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The importance of seduction and curiosity

Blondegirl_2

Part of creating passionate users starts with building curiosity. Inspire them to want to learn, know, and do more. A comment from John Mitchell on my motivated to learn blog reminded me about this--he mentioned the importance of being passionately curious about the topic (and I couldn't agree more).

So can you inspire curiosity? Can you seduce the user into actively wanting more, even if that user didn't start out with their own intrinsic intellectual curiosity?

Sure. It won't work for everyone and every topic... but think about things that you know have worked for you in the past:

1) Be passionately curious yourself (good point John!)
The brain is tuned to mirror the behavior of others, so if your passionate curiosity is stronger than the other person's passive disinterest, you have a chance to "infect" the other person. It's not just that you know what's exciting, wonderful, fascinating about a topic--it's that you genuinely feel it, and this is reflected in the way you talk about it, not just the actual content of your words. Passion breaks through.

2) Be seductive
That means knowing when--and what--to hold back. Don't hand them all the answers... take them part way and tease and tantalize them into going the rest of the way. The brain wants to find out what happens next. It's what keeps you watching the movie until the end, staying up late at night with a page-turner, tuning in next week (especially if last week's episode was a cliff-hanger), and hoping for that second date... NPR refers to the phenomenon of wanting to hear the end a driveway moment--where you're listening to an engaging story (like on This American Life, or a radio diary) but arrive home before it's over. You can't get out of the car. You just have to hear how it all turns out.

3) Make them curious by doing something unusual, without an obvious explanation (a variation on #2)
In the Parelli natural horsemanship program, I learned a new way to "catch" a horse. I walk into the big pasture holding the halter and instead of walking straight toward my horse, I kind of meander around not even looking at her. Then when I come close enough for her to know I'm there, I stop and turn around so my back is to her... and I might even start walking away while fiddling with whatever I'm holding. Eventually, she can't stand it and has to know what I'm up to and why I didn't try to catch her. So she'll come over and "catch" me. (For all you pet people, I'll mention that we are not allowed to use treats as an enticement. They're coming because they're curious and it triggers their play--rather than fight or flight--instinct). In the book/movie "The Horse Whisperer", Robert Redford's character spends hours sitting in an open meadow until the terrified, escaping horse finally walks up to him. Curiosity can beat fear.

4) Offer a puzzle or interesting question... without giving them the solution.
It's almost impossible to turn away from a TV game show when a question has been asked but not yet answered. But it works for almost anything that engages the brain's strong desire to find solutions.

Here's an example (many of you will know the answer to this already, so it won't work for you... but it works quite well on most people who don't know this problem):
===================
Kevin, a college student, walks into the cafe where he spots Reese, the gorgeous math whiz he's seen around campus. He works up the courage to walk over to her, "Hey Reese, I'm Kevin, and I heard you're the only one to ever get an A in Bozeman's Stats class... I'll buy you dinner if you help me study for the exam."

Reese look up skeptically then says, "I need to know if you're worth helping. Tell you what, I'll write my phone number on the back of one of these three business cards. I'll mark them on the front A, B, and C, but you won't know which card has my number on the back. If you pick the right card, you can call me and we'll schedule a study/dinner date."

Kevin's not happy, "But what does that prove about me? You're not even giving me a 50/50 chance... but OK, if that's the best I can do... I'll pick card 'B'".

Reese is left with cards 'A' and 'C', and says, "Before we look at your card, I'll give you another chance. I've just turned over card 'A', so you can see it doesn't have my number. That means my number is either on the card you picked, 'B', or the card I haven't turned over, 'C'. Do you want to switch your card 'B' for my card 'C'?"

Kevin cocks his head and thinks to himself, Ah... she's trying to see if I recognize that the odds are the same for both cards (duh), since they both began with a 1 in 3 chance. She probably wants to see if I'm decisive and confident... He looks at Reese and says, "No thanks; I'll stick with my original choice 'B'. It's just as likely to be the winner as your card 'C'."

