# Interactive problem to demonstrate non-intuitive probability results to a large crowd?

I will be presenting to a large crowd of about 2,000 people on how our intuition does not align with reality in many situations, as I've always been impressed with how statistical analysis can show us how the world really works, to get us to accept that our intuition on probability can be way off.

What I'm looking for is

1. Suggestions on specific questions/riddles or situations that I can give to this crowd that would demonstrate this in the most impressive way by them actually answering the questions & the result being counted right there

2. Some suggested Statistical Mathematics applied to that problem, to demonstrate a likely expected (with, let's say 80% confidence, or whatever would be more appropriate) end-result given a sample size of N, so I can know ahead of time the number I should expect the result to be close to.

The tone of this event doesn't allow me to give any Mathematical explanations, so a demonstration would be more powerful to illustrate the point. As such, questions I'd ask the crowd would need to be straightforward for most people to follow and to allow for a show of hands to be counted, or for them to have discussions with those around them (in groups of x, that could be defined in outlining the problem). I'll have about two dozen helpers to potentially count hands or to ask groups what the result is, and we'll be doing it quickly, so precision can be sacrificed for the sake of simply getting an impressive final result.

The classic example that demonstrates this well for explanatory purposes (without relying on an actual live implementation) would be the birthday problem, but that only needs 23 people for the odds to be over 50/50 of the same birthday, and even if you tested it on a group of 23 people, doing it once will be a bad way to demonstrate your point because of how probability actually works meaning that maybe there won't be that shared birthday this time. But with a large group of people, we can be more confident to get close to a particular number.

At first, I thought that I could scale up the birthday problem and simply ask a small number of questions to get to the most common birthday for the entire crowd, but in posing another question here to see what the expected number may be, I'm reminded why I'm asking here in the first place in how rusty my statistics knowledge is, given that doing so would face the law of big numbers and actually give the intuitively (much less impressive) expected result.

I could, of course, give a simple twist to a classical probability situation, such as the Monty Hall problem, by getting a show of hands for who would do what, or who would expect what odds. That result won't be as impressive though since the large number isn't really a result, so much as psychological proof that most people's intuition is off. If nothing else works, I'd do this and simply explain the issue, but that would work for a small crowd just as well, and I'd like to take advantage of having such a big audience, and having the law of big numbers work for me to be confident on getting a particular result or very close.

Here are a couple of my quick ideas on doing this, but I'd like to see if someone can give me a demonstrable proof of one of these or another demonstration you may have in mind, which would give a large final result, where intuition (for the layperson unfamiliar with statistics) may not expect it:

• Attempt to split the crowd into groups of approximately 23 (or slightly higher if seating arrangements make that easier), and to give them 2 minutes to figure out if there are any shared birthdays. When they are finished, count the number of successes, which should be approximately half (about 43 of ~87 groups that would form). [Would this work as expected, or am I forgetting something?]
• The top answer on this question has multiple interesting suggestions that I'll be looking into to see which could be most entertaining, but can we find an expected answer given my criteria, and from that, know which one is going to be most impressive?
• Maybe there are some idea in the answers to this post: math.stackexchange.com/questions/2140493/… – Bram28 Jun 22 '17 at 18:06
• a couple of thoughts - with 2000 people you can guarantee at least 6 have the same birthday. also ask everyone who thinks more than half the crowd will raise their hands to raise their hands - kind of Nash equilibrium. but neither are probability. sigh. – JMP Jun 25 '17 at 7:53
• What does "with large end-result" mean in you title? Judging from the body of the question, maybe you mean a result obtained from a sample with large size. I suggest simplify the title by just making it "Interactive problem to demonstrate non-intuitive probability results to a large crowd?". Also add the soft-question tag. – Thanassis Jun 29 '17 at 1:36
• @Thanassis Thanks for those suggestions - I've just edited the question as suggested! The "large end-result" is essentially a number of people in the audience who will find a birthday match (not the number of matches, which would be about half as much), or some other number that is as large as possible given a crowd of 2000, since that higher number will potentially demonstrate impressive unintuitive probability results to the crowd with more potency. Apologies if my question has been phrased confusingly! – Benny Lewis Jun 29 '17 at 1:45

