The Logic of Probability

[Halleck]

This one reminds me of Monty for some reason. I think I get it, but I was hoping you guys could help me out and get some enjoyment from it. I apologize for the theme. My teacher really likes potter.

6 wands on a table. Each of the four wizards pick a wand one after each other. 4 wands will produce a dragon. 2 will produce a toad. When your wand conjures each animal you can't see it, but everyone else can. Wands are removed after being used.

1st wand Dragon
2nd wand (you) unkonwn.
3rd wand Dragon
4th wand Dragon

What is the probability you conjured a toad.
What is the probability you conjured a dragon.

Have fun.

edited for pedantry of respondents. -jr

[rigwarl]

Spoiler:

Out of the 3 remaining possible wands you could have taken, two produce a frog and one produces a dragon. P(frog) = 2/3, P(dragon) = 1/3.

Note that the ordering is a red herring and completely irrelevant.

[splidge]

Spoiler:

Maybe I'm missing something, but I want to say P(toad)=2/3, P(dragon)=1/3.

At the time you picked it, obviously P(dragon)=3/5 and P(toad)=2/5, but since your choice hasn't had an observable effect (to you) I think the subsequent events change the probabilities. For instance, if the 3rd and 4th wands had both produced a frog then P(dragon) would surely now be 1.

This is different from the Monty Hall problem because the 3rd and 4th picks were not influenced by your decision (at least, it doesn't say so in the description). If the 3rd and 4th wizards knew where the dragon wands were and have chosen them deliberately then P(Dragon)=3/5 still - so if this was a Monty Hall like situation where there were some form of reward for picking out a dragon you'd be better off staying where you were than picking one of the two remaining wands.

[curtis95112]

Spoiler:

I'd say that your choice influences the others because it changes the number of wands left. However, it seems the numbers work out in a way that it doesn't matter. Maybe this will change if you change the starting conditions, but I cba to try to find a counterexample or to prove it doesn't matter.
Possibility of you choosing dragon: (3/5)*(2/4)*(1/3) = 6/60 =10%
Possibility of you choosing frog: (2/5)*(3/4)*(2/3) = 12/60 = 20%

But since we're not considering the remaining 70% of cases in which the 3rd and 4th wizards don't both pick dragon.
Possibility of you choosing dragon: 1/3
Frog: 2/3

[Scuttlemutt]

Spoiler:

I think I see the similarities to a Monty Hall problem. Imagine the reverse of the usual scenario; There's 6 doors with 5 cars behind it, and one door is empty. You pick one door, and then he reveals the other doors until it's down to just your picked door and the empty door. Should you switch doors? The problem with 5 empty doors and 1 car suggests that to get a car, your chances are greatest if you do (original pick is 1/6 of being right); But doesn't this imply that the reverse scenario has the reverse solution too (don't switch, original pick is 1/6 of being wrong)?

Now just substitute doors with wands, cars with dragons, and empty doors with toads, and remove the option of switching to just figure out if your original pick was a correct choice after all. Doesn't the situation imply that it is still more likely that you picked a Dragon? (3/5 odds)

Or are there factors I'm not considering? I'd love an explanation of why this train of logic is false.

[splidge]

Spoiler:

[spoiler]I think I see the similarities to a Monty Hall problem. Imagine the reverse of the usual scenario; There's 6 doors with 5 cars behind it, and one door is empty. You pick one door, and then he reveals the other doors until it's down to just your picked door and the empty door. Should you switch doors? The problem with 5 empty doors and 1 car suggests that to get a car, your chances are greatest if you do (original pick is 1/6 of being right); But doesn't this imply that the reverse scenario has the reverse solution too (don't switch, original pick is 1/6 of being wrong)?

Yes, it does. This is all correct.

Now just substitute doors with wands, cars with dragons, and empty doors with toads, and remove the option of switching to just figure out if your original pick was a correct choice after all. Doesn't the situation imply that it is still more likely that you picked a Dragon? (3/5 odds)

It depends on what exactly the 3rd/4th wizards did. In the Monty Hall problem, the host *knows* he is opening a goat door (he knows where the car is and will never open the door with the car). In this scenario, if the 3rd/4th wizards knew where the dragons were and picked them deliberately then you are right P(dragon)=3/5 when you picked your wand and the other wizards picking dragons has not provided new information (they knew they were picking dragon wands).

However, if the other wizards have picked randomly and happened to have picked the dragons, then P(dragon)=1/3. As Curtis points out it doesn't matter which way round you work the numbers.

The original problem spec does not say which one is the case; in the absence of other information I would assume a random choice.