Reese flips his card 'B' over and shows that it's blank. Her phone number was on card 'C'. Kevin laughs and says, "Well, you didn't give me a fighting chance. All the cards had a 1 in 3 chance of being the right one, and at least I didn't fall for your little swapping trick... you should still give me your number."

Reese rolls her eyes, shakes her head slowly, and says with a frown, "You know Kevin, if you would have agreed to swap cards when I gave you the chance, I would have given you my number even if it wasn't the card with my number. Switching cards would have shifted the odds in your favor, and I was really hoping you knew that. Sorry, but I don't want to waste my time trying to help you."

Unhappy and a little angry with Reese, Kevin leaves the cafe and tells the story to his roommate Manny. Kevin says, "Reese is just wrong... there's no way that switching my card for hers at the point would have made any difference. Both cards started with a 1 in 3 chance, and nothing changed that."

Manny looks at Kevin and says, "Dude... Reese is right. Switching cards would have changed your odds from 1 in 3 to 2 in 3. You blew it."

So... are Reese and Manny right?

Yes, they are. Swapping would have changed his odds of having the winning card.

Your job is to figure out why it works that way... and if you want to learn it here, you'll just have to wait for another blog entry (later this week, I promise) to find out.

[Note to those who recognize what the Kevin/Reese puzzle is about: don't reveal the true "name" of this problem, so we can make googling for the answer a little less easy for everyone else ; )]
===================

People who don't immediately understand the problem and don't believe it, will often set out trying to disprove it (they're curious to find proof that Reese, Manny, and you are wrong), or they believe it but they're so puzzled by it that they try to find out what's going on (curious to understand).
The brain wants it to make sense. : )

There are obviously lots of ways to get people curious, but it's been a highly underrated strategy for getting people engaged, hooked, motivated... all prereqs for passion. Think about ways you might use curiosity as a technique in everything from user documentation and tech writing (including books) to teaching to dating to product development to marketing to horse training to parenting and even delighting your significant other.

Having a passionate curiosity is a true gift, and anything you can do to help give a little of that to another person is enhancing their life. If you want to truly delight someone, then seduce them (not the same as coerce or manipulate, if we make ethical distinctions--I realize some people don't like the word "seduce", but we love it) into being curious to learn more, grow more, stretch their mind, become more skilled, or just find out what it's like to be better at something.

Posted by Kathy on March 30, 2005 | Permalink

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Comments

About the only way to convice people about the M..., er, "phone number" problem: have them try it themselves a bunch of times.

Once they are convinced that it is true, then (and ONLY then) is it worth talking about WHY it is true.

Posted by: BradC | Mar 30, 2005 3:21:29 PM

I can hear [name edited by Kathy] saying, "are you sure you want what's behind curtain number 3 or would you like to switch?"

The poor bastards should have always switched.

[note from Kathy: T-Bone, sorry but I edited out the name so it wouldn't be so easy to google on the problem... ; ) You're right, of course!]

Posted by: T-Bone | Mar 30, 2005 4:11:46 PM

The "Horse Whisperer" was loosely based on the ideas and horse handling techniques of Monty Roberts. If you're serious about horses, or just interested in guys thinking out of the box, he's got a great autobiography, "The Man Who Listens To Horses".

Posted by: Eric Lechner | Mar 30, 2005 4:29:06 PM

Brad, good idea on trying it!! (And thanks for resisting the urge to use the real name, yes, let's call it the "Phone Number Problem.")

You just gave me the idea to put up some Java code that duplicates exactly what Reese and Kevin are doing. I'll try to make it human readable even if you're not a Java programmer.

Eric, thanks for the Monty Roberts recommendation. I haven't read that book, but it's going on my list now. I'm new to horses, but taking it very seriously (I have a four-year old Icelandic mare)

Posted by: Kathy Sierra | Mar 30, 2005 4:33:23 PM

People who don't immediately understand the problem and don't believe it, will often set out trying to disprove it (they're curious to find proof that Reese, Manny, and you are wrong), or they believe it but they're so puzzled by it that they try to find out what's going on (curious to understand).
The brain wants it to make sense. : )


The vast majority of people I have explained this problem to don't believe the answer. Some have puzzled over it all night to try and counter the (obvious once you grasp it) truth.