Maybe you gave up on the birthday problem too quickly. If you divide the audience up into groups of 50 (probability .97) then almost every group will have one or more birthday matches; (and probability .994 for group of 60). Intuitively, it seems to me almost as unlikely to get a match among 50 as among 23. [Computation below from R statistical software.]

n = 1:60;  p = numeric(60)
for (i in n) {
q = prod(1 - (0:(i-1))/365)
p[i] = 1-q }
plot(n, p, pch=20);  abline(h=0:1, col="darkgreen")
p[n==23]; p[n==50]; p[n==60]
## 0.5072972  # just over 1/2 for 23 people
## 0.9703736  # above 97% for 50
## 0.9941227  # almost sure for 60


Why do people misjudge the probability? (a) Thinking of the probability their own birthday will be matched, but there are ${n \choose 2}$ pairs of people that might have a match. (b) It would take 366 people (ignoring Feb 29) to be absolutely sure of a match and 50 is a long way from 366, but the probability does not increase linearly (as shown in the graph).

Notes: (a) The model is not exactly right because in the US there are more births in late summer than in winter. However, simulation studies with unequal birth frequencies as in the US show very little change from the theoretical values (and the change is to increase the probability by a little). There is of course a breakdown point. You wouldn't want to try this at a convention of twins or at a meeting of the Sagittarian Society; then obviously you'd get matches.

(b) One can also find the expected numbers of birthday matches; mean about 0.8 in a group of 23. I could estimate it for a group of 50, if you like. (Three people born on 7/4 counts as two matches.)

(c) How for a group of 50 to determine quickly if it has matches? Start by finding matches in Jan, Feb, etc. Then check for exact days within those groups. I have tried it in a class of about 50 without chaos.

Addendum: In case it is of use, here is a paper from a conference presentation given by one of my former students based on notes from classes by me and a colleague. Somewhat similar material appears in Ch 1 of E. Suess et al.(2010) Springer.

• In a million simulated groups of 50, assuming equally likely birthdays, the average number of birthday matches was $3.213 \pm 0.003$ per group. – BruceET Jun 29 '17 at 0:43
• Interesting! So with an expected 3.2 matches per group, that could mean a pretty high confidence level of 3.2x40 (groups) = 128 birthday matches in a crowd of 2,000. Since I'm going for impressive numbers here, instead of "matches" if I say "individuals who share a birthday with anyone in their group" (IWSBWAIG), any idea what the expected outcome would be? ~x2 # of matches? And is there an ideal group size (even by testing just in increments of 10) such that IWSBWSIG results in an overall higher number after multiplying by # of groups from pool of 2000? Hopefully my phrasing isn't confusing. – Benny Lewis Jun 29 '17 at 1:02
• There are at least two people involved in each match. There are lots of birthday matching Q&As on this site. As I recall one of them found a combinatorial formula for the number of people involved in matches. // I think you need to choose group sizes you think will be manageable. I don't immediately see an optimal size based on a statistical argument, if I understand your intent. – BruceET Jun 29 '17 at 1:08
• For 2000 people I got over 1630 matches. There are only 365 unique birthdays to go around. After you've accounted for those everyone else has a match. (Also remember if $k + 1$ people have the same birthday that counts as $k$ matches.) // Whatever you decide to do, I wish you great success with your talk! – BruceET Jun 29 '17 at 1:31
• Tversky has lots of interesting examples on how people make irrational probability assessments. Recent reference. – BruceET Jun 29 '17 at 2:23

This is probably not practical for an audience of 2000, but suppose you could prepare enough envelopes so that each person in the audience receives one. You could then do a simulated cancer-screening demo, along the following lines.

You tell people there is a screening test for a form of cancer that occurs in one percent of the population. The test is 95 percent accurate, meaning in this case that it never gives a false negative but it gives false positives 5 percent of the time. Inside your (white) envelope is the result of your test. If it's a green envelope, your test result was negative, and you are cancer-free. If it's an orange envelope, your test was positive.