[mfb]

Spoiler:

I assume that all wizards choose at random and do not know the wands.

The intuitive 1/3 for the dragon is correct.

Here another way to show that "wizard x has dragon" does nothing than removing this dragon:
4 wizards pick a wand. The wizards have (6 choose 4)=15 possibilities to choose wands, one gives 4*dragon (and you have one as well), 8 give 3*dragon + 1*frog (where you have the frog in 2 cases), 6 give 2*dragon 2*frog.
From the observation of 3*dragon, we can rule out all but 3 options, in 2 of them you have the frog.

[undecim]

Spoiler:

Since nothing here is changing (unlike Monty Hall), then it's the intuitive 1/3.

The reason for Monty Hall's subtlety is that you learn a little bit more than just "this door doesn't have a prize", because the door is picked based on what's behind each door. Here we have no such thing going on.

[EricH]

Spoiler:

Probability of conjuring a toad = 0
Because the other wands produce frogs, you see...

[jestingrabbit]

Spoiler:

Probability of conjuring a toad = 0
Because the other wands produce frogs, you see...

*groan*

[Yakk]

The usual problem with this kind of logic puzzle is that it doesn't include information required to solve it.

Describing the sequence of events that happen is not sufficient to determine the distribution of probability.

As an example, suppose you know (are 100 certain) that the last wand is a frog wand. And when you picked a wand, you picked the last wand. This fits the described events, yet in this case the probability that you summoned a frog is 100%. Similarly, if you knew the first wand was a dragon wand, and you picked it, then the probability you summoned a dragon is 100% -- and this also fits the described events.

So to solve this problem, we need a description of the algorithm that each of the wizards used to select their wand. Now, we could invent an algorithm -- but that makes this a mind-reading puzzle rather than a logic puzzle. The Monte Hall puzzle has a similar problem.

It is true that entire classes of algorithms that pick wands (or doors) lead to the same result. But there are classes of algorithms that are not restricted by the description that lead to results that don't agree. So short of assigning a probability to each algorithm (either explicitly, or implicitly by picking one and saying that is the one that is used), the question lacks sufficient information to answer.

This is also the reason why you don't place bets with magicians.

[undecim]

The usual problem with this kind of logic puzzle is that it doesn't include information required to solve it.

Describing the sequence of events that happen is not sufficient to determine the distribution of probability.

As an example, suppose you know (are 100 certain) that the last wand is a frog wand. And when you picked a wand, you picked the last wand. This fits the described events, yet in this case the probability that you summoned a frog is 100%. Similarly, if you knew the first wand was a dragon wand, and you picked it, then the probability you summoned a dragon is 100% -- and this also fits the described events.

So to solve this problem, we need a description of the algorithm that each of the wizards used to select their wand. Now, we could invent an algorithm -- but that makes this a mind-reading puzzle rather than a logic puzzle. The Monte Hall puzzle has a similar problem.

It is true that entire classes of algorithms that pick wands (or doors) lead to the same result. But there are classes of algorithms that are not restricted by the description that lead to results that don't agree. So short of assigning a probability to each algorithm (either explicitly, or implicitly by picking one and saying that is the one that is used), the question lacks sufficient information to answer.

This is also the reason why you don't place bets with magicians.

It's generally assumed that these things are chosen uniformly at random, unless stated otherwise. Also... your example with picking the last wand knowing it will conjure a frog is basically saying "If your wand will conjure a frog, then your wand will conjure a frog".

[Yakk]

Yes, I agree, things about these kind of probability problems are often left to be assumed by the reader. As noted by other posters, different things are assumed by different people, leading to different results in this case. And then people yell at each other implicitly over whose assumptions are the right ones (be it with Monte Hall or other kind of problem".

If you qualify the assumptions -- ie, insert "each wizard picks a wand uniformly and randomly from the ones present", which is pretty easy, some major assumptions are removed. Or you could mention it in the answer.

The point is that you cannot, from the observations described, exactly determine the probability you summoned a dragon. Ie, if you where in an equivalent situation, making a bet with real cash at 1:3 or 1:5 odds would be foolish, as the described events aren't sufficient to generate enough knowledge about the odds of the situation. If you disagree with this and think you can build a strong probability model from a description like the above, would you like to pick a pea from under a shell?

[Adam H]

The point is that you cannot, from the observations described, exactly determine the probability you summoned a dragon. Ie, if you where in an equivalent situation, making a bet with real cash at 1:3 or 1:5 odds would be foolish, as the described events aren't sufficient to generate enough knowledge about the odds of the situation. If you disagree with this and think you can build a strong probability model from a description like the above, would you like to pick a pea from under a shell?