Posted by: Anon | Mar 30, 2005 4:39:01 PM

Sorry, I don't do Java, maybe I need to get a book. Any suggestions? :) Here's a version in VB.NET:

Function PhoneNumber(ByVal tries As Integer) As Single
  'Calculate % of wins by switching
  Dim wins As Integer, rand As New System.Random

  For i As Integer = 0 To tries
    'Reese writes her number on card 0, 1, or 2
    Dim phone As Integer = rand.Next(3)
    'Kevin chooses a random card
    Dim chosen As Integer = rand.Next(3)
    'Reese reveals a losing card
    Dim reveal As Integer
    If chosen = phone Then
      'reveal either one of the others
      reveal = (chosen + rand.Next(1, 3)) Mod 3
    Else
      'Only one possible card to show
      reveal = 3 - chosen - phone
    End If
    'Now switch your choice to the other card
    chosen = 3 - chosen - reveal
    If chosen = phone Then wins += 1
  Next
  Return (wins / tries)
End Function

Posted by: BradC | Mar 30, 2005 9:29:24 PM

I'm a student in college right now and I'm taking a Database Systems course. I'm interested in databases and I like the class, but the book is full of things like "Brain teaser: What does X->Y mean if X is empty?" and I hate them, because the book doesn't contain solutions for them, and even though I follow the material and read the textbook, I don't even understand the question half the time.

So, I think authors and teachers should be careful not to try to be too seductive, because maybe I'll hate it.

Posted by: Keith Lea | Mar 30, 2005 11:07:58 PM

@Keith: Is it really seductive though if they intentionally frustrate the user?

The questions Kathy raises are out to prove a point through their solution rather than to test if the user has been following along. I.e. She's generating the user's passion rather than checking for it.

Anyway awesome way to illustrate a point. You had me hooked with all that horse stuff and I'm not even much of an animal person.

Posted by: Jack | Mar 31, 2005 12:07:11 AM

BradC's VB translates, I think as:
(excuse the hideous lack of indentation - is there any way to do pre or code tags in typepad?)

public class KevNReese {

public static void main(String[] args) {
KevNReese knr = new KevNReese(); // Set up our KevNReese object
System.out.println("Wins/Tries = " + knr.go(10)); // Let's let it run for 10 tries
}

double go(int tries) {
int wins = 0;

for(int i = 0; i < tries; i++) {
int phone = (int)(Math.random()*3);
int chosen = (int)(Math.random()*3);
int reveal;
if (chosen == phone) {
reveal = (chosen + ((int)(Math.random()*3)+1) % 3);

} else {
reveal = 3 - chosen - phone;
}
chosen = 3 - chosen - reveal;
if (chosen == phone) {
wins += 1;
}
}
return (double)wins/tries;
}
}

Posted by: Matt Moran | Mar 31, 2005 12:13:53 AM

Except, Matt, that
reveal = (chosen + ((int)(Math.random()*3)+1) % 3);
should be
reveal = (chosen + ((int)(Math.random()*2)+1) % 3);
Because you don't want it adding 3 then modding to the same value--that will "reveal" the "chosen" card, which you can't do.

To do some (slight) indenting, I manually put some &nbsp; in front. I tried [font] and [pre], but nothing seemed to work.

Posted by: BradC | Mar 31, 2005 12:43:51 AM

Ah, thanks Brad - I did wonder about that bit but the docs I found online about VB.Net were a bit hazy about it.

Posted by: Matt Moran | Mar 31, 2005 2:08:38 AM

Monty Python is good comedy.

Posted by: Umesh Persad | Mar 31, 2005 7:28:15 AM

I prefer watching Monty Python in a concert hall!