Before you open your white envelope, how worried are you that you have cancer? Discuss with the friends around you.

Now open your white envelope. Are you relieved by the result? If your inside envelope is orange, what are your chances of having cancer? Discuss with the friends around you.

Everyone with an orange envelope, please stand up. Look around you. Does this make you more worried or less? Discuss.

As it happens, your orange envelope contains a card that's either green, for healthy, or red, for, well, you know. Go ahead and open them.

A demo like this may or may not work well, but it's well known that even doctors vastly overestimate the odds that a positive screening test portends a diagnosis of cancer. (The same is true for other diseases as well.) It's common to disregard the stipulated rarity (in this case one percent) and simply think that if a test is 95 percent accurate, then a positive result means your chance of having cancer is 95 percent. The demo (and discussion) would hopefully show that's closer to 1 in 6. Bayes' Theorem, it seems, is hard to wrap your head around.

• Nice idea. No reason all 2000 need to get envelopes. Maybe just a few hundred in the front of the auditorium. Maybe envelopes are taped under seats to avoid confusion on entry (+1) – BruceET Jun 30 '17 at 5:54

I still like your idea to have some variant of the birthday problem:

What is the chance that there is at least one day of the year that no one in the audience has their birthday at?

Unless I am mistaken:

$A$: There is at least one day where no one has their birthday

$B$: There are no days where no one has their birthday = for every day, at least one person in the audience that has that as their birthday

$C$: For some specific day, at least one person in the audience has their birthday on that day.

$D$: For some specific day, no one in the audience has their birthday on that day.

$P(D) = \frac{364}{365}^{2000} \approx 0.004$

$P(C) = 1-P(D) \approx 0.996$

$P(B) = P(C)^{365} \approx 0.22$

$P(A) = 1-P(B) \approx 0.78$

So .. there is a pretty good chance (78%) that there is at least one day where no one in the audience has their birthday ... which may come as a bit of a surprise ... especially after you talk about the law of the big numbers!

Indeed, you should have some interactive demonstration of the Law of the Big Numbers given the very big audience you have ... so for that, you could ask:

How many days in the year should I pick so that at least one half of the audience has their birthday during one of those days?

Now, frankly I don't know the expected number of days that you need to pick (maybe someone better at math and statistics and probability can find that out ..) but I am pretty sure that if that expected number is $X$ (say, 100 days), then with an audience of 2000, you are very likely to be very close to $X$. That is, you can proclaim 'I think it is about X days!', and I am guessing you have a very good chance that you are within 5 days when you do this with your audience.

So .. how do you do this with your audience? Assuming they have their phones on them, they could go to a website you give them, and select their birthday, and a computer does the rest ... do you have someone who could put that together?

• P(A) = .78 confirmed by simulation. With 2000 participants, avg number of 'unused' birthdays among 365 equally likely birthdays is about 1.5. Maybe somewhat greater for actual US birthday dist'n. // Can auditorium wi-fi support 2000 almost-simultaneous connections to a website? If so, can website handle them? – BruceET Jun 29 '17 at 8:32
• @BruceET Thanks for the confirmation!! I was wondering: would you be able to run some simulations for the second question as well? I am just wondering what that expected number of days 'X' is ... and if there is indeed relatively little variation in that between simulations. But yes, the tech/wi-fi will be a weak link in this kind of thing ... maybe the audience could fill out a card as they walk in and volunteers tally the results? – Bram28 Jun 29 '17 at 11:51
• I'm recovering from a 'minor' medical procedure today, and not at my best. Suggest you pose this as a specific separate Question (linking to here). A combinatorial solution would be preferable. And giving a chance to see a variety of methods to answer is always a good idea. 'Birthday questions' usually get good attention. (Minor procedure: one done on somebody else.) – BruceET Jun 29 '17 at 19:49
• @BruceET OK, good idea, thanks! Speedy recovery! – Bram28 Jun 29 '17 at 19:52

