’ve been trying to think through a clever way to get to a compelling argument for the answer and it hasn’t hit me, which I blame on my deep inner emotional turmoil.


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