Math update: The Lady or the Lions

ALERT: If you have come directly to the page for this post (say, from a feed reader) you are going to get the whole thing. It includes solutions to the problem. I tried to put this after a break that you would have to click through to get but that only seems to work if you read it on the main page. Click here to go there now if you aren’t sure you want the solution.

Yesterday AnimalGirl came around to tackle this problem with Tigger, which I blogged about a couple of weeks ago. I thought I’d provide some detail of how they worked on it because I know some folks are interested. First go look at the problem. The girls read through the problem and then focused on the map. They weren’t sure where to start but one of them suggested working out what all the possibilities were and going from there. This was probably the best thing to do and they worked out what all the possibilities were pretty quickly and made a decision about where the princess should go. I thought that looked like a reasonable solution but something was bugging me that I couldn’t quite put my finger on. I know that there is a solution provided on the nRICH site, so I went and got that for them. (There is a link at the top of the page I linked above.) Their solution was somewhat different from what the girls had come up with so the 3 of us tried to work out how and why they differed. In the end, they agreed that the other one was probably a better solution (and thought they should change their view of which room the princess should go in).

I pointed out that the difference between the two rooms wasn’t that great and talked a bit about how probability problems often come up with answers that require some judgement. Whichever room you choose, there is still a pretty good chance that the poor peasant is going to get eaten by the lions. This problem didn’t take them very long but they enjoyed it and it did make them think.

For those who want to know what their solution was and how we differed from the published solution, I’ve got that after the break.
The girls looked at the map and found 7 possible paths the peasant could take:

  • Directly to room A.
  • via the room on the right, then directly to A by the left door
  • via the room on the right, then directly to B by the middle door
  • via the room on the right, then directly to B by the right door
  • via the room on the left, then directly to A by the right door
  • via the room on the left, then directly to B by the middle door
  • via the room on the left, then directly to B by the left door

3 of 7 options lead to A and 4 of 7 options lead to B so they decided that the princess should be in room B.

The solution posted with the problem comes to the opposite conclusion, that A has a higher probability of being found by the peasant and thus the princess should be there. It talks about chances in 9. We looked for 2 more routes that we missed and couldn’t find any.

This is the hitch: If you go directly to A on your first move, you won’t take a second move. To get to B, you must take more than one move. If you think of the problem as a two stage decision and compare it to other probability problems with two stages, you know that you multiply the chances at the first stage by the chances at the second stage.

The chance of A at stage one is 1/3

The chance of not-A at stage one is 2/3

If you don’t choose A at stage one, your chance of A at stage 2 is 1/3 (for the second choice only), multiplied by the chance from stage one that got you there (2/3) produces a total chance of getting to A at stage 2 instead of stage 1 of 2/9.

Adding the chance of getting to A at stage 1 (1/3) to the chance of getting there at stage 2 (2/9) produces a total chance of getting to A at either stage of 5/9.

If you don’t choose A at stage one, your chance of B at stage 2 is 2/3, producing a total chance of getting to B at stage 2 of 4/9. You can’t get to B at stage 1 so you don’t have to add anything.

So even though you can’t see 9 total paths, the probabilities are out of 9 anyway because of the multiplier effect of needing a particular choice in the first round to open up (or close down) choices in the second round.

I found the diagram included with the published solution a bit confusing precisely because there are NOT 9 valid choices at the outset but the chance of getting to A is greater because you can do it with only one choice rather than a combination of two.

When we were working out whether to add or multiply we used examples of rolling a die twice in succession to think through the logic. I’m pretty sure the girls understood what we did and why. I suspect they’ll need more problems to really get comfortable with it, though. As might I πŸ™‚ But I was impressed that Tigger is already comfortable with the idea that in this kind of problem you need to have some sense of the total number of possibilities.

As you can see, the peasant doesn’t have a great chance either way. I pointed out to the girls that 4/9 and 5/9 were both pretty close to 1/2. He’s basically not much better off than with a coin toss. Poor lad.

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6 thoughts on “Math update: The Lady or the Lions

  1. I don’t know if you know, JoVE, but the problem is based on the old Frank Stockton short story, “The Lady or the Tiger”. Tigger might have fun reading it, between her name and the math problem and all :). I have it in a short story collection book, but will see if I can find an online link for it, since it’s from the late 1800s.

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  2. I think the 7 vs 9 issue is related to the most common misconception about probability that if you have n outcomes, your probability of each outcome should be something/n.

    In other words, we tend to think that the denominator of the probability should match the total number of outcomes, but this is ONLY the case when each outcome is equally likely to happen, something that happens in basic textbook probability questions much more often than real life! πŸ˜‰

    So yes, if you roll a die, the probability of each event is 1/6, and we’re taught that this is because there are 6 “outcomes” and there is 1 “way” to roll each individual outcome.

    But, tomorrow, it could rain or not. Is the probability of it raining tomorrow 1/2?

    Clearly not, but most people still have to “think about it” for a split second because we naturally want to say that 2 possible outcomes means probabilities with denominators of 2.

    Since you mentioned the visual being confusing, here’s how a math teacher would probably represent this question visually, for better or worse! πŸ˜› (hope I can get the idea across in text)

    —–LEFT—–A—–RIGHT
    ——1/3—-1/3—–1/3
    —A–B–B———A–B–B
    -1/3-1/3-1/3—–1/3-1/3-1/3

    (Note: my “paths” aren’t concerned at the outset with whether they’re going to hit A or B. I’m simply being mechanical about recording the possible outcomes of your first decision, then from that room, the possible choices available to you during your second decision.)

    Therefore the probability of each possible path respectively is:
    1/9–1/9–1/9—1/3—1/9–1/9–1/9

    The total of the paths that include A is 1/9 + 1/3 + 1/9 = 5/9
    The total of the paths that include B is 1/3 + 1/3 + 1/3 + 1/3 = 4/9

    In fact, if you want to get even MORE comfortable with the idea that it’s only outcomes of decisions that matter and NOT physical paths, a more “efficient” tree diagram would look like this:

    —–LEFT—–A—–RIGHT
    ——1/3—-1/3—–1/3
    —A—–B———A—–B
    –1/3—2/3——-1/3—2/3

    (This makes it look like there are only actually 5 possible paths, instead of the 7 that there are, and yet, it’s considered a “better” way to solve the problem!)

    So in this case the probabilities of each branch would be:

    1/9—2/9—-1/3—1/9—2/9

    So, you can see that our tree diagram doesn’t need to represent that there are 2 ways to get to B via different paths… that information is encoded when you note the probability of moving to B from one of the initial rooms.

    In fact, while there are 7 paths the guy could take, are there seven “outcomes” to the maze or really only 5: straight to A, A from left, A from right, B from left, B from right?

    … or, are there only 2 outcomes: end at A end at B?

    You can see how thinking in terms of countable outcomes/paths etc. can easily mess us up, and neither 7 nor 5 nor 2 instinctively leads us to probabilities of 9ths.

    Hope the kids have more fun with this. There are so many ways to think about these questions! Thanks for sharing!

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  3. I like how Sarah explains it and I think that is where we got to, though I might have written it out differently. I guess this is another situation where the standard textbook approach actually simplifies too much. There are a lot of coins and dice in the math books I’ve seen πŸ™‚

    I need to think about how we move forward with this and find more problems like this one where the chances and the possible outcomes are different.

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