First let me say it's great that you are trying to engage such a large audience with an interactive question. It's hard, but can be very effective. I am not sure what's the topic of the talk, or how mathematically versed the audience will be, but challenging our intuition can be a great engagement technique. You mention that you will not have the time to show/prove to the audience why the correct results are correct. Using the experiment(s), you will just suggest to them that their intuition is wrong. It would be good to give them some kind of intuitive "why" to help them absorb these new results. Or maybe your message can be "Math can help you build a better intuition, find out how". I am not sure what your goal is. All these issues (topic, your goals, your audience) are important and pertinent to this soft question, so maybe you can amend the description of the question with these details.

As you've probably figured out, the hard question is not so much what problem to choose but how to adapt it theatrically to make it work in terms of logistics, time, tediousness, and dramatic suspense.

The birthday problem is really good in terms of counter-intuitiveness, but the details of how you showcase it are very important. Your suggestion of splitting the whole audience in groups of $23$, have them do the test, report back success or not, and then see that about $50\%$ of the groups were successful, while correct in (mathematical) principle, seems like a nightmare logistically and theatrically. First of all, $23$ is a large number for people to form groups, especially while seated. But more importantly after the experiments are done, it's hard to immediately see how our intuition is wrong. We can have group leaders (oh right you need to select these too) raise their hand on success, but it's hard for someone to immediately interpret the result visually. They'll see about 40 hands up, and you'll have to explain what this means. In other words, you don't get the "Wow!" immediate reaction from the audience.

## The birthday problem (redux)

I think for the birthday problem you'll need a representative group to bring up to stage (or somehow bring up front). $23$ people are too few, because it's only $50\%-50\%$ to get a hit. I'd go with a group that gives you 90% or more. This is how I would setup the experiment. First ask: "How many people would we need to guarantee 100% that at least two people in the group share the same birthday?" $366$ (or $367$) are acceptable answers. It makes intuitive sense.

Then follow with the more interesting question: "How many people do we need to be almost certain, say with $95\%$ probability that we will get at least one pair matching?" Usually people respond with numbers upwards of $300$. Since it's a huge audience, you can offer them options: Show of hands who thinks it's less than $350$... less than $300$... less than $200$...?

"What if I told you that it's only $46$ people! We can make an experiment, do you want to make an experiment? Let's get $46$ people on stage, I bet we'll find a match!". You can also raise the stakes by risking to lose something (e.g. $46$ one dollar bills, or paint a red X on your forehead for the rest of the talk, be creative). Then it's also important how you setup the "reveal". Have them on a line, give them one mic, and let them announce their birthday one by one. Also instruct them that if they hear their birthday said by someone else to raise their hands in the air and say "Me too!". You can also have them shake hands (or high five, or hug) and the experiment will be over! You won the bet [hopefully :)]. Note that you will need fewer than $20$ people on average to talk into the mic, before someone else from the $46$ total yells "Me too!". It can all be done $3-4$ mins (the slower part is getting people on stage). You can then even offer some intuition why such a low number is right (BruceET's answer describes the basic concept).

## Coin problems

It's also worth exploring some coin flipping problems. Since it's hard to rely on everyone having coins, or giving coins to everyone, I would again go with a representative group of around $20$ people that you can invite on stage and give them coins.

Flip a coin 5 times. Which is more probable?

• You will get the same result at least 3 times in a row
• You will not get the same result 3 times in a row

People usually think that it is improbable to get the same result $3$ times in a row. Again you can give them options to have an estimate: "What percentage of the $20$ people doing the experiment will get $3$ in a row?" Show of hands for $50\%$, $30\%$, $20\%$, less than $10\%$. Do the experiment. People that got $3$ in a row raise your hand (or step forward). How many do we expect to get? Mathematics says that the answer is $50\%$ ($16$ out of the $32$ possible flip outcomes get the same face $3$ times in a row or more). So you'll get the surprising number of around $10$ people stepping forward.

