The Probability Paradox

Today's weather: High = 23 Low = 18
Cloudy, windy, approaching storms

Lately I'm wrapping up the final unit in the math 12 course I'm teaching -- about probability. Needless to say it's been a very interesting topic. We stumbled upon a very fascinating paradox that. Regretably, this may have implications for a Tibet trip I'm considering this summer.

The short end of it is that my research into the Tibet trip points to the fact that independent travel, especially cycling, is not allowed. But many people have actually done it, including a guy's blog I'm reading right now who just pedalled into Lhasa on Friday.

As with anything in China, the motto is that semi-legal is better than illegal. That is to say, there are ways to do such a trip in such a way that still involve the proper paperwork, but is not 100% proper. The main difficulty is that getting a travel permit into Tibet requires that you join a tour group, hire 4x4 landcruisers, etc. and it could end up costing $100 a day. I don't think so.

The downside to a genuinely independent trip in Tibet is that it will involve calculated risks. This is exactly what we're going to calculate right now in this post.

For the sake of a crude analogy, let's say that you're traveling along a road with a series of 10 police checkpoints to pass through in series. Suppose that rolling a dice and getting a '2' means you fail at at particular checkpoint and get sent back. Rolling a dice and not getting '2' means you pass that particular check.

The key paradox is that any given roll has an equal chance of success (1/6 or 83%) since it is independent of the roll that came before it. But you still had to roll the dice 'n-1' number of times previously in order to perform the 'nth' roll

In other words, it means you had to pass checkpoints A through E before rolling the dice at checkpoint F. This only makes sense, because failure at checkpoint F means you get sent back to the beginning, and all former successes are rendered invalid.

Is this simply a matter of looking at the glass and saying it's half-empty or half-full? That is, can you put on the rose-colored glasses and say there is always an 83% chance of passing the next checkpoint and continuing on without being sent back? Let's calculate and find out.


Pass up to #1: 5/6 chance = 83%
Pass up to #2: 25/36 chance = 69%
Pass up to #3: (5/6)^3 chance = 58%
Pass up to #4: (5/6)^4 chance = 48%
Pass up to #5: (5/6)^5 chance = 40%

Pass up to #10: (5/6)^10 chance = 16%

To calculate the failure at a particular checkpoint, you can correctly reason that the previous checkpoints had to be passed first -- with the exception of the first one.

Fail at #1: 1/6 = 17%
Fail at #2 = 83%*1/6 = 14%
Fail at #3 = 69%*1/6 = 12%
Fail at #4 = 58%*1/6 = 9.6%
Fail at #5 = 48%*1/6 = 8.0%

Fail at #10 = (5/6)^9 * 1/6 = 3.2%

Your chances of success or failure both decrease by the same amount in this sense. So your relative success to failure ratio remain the same. 5 to 1 odds for passing whatever the next checkpoint is. So we really haven't concluded that much. We can still be optimistic. Or can we? Put on your thinking caps, it's gonna get a little wild.

Let's suppose that you didn't pass checkpoint C. Does this mean you failed at checkpoint C (12% chance, see above)? Maybe it does, but maybe you failed before C, in which case you still would not have passed C !!!

You could have failed at C (12%)
You could have failed at B (14%)
You could have failed at A (17%)

Add these cases: 12 + 14 + 17 = 42%.

The key to this paradox is understanding that not passing a particular checkpoint is not the same as failing at that particular checkpoint. Not passing the checkpoint means failing there, or prior to there.

We can prove this by the following logic: 100% - (pass up to C) = 100% - 58% = 42%

So what if you've made it past a certain number of checkpoints? Does this increase the chance you'll make it through the remaining ones? We can also use simple logic to figure this out out:

Pass the last 5 if you passed the first 5 = pass the first 5 = 40%
Pass the last 4 if you passed the first 6 = pass the first 4 = 48%
Pass the last 3 if you passed the first 7 = pass the first 3 = 58%
Pass the last 2 if you passed the first 8 = pass the first 2 = 69%
Pass the last 1 if you passed all 9 before = pass the first 1 = 83%

Conclusion? The mathematics backs up what we already know to be true intuitively. Your chances of successfully passing the remaining checks increase the further along you go. But it's also harder to progress further and the stakes of the game get higher and higher as you do so.