Three probability puzzles that will fool you for sure.

One of the biggest areas where Occupy Math hopes to help readers is improving their sense of how likely things are. The old adage “If it sounds too good to be true it probably is” is a good starting point, but examples are a big part of developing your intuition. This week Occupy Math has gathered two old standards and a new activity that can help you develop your sense of probability. This is not Occupy Math’s first visit to this topic. A critical example is the way that doctors’ inability to assess simple probability often terrifies their patients. Occupy Math also worked through an example of cashing in on people’s inability to do probability. This week, the Monty Hall problem, the birthday conundrum, and monotone runs of flipped coins.

The key point is that humans – even ones with PhDs in math – have bad intuition about probability.

Monty Hall was the host of a game show called Let’s Make a Deal. The show was composed of a series of chances for contestants to keep or trade items they had (very) incomplete information about. The Monty Hall problem is in a similar format. The rules of the puzzle are as follows:

  1. You start with three closed doors. Behind one of the doors is a car, behind the other two are goats (assume the contestant would rather have the car).
  2. The contestant picks a door.
  3. The host then opens a door the contestant did not pick, showing a goat.
  4. The contestant may then keep the door chosen in the first place or switch doors to the other unopened door.

The puzzle is this: is it better to switch doors or to keep the door you started with, or does it not matter? Marilyn vos Savant popularized this problem. The interesting thing is that about 90% of people with doctorates in mathematics got the problem wrong. They thought that it did not matter if you switch doors. The link in this paragraph has a detailed explanation of the solution, but Occupy Math will try a simple one.

The first door picked had 2/3 of a chance of having a goat behind it. Once the host eliminates one goat, that means there is a 2/3 chance the “switch” door has a car behind it.

A lot of people think that since there are two doors, one goat, and one car, the odds are fifty/fifty. The error is that there were not two doors when you made the choice, there were three. What you want to avoid is estimating the probability of an event that occured before something changed (the host showing you a goat) after the change happened. Pinpointing the exact time when the choice was made can help avoid this. To aid understanding, let’s look at the hundred-fold version of the Monty Hall puzzle.

  1. You start with 100 envelopes. One contains $1000, the others contain cut-up newsprint that feels like $1000.
  2. The contestant picks an envelope.
  3. The host then opens 98 envelopes containing newsprint.
  4. The contestant may then keep the envelope chosen in the first place or switch to the other unopened envelope.

Humans sometimes find this “hundred-choices” example easier to
follow. You had a 1% chance of picking the good envelope in the first
place and the host showed you 98 bad envelopes, leaving one. It’s
pretty clear that your odds are better if you switch.

We now turn to the birthday conundrum. The puzzle is:

How many people need to be in a room to generate even or better odds that two of them have the same birthday?.

Start by jotting down your own, personal guess. Now, if we think of odds as numbers between 0 and 1 (e.g. 0.5 is a 50% chance), then the chance of something not happening is one minus the chance it happens. We need this idea to solve the birthday conundrum. You see, computing the odds that no two people have the same birthday is way easier than (directly) computing the odds that two (or more) share a birthday. All we have to do is go through the people and compute the odds their birthday misses the birthdays of people we listed earlier. The odds evolve like this:

  • The first person has 365/365=1 chance of missing the nobody at all that have come before him,
  • The second person has to miss the first one, giving him 364 days and odds of 364/365 of not sharing the first person’s birthday,
  • Similarly the third person has odds of 363/365 of missing the birthday of the first two,
  • and so on…

That means we need to find the first time that:

birthday
which is not too hard to do with a spreadsheet. Occupy Math’s open office calc tells him that the odds of sharing a birthday first pass 50% when there are 23 people in the room – a 50.7% chance. At 32 people the odds increase to 75.3%! This puzzle can accompany an activity where a class checks for common birthdays – which is more likely to work if several classes do it.

Caveats: we are ignoring leap years, twins, and the information that birthdays are not evenly distributed through the year. This latter fact means that the odds of two people sharing a birthday in a given group are slightly better than the ones computed above.

Most people substantially overestimate the number of people needed to satisfy the birthday conundrum.

The last of our three probability vignettes involves flipping 20 coins. Occupy Math has used this as a classroom exercise many times and it is pretty reliable. Have the students in your class do the following:

  1. Write down the way you think twenty coin flips in a row would come out, e.g. HHHTTTHHTTHHTTHHTHTT,
  2. Now flip a real coin twenty times and write down the coin’s results,
  3. Find the longest consecutive run of heads or tails in each sequence of heads and tails. In the example above the longest run is 3: HHH or TTT,
  4. Separately tabulate the lengths for human belief and real coins: compare.

The key to this exercise is that humans find long runs of heads or tails offensive to their sense of probability. The best result Occupy Math had with this exercise was a seventh grader who got a run of 11 tails. Her take on the matter?

“Can I have a new coin? Mine is broken!”

Another important fact is that the way a probability problem is presented can change how hard it is – quite a lot. If you would like to see more exercises and puzzles like this, or if you have ones that Occupy Math should know about, let Occupy Math know with a comment or a tweet! You can improve your probability sense with practice, but just knowing how unreliable it is can protect you from some scams and cons. Remember that math, like not geting ripped off, is the right of all free people.

I hope to see you here again,
Daniel Ashlock,
University of Guelph,
Department of Mathematics and Statistics

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2 thoughts on “Three probability puzzles that will fool you for sure.

  1. The Monty Hall problem certainly puts many math phD to shame, but the mistake is rooted in the belief that knowing what’s behind one of the door couldn’t possibly change the outcome. On the other hand, the birthday problem is a different beast, and speaks to the difference between subjective and theoretical probability. Either way, it’s fair to say that probability does defy common sense sometimes.

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  2. The Monty Hall problem contains a notorious ambiguity that is a good example of how an unstated assumption can change the outcome of a word problem. Consider a “random Monty Hall” who flips a coin to determine which of the two curtains to open. You (at first) have 6 equally probable outcomes. But two of these outcomes would have resulted in random Monty revealing the car. This clearly didn’t happen. So now you are faced with 4 remaining outcomes each of which is equally probable. In two of these you have already chosen the cutain in front of the car. In two of these you have not. And this analysis all hinges on one unstated assumption.

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