Flip a coin four times. Which is more probable?

• You will get [Heads] exactly two times
• You will not get [Heads] exactly two times

Seems intuitive that it is more probable to get exactly two head and two tails. This is the expected result after all. But this is true only in the average. It is not that probable for the coins to be perfectly divided in equal numbers.

For 4 total flips the probability of getting exactly two H and two T is ${4 \choose 2} 0.5^4 = 0.375$

For $10$ flips a perfect division has probability ${10 \choose 5} 0.5^{10} \approx 0.246$

For $20$ flips a perfect division has probability ${20 \choose 10} 0.5^{20} \approx 0.176$

## Dice or cards problems

You have four cards: A, K, Q, J. Your friend picks two cards at random. What's the probability that they did not get either the K or the Q?

The answer is just $1/6 \approx 16\%$

Roll a 6-sided die. Then roll a 10-sided die. Then roll a 20-sided die. Which is more probable?

• The three results are in increasing order
• The three results are not in increasing order

The answer is that the results are not in increasing order with probability around $57\%$.

This problem/experiment might not be as successful, because the intuition behind it is weaker. People might think increasing order is more probable, but my gut feeling says they will not care too much one way or the other. It is also harder to do since you need to provide the not-so-common 10-sided and 20-sided dice. You can easily order them online, but it requires more preparation on your part. It's not as easy as providing $20$ coins.

I hope my ideas helped you a bit. Please let us know how the talk went!

• Birthday redux with 60 people might work very nicely. (+1) – BruceET Jun 29 '17 at 8:48
• @BruceET, thanks. Is there a particular reason you are suggesting 60 people? I was trying to minimise the number of people (already 46 will be cumbersome to handle) and the need for some certainty. I went for 46 (and 95% probability). We could go down to 40 (giving us 89%), but I think 95% sounds better costing us just 6 people more. 60 people would give us 99.4% certainty, which I think is needlessly high. Did you have another criterion in mind? Maybe multiple matches? – Thanassis Jun 29 '17 at 9:01
• Just cautiously going for the 'needlessly high' probability. If 40 are manageable, why not 60? I have had failures with 40, quite possibly because not everyone can hear the birthdays being announced or is paying attention. – BruceET Jun 29 '17 at 9:08
• @BruceET Sure it's not a huge step up, but my personal preference would be to keep the number of people on stage as low as possible. I have worked with facilitating groups of people (in theatrical activities) and there is a big different between 30 people and 60 people. Not exactly our situation, but I would take the 5% risk, for the benefit of fewer people. I guess the OP will have to decide what they feel comfortable with. – Thanassis Jun 29 '17 at 9:18

Here are a couple of suggestions:

Suggestion 1:

People are often surprised at how long the longest run is in a series of coin flips. One way to get at this is to have each person write down what they think might be a reasonable sequence of coin flips (often 200 flips is used) and then find the longest run of this sequence of (faked) coin flips. Then have the person actually flip a coin that number of times and check the longest run. Usually, for 200 coin flips there will be a run of length at least 7, while this is relatively rare for the made up sequences.

To get the actual sequence of 200 flips, you could break into groups of 10 and have each person flip 20 times, and then paste together their results to form one long sequence.

Or maybe you'd want to just look at sequences of say 50 flips (which makes it easier to produce the sequence and to identify the longest run). With 50 actual flips, there seems to be a better than $50\%$ chance of a run of at least 6 (based on a computer simulation I ran 100000 times).

Suggestion 2:

Imagine that each day you and your 9 office mates (total of 10 people) have a random drawing to see who has to make coffee that day. How many days would you imagine until everyone has made coffee at least once? [I think people tend to low-ball this when guessing, but on the average it will take about 29 days for each person to draw coffee duty at least once.]

This is something you could simulate in groups of size 10, where perhaps each person writing their name on a slip of paper, and then drawing randomly, with replacement, until each name had been seen at least once.