Making a bet with real cash would not be foolish at all. It would be foolish if you made the bet with the wizards who were picking up the wands, or anyone who could control the outcome.

If I flipped a coin what would you say the probability is that it was heads? Keep in mind that it MIGHT be a double sided coin. I suppose you could take into account that double-headed coins are more common than double-tailed coins (or something), but 50/50 is PROBABLY the right answer, especially if I didn't have a motive for tricking you.

You don't have to treat logic puzzles as if someone is trying to "get you".

[splidge]

would you like to pick a pea from under a shell?

Sounds good, as long as I get to be the guy with the suspiciously similar accent who has the first go.

[Yakk]

Adam H: No, I treat claims of fact and proof as if I'm trying to "get it". This helps find errors in the chain of reasoning.

Which is why things like "by convention, we will assume", which admits an arbitrary choice, are less problematic than simply assuming it and then thinking that the results are actually facts. Conditional facts that admit it are more true than those that don't.

I think this position is wise, because I see people arguing over logic puzzles like this one and Monte Hall over basically this very problem: they make different assumptions, don't state them clearly, and then get different results. Both positions often end up with answers which "fit" the description of the problem, yet disagree on the answer. If you add in the assumptions, it becomes clear why they disagree -- because the question isn't clear. (Not to say that there aren't also people who answer differently because they are just wrong!)

[Jaywalk3r]

The problem has a lot of extraneous information that seems to be throwing people off.

Spoiler:

The probability that a toad was conjured is either 1 or 0. The probability that a dragon was conjured is either 1 or 0. The only relevant information is that the experiment has already been performed, and each resulting datum has already been observed by three wizards. Nothing random remains. That the result of your wand is unknown to you doesn't change that.

The puzzle reminds me of the explanation of confidence levels. Suppose an experiment is performed, the data are observed, and a 95% confidence interval is produced. The population parameter is either covered by that interval or it isn't. The parameter is unknown, not random. The probability that it is covered is either 1 or 0, not 0.95.

[mfb]

That depends on your interpretation as frequentist or bayesian.
An interpretation which cannot give you a probability that you picked a dragon wand, given the information you have, has bad limits to its applicability.

[Yakk]

More importantly, there are uses of the mathematics of probability that describe levels of certainty that something is true given limited information.

It doesn't matter if you are a frequentist or bayesian, the bayesian approach can be used to analyze certain situations, and generates mathematically valid results. You might be uncomfortable using the world "probability" to describe what it gives you, but that is a problem with semantics, not the math.

[Jaywalk3r]

An interpretation which cannot give you a probability that you picked a dragon wand, given the information you have, has bad limits to its applicability.

We are not constrained by such limitations in this case. We can easily determine the probability that the experiment will (note future tense) result in a toad or dragon for a particular wizard in any position in the choosing queue, given that the results produced by previous wands in the same experiment.

If a wizard has a betting partner who can also see the results of the other wizards' wands, but not the results from the wand of his betting partner, fair bets can be made between them, even after the experiment is complete and the results observed by the other wizards. That does not imply that the probability of the wand producing a dragon is different from what I previously stated.

Consider:

Spoiler:

Suppose I want to bet with my roommate on the 345,876,912th digit (to the right of the decimal point) of π. Given that neither he nor I have the 345,876,912th digit memorized and that the distribution of digits of π is approximately uniform, 9:1 against guessing the correct value for the digit would be fair betting odds. We could then make our bet and look up the value of that particular digit to see who wins. Suppose the bet is made and the guess is 3. We cannot say that the 345,876,912th digit has a ten percent probability of being a 3. The value of that particular digit is fixed, even if it is unknown to the two of us. It is a 3, or it isn't. It isn't going to be a 3 ten percent of the time, and ninety percent of the time not be a 3. The probability that the digit is 3 is exactly 1 or exactly 0. On the other hand, we can say that the probability of guessing the correct value of the digit is 0.1, since we can treat the guess as a random variable, as long as we do so before the guess is made.

[mfb]

As I said, it depends on the interpretation.

We cannot say that the 345,876,912th digit has a ten percent probability of being a 3.

The 345,876,912th digit has a ten percent probability of being a 3 (for me, with my current knowledge).
I can.

[Tunga]

4 wands will produce a dragon. 2 will produce a frog.

What is the probability you conjured a toad.

Zero!

V V V I'm sorry, was just amazed it went this far without anyone noticing!

[jestingrabbit]

You have moved me to edit the OP.