Posted by: T-Bone | Mar 31, 2005 7:42:41 AM

This example would be an antimotivator for me. I would get caught up in what a jerk Reese is. She is a bully and I didn't want to find out anything more after this:

"You know Kevin, if you would have agreed to swap cards when I gave you the chance, I would have given you my number even if it wasn't the card with my number. Switching cards would have shifted the odds in your favor, and I was really hoping you knew that. Sorry, but I don't want to waste my time trying to help you."

What a bully. Kevin has said he has problems and is asking for help. Reese gives him a challenge that many people get wrong.

What's the message there? Helping someone who needs help is a waste of time? I'm sure I'm reading more into this than I need to, but as a recovering academic, this example really rubbed me the wrong way.

Also - I have a problem with appropriating a problem and hiding references to the original. We all stand on the shoulders of those who came before us and should credit these wonderful sources of problems and explanations. Let's not call this the "phone number problem"

Posted by: Daniel | Mar 31, 2005 7:57:23 AM

Hey Daniel, yeah I think Reese is a bit harsh with poor Kevin, but there's always more to the story... I think she's jaded after all those guys *pretending* to want help but who were really just hitting on her ; )

But that's their story... and we're only hiding the references (there are many, depending on how you choose to interpret the problem) to where this really came from *temporarily* to make it harder to google on the explanation. As I mentioned in the post and the comments, we'll use "the phone number problem" as a kind of code name until it's time for the explanation. But don't worry -- this weekend we'll give the rest of the story with references to some of the other ways in which this has been discussed/argued/debated/explained. Again, the point of hiding the more well-known names (two of them that come to mind including the name typically given to this problem) is because we're assuming that a lot of the younger folks here won't automatically remember the, um, "show" that started it all...

As for being demotivating, well, that's why some people love our books and others hate them with a passion. This approach doesn't work for everyone, and when it doesn't work it REALLY pisses people off. But we follow Don Norman's mantra, "If some people don't really hate your product, it's probably mediocre." And part of our metacognitive approach is to up the gain on the emotional content, just a bit more than you'd normally find. This is not a straight "Dick and Jane" style of story.

But we look at it this way... if the reader/learner gets pissed off at Reese, or even the example itself, then he's just told his brain that this is worth paying attention to. And that's why we do what we do! These are the kinds of tricks that tell the brain that This Thing Matters--through those subtle shifts in chemistry. If the story is an emotional flatline for the reader, then even if he DOES keep going out of intellectual curiosity, he might not break through the power of CREB-2 (the agent that *represses* your ability to create memories).

Of course, it's a lot more risky to do it this way. Because there's always the danger that a few people will be SO pissed off that they simply won't look at the rest. That's the risk we're more than willing to take, and as I've said, it definitely doesn't work for everyone.

Posted by: Kathy Sierra | Mar 31, 2005 9:50:08 AM

On the other hand, I stopped reading this blog entry to google for the answer. It didn't take me too long, and I did come back to finish this post, but how many people might have stopped reading altogether because of curiosity getting the better of them? B-)

Posted by: GBGames | Mar 31, 2005 10:32:20 AM

"...how many people might have stopped reading altogether because of curiosity getting the better of them?"

Good point : ) Then we've still succeeded. It doesn't matter if they don't come back *here*. All that matters is that they were motivated to find out more... and we don't really care what their ultimate reason was ("you're wrong and you suck, and I'll prove it." or "hmmmm... that's weird..."). If passion is the goal, and curiosity is one possible path toward reaching that goal, then that's all we care about. Of course we're normally doing this in a book, so we hope the book is still sitting there on their desk after they finish their googling or talking to friends, etc., and that even if they're pissed off about it, they come back to see what we said. We often don't finish things with nice, tidy, neat, group hug feel-good endings, though. Sometimes we just want them to choose their own point of view, and we might present more than one that could be equally valid.

Of course, some people simply want to know if Reese eventually agrees to hook up with Kevin (she does... she gives him a second chance when she realizes he's not just another guy scamming her with the whole, "Oh, this is just about help with the class" ploy ; )) and he eventually gives HER an interesting problem to solve which she loves him for. (He had her at "... imagine you're at a party... how many people have to be there before there's a greater than 50% chance that two of them have the same birthday?" ) [not that you'll want to figure THAT one out... ; )]

Posted by: Kathy Sierra | Mar 31, 2005 11:45:34 AM

Ooo, ooo! That's a good problem, too!

One answer is: much fewer than you would initially think. :)

Posted by: BradC | Mar 31, 2005 1:00:50 PM

Well I think the activity in comments alone proves your point. Everyone loves a good mystery.

... like who exactly is Blondegirl_2 and what ever happened to Blondegirl_1?

Posted by: Shaded | Mar 31, 2005 1:50:57 PM

Here's some Java code... two versions, one where he doesn't switch, and one where he DOES. It's not the best code, but my goal was to make it somewhat readable even for non-programmers (so don't flame me for writing it in main()!!)

========

// Kevin does not switch
import java.util.*;

class TestProblem1 {
public static void main (String[] args) {

int kevinWins = 0;
int kevinLoses = 0;

String[] cards = {"A", "B", "C"};

Random randomPicker = new Random(System.currentTimeMillis());

for (int i = 0; i < 500; i++) {
int winningNum = randomPicker.nextInt(3);

String kevinPick = "B"; // this is card B
String winningCard = cards[winningNum];

// display the non-winning card that Reese exposes
if (! winningCard.equals("A")) {
System.out.println("Reese turns over card A");
} else {
System.out.println("Reese turns over card C");
}


System.out.println("Winning card is " + winningCard);

if (winningCard.equals(kevinPick)) {
kevinWins = kevinWins + 1;
System.out.println("Kevin won");
} else {
kevinLoses = kevinLoses + 1;
System.out.println("Kevin lost");
}


} // end loop

System.out.println("Kevin won: " + kevinWins);
System.out.println("Kevin lost: " + kevinLoses);

int percentWins = (int) ((kevinWins * 100) / 500);
System.out.println("Kevin won " + percentWins + "%");
}
}

=====
// Kevin DOES switch

import java.util.*;

class TestProblem2 {
public static void main (String[] args) {

int kevinWins = 0;
int kevinLoses = 0;

String[] cards = {"A", "B", "C"};

Random randomPicker = new Random(System.currentTimeMillis());

for (int i = 0; i < 500; i++) {
int winningNum = randomPicker.nextInt(3);

String kevinPick = "B"; // this is card B
String winningCard = cards[winningNum];

// this is the ONLY part that changed from the version
// where Kevin does *not* switch

if (! winningCard.equals("A")) {
System.out.println("Reese turns over card A");
kevinPick = "C";
} else {
System.out.println("Reese turns over card C");
kevinPick = "A";
}

System.out.println("Winning card is " + winningCard);

if (winningCard.equals(kevinPick)) {
kevinWins = kevinWins + 1;
System.out.println("Kevin won");
} else {
kevinLoses = kevinLoses + 1;
System.out.println("Kevin lost");
}

} // end loop

System.out.println("Kevin won: " + kevinWins);
System.out.println("Kevin lost: " + kevinLoses);

int percentWins = (int) ((kevinWins * 100) / 500);
System.out.println("Kevin won " + percentWins + "%");
}
}




Posted by: Kathy Sierra | Mar 31, 2005 2:41:53 PM

Hello there!

This is a classic problem in disguise and the first time I came across it I just didn't believe the answer. I know enough about probability to know that I don't really understand probability! I knew therefore that I couldn't find a proof mathematically.

What finally did it for me was modelling the problem in a spreadsheet. I just used the random number generator to pick a number between one and three, another column had a random guess, then two column; one for keeping the first guess and another for changing. (I can't remember the exact details, but it was something along those lines.) I then duplicated the formulae across umpteen lines and worked out the percentage of times I won by staying the same and the percentage I won by changing. The answer was consistent and clear no matter how many times I tried it. I didn't understand the maths, but the spreadsheet was easy enough to set up and understand, and visual enough to convince me!

Thanks for this post... and the one on Just-in-time which seems to me to be about the same kind of thing. It kept me curious and motivated. :-)

Posted by: David Muir | Apr 1, 2005 7:25:42 AM

Now, this is an interesting problem. Your code is actually modeling a different problem. If Kevin knows Reese's algorithm (which is consistent with the original problem) and sees the card C then he knows for sure that A is the winning card. And if he sees the card A - what is it that has been revealed and what are the probablities now that Kevin should stay or change? It is interesting that the strategy can remain the same although the conditional probabilities change.

Posted by: daniel | Apr 1, 2005 9:34:01 AM

Hmmm... I see what you mean about the code, but my assumption is that Kevin never *does* figure out the algorithm and that Reese just always turns over a blank card. But you're right, if it really was Kevin repeating that over and over (as opposed to a string of different people who don't talk to each other or ever get to *watch* her little trick), and he really *is* paying attention (might depend on how much he's had to drink, etc.), I'd probably want to change the code so that if she does *not* have the winning card, her choice of which card to flip over is random ; ) Otherwise, as you said, he *might* start to figure it out and that would indeed change the game.
But no, for this problem -- when I said that I left something ambiguous, it was to get people to speculate on whether her motivation had anything to do with it. And although this is different from the original, my answer is NO -- her motivation doesn't change. In my game, she ALWAYS gives the other person a chance to switch, *regardless of who has the winning card.* So it's actually a simpler (although still counterintuitive for people like me who did NOT want to believe this the first, oh, thousand times I thought about it) problem: "Given the opportunity to switch, and assuming the choice is offered EVERY TIME, will switching change the odds, and if so, in what way?"

Anyway, I'll put up one possible explanation for this tomorrow, along with links to some discussions of the original (of which the "phone problem" is a "dating variant" ; ))
Now... to the REAL mystery... "who is Blondgirl_2 and what happened to Blondgirl_1?" It's a sad story, actually. Turns out Daniel is right, and this "pick the card" girl has serious issues. She's actually Reese's evil twin Katherine, who has the *real* Reese locked up in a shed somewhere. Katherine is merely *pretending* to be Reese because, well, the real Reese is a sweet and lovely person who the guys find impossible to resist... because of her math abilities, of course.

I don't actually *have* the real birthday party solution -- although it's one of Bert's favorites. So if anyone else wants to post their explanation of the birthday thing, go for it! Otherwise I'll convince Bert. We heard someone on NPR the other day discussing it, and I also just finished reading a novel "Improbable" where the protagonist is a former stats teacher, and he does a variant of the birthday thing in a classroom. But his solution was more elaborate then the one Bert uses... so I don't really know for sure.

Posted by: Kathy Sierra | Apr 1, 2005 11:09:11 AM

Regarding the birthday issue:

Its easy to calculate the % chance that a second person has the same birthday as the first:
1/365 (ignoring leap year... that's just too messy).

But what about 3?
Is it 1/365 + 1/365 = 2/365? NO!
Just like flipping a coin twice doesn't DOUBLE your odds of getting heads! (1 flip = 1/2, 2 flips = 3/4, NOT 2/2)

We have to do this another way. For each new person that comes in the room, we have to calculate the chance that they have a DIFFERENT birthday than everyone else.

1 person in the room = 0% chance of two people having same b-day
2: 364/365 chance they don't match, 1/365 chance they do
3: 364/365 * 363/365 = 99% they don't, 1% they do
...
5: 364/365 * 363/365 * 362/365 * 361/365 = 98.9% , 1.1%
...
23 people: 364/365 * 363/365 * ... * 343/365 = 49.3% they don't share a birthday, 50.7% they do.

(with a little help from Excel.)

So the answer is 23. And MS Excel.

Posted by: BradC | Apr 1, 2005 4:45:47 PM

Or, functionally speaking:

Chance of everyone having DIFFERENT birthdays with n people in the room:

364! / ((365-n)! * 365^(n-1))

In MS Excel, you'd have to do
=PERMUT(365,n)/POWER(365,n)

which gives the same answer: when n=23, the chance is around