Note: If you have baseball fans in your audience, they might be amused to consider how this idea applies to the Chicago Cubs recently-ended 108-year World Series drought: In a league of 30 teams (i.e, MLB), assuming (unrealistically) that each year each team has a $\frac{1}{30}$ chance of winning the World Series, how long would you expect the unluckiest team to have to wait for a championship (equivalently, how many years until each team has won at least once). It turns out that on the average, the unluckiest team has to wait about 120 years for a championship. With this as background, a 108-year drought isn't at all unreasonable for some teams to have to endure; although in fairness, the league didn't have 30 teams for all those years.

See this discussion of intuitional betting strategies versus calculated probabilities. Doing something along these lines would be 1) relatively easy logistically (similar to a door prize), 2) easy to understand and relate too, 3) naturally introduces expectation into the picture, which is often more important than raw probability.

Here a couple of rough idea outlines:

1. Hand out lottery tickets to people entering the events (as if there will be a door prize). Design the tickets so that there are at least a couple of alternative ways for the ticket to win. Each audience member can select one of the alternatives (similar to the linked article giving a choice of different ways to bump chance of winning by 5%). Then draw a few tickets and tabulate the winnings by strategy.
• You probably want to design this so there's little motivation to "cheat". If you gave out real prizes, that would be a motivation to cheat. Even if some part of the experiment just involved raising hands, there's a motivation to "go along with the crowd". (Although going along with the crowd could be part of the intuition you want to highlight)
2. Have an experiment similar to #1, but without lottery tickets. Instead just have folks raise their hands to select which strategy they choose and have your team count (approximately) hands. Less logistics, but there's less emotional attachment (as mentioned in the article) to the outcome.

Here's a more concrete suggestion:

1. Give every audience member two ordinary door prize tickets. One ticket is odd numbered, the other is even. If asked what tickets are for, your team says there will be a drawing as part of an experiment.
2. During course of talk, give the setup: each audience member can choose one of two strategies, by keeping one ticket and passing the other ticket to the end of their aisle (this part is to minimize cheating and could be dispensed with).
3. The first strategy, represented by an even number, is an imaginary \$50,000 prize, divided evenly among all even ticket holders, if an even number is drawn. The second strategy, represented by an odd number ticket, is an imaginary \$10,000 prize to holder of the matching ticket if an odd number is picked during drawing.
4. In describing the two strategies you could slant things in a number of ways. You could describe first as kind of the ho-hum, not so bad alternative, but subsequently get more excited when you describe the second. You could emphasize the second strategy is "more likely" to have a higher upside, since winner doesn't split the pot (note "more likely" can be deliberately confused with probability-based-on-statistics, vs it's other meaning of a common sense sort of belief).
5. Then draw a few tickets and see how everyone does.

The expectation of the second strategy is five bucks $(\frac{\$10,000}{ 2000})$. The expectation of the first strategy is$(\frac{\$50,000}{2n})$ where $n$ is the number of people who kept even tickets. Since $n \leq 2000$, then expectation is $\geq \$12.50$, and could go as high as$\$25,000$.

First strategy has highest expectation and highest potential upside. But I would still guess a significant number of people would go with second strategy.

And of course, you could change the strategies, the amount of money in the pots, etc. to get to a point where you think a significant number of people will choose the alternative where the expectation (or another statistic) is less favorable.

Maybe you could do the airplane problem, where you have 100 passengers, and all of the passengers have a ticket with seat number, except number 1, because he/she forgot his/her seat number. So Passenger 1 gets on the plane, chooses a random seat. Passenger 2 gets on the plane, and if his/her seat is not taken, he/she sits in it. Otherwise, they have to choose a new seat. It continues like this until passenger 100 either finds his/her seat empty and sits in it, or finds it with someone in it, and thus chooses the only single spot remaining. What is the probability that the 100th passenger will be in the right seat? Here is a post with answers: Taking Seats on a Plane.

You can do this with your 2000 people by splitting them up in groups of 50-100 or something and give each group a bunch of numbers going from 2 to the number of people in the group, and play the game with them, and at the end ask the last passenger of each group to say whether or not they ended up in the right seat.