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Probability question

October 19, 2013

Our current PGCSE student passed on an interesting little problem.

You have 20 counters and the numbers 2 to 12.

You may distribute your counters anyway you choose, placing them on the numbers.

Once you’ve decided, you roll two dice and add the totals. If you have any counters on that number, you remove one of them.

Repeat until they’re all gone.

What’s the optimal distribution that means you expect to use the least number of rolls?

 

I tried a similar problem with my year 12 stats students.

You have 10 counters and the numbers 0 to 5.

Same rules again but you roll two dice and find the difference.

 

I had four teams and the winning one put 9 of their counters on 1 and the remaining one on 2.

The team in second place put all of theirs on 1.

The winning team confessed that that was also their plan but since the other team had already done that, they made the single change of putting one on 2.

 

What would you do in each game?

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4 Comments
  1. Ronald Stewart permalink

    My first thought would be to minimise the wait time.
    So, initially I’d have something close to the probability distributions:

    2:0, 3:1, 4:2, 5:2, 6:3, 7:4, 8:3, 9:2, 10:2, 11:1, 12:0

    and in the differences game:
    0:1, 1:2, 2:1, 3:1, 4:0

    Intuitively, there are some advantages to distributing the pieces rather than stacking them – i.e. so that you’re not just waiting for all the 1s.

    As a result, your first place team is very likely to beat the second place team almost always (I think 94.7% of the time).
    The second place team is strategically dominated: good work, first place team.

    Even placing a counter on 4 could help you some of the time – usually you’ll be in trouble, but if a 4 happens to come out then you’ll be in a great position to take victory even against a large number of teams and your strategy surely won’t be dominated by the other teams.
    If I was facing dozens of opponents in a winner-takes-all game, I might think more about a strategy like that.

  2. I think I’d be inclined, without too much thought, to stack on 7 and 1, as these are the most likely, but then after more thought I’d be likely to spread it a bit more.

  3. Stevie D permalink

    A quick bit of modelling in Excel suggests that (in the addition game) the best tactic is to spread your counters out in a flattish pyramid shape, pretty much as Ronald suggests. Whether you spread them right out to 2 and 12 or leave the outer edges empty and put a few more in the middle doesn’t, from an anecdotal impression point of view, produce a hugely significant variation – any of the patterns I used could be cleared after about 30 rolls or still going at 60, but tended to clear at about 40–50. Stacking them all on 7 more often than not left the game still going after 60 rolls.

  4. Stevie D permalink

    For the difference game, the optimal pattern appears to be something along the lines of:
    0:2, 1:3, 2:2, 3:2, 4:1, 5:0
    based on the probability distribution of each of those numbers coming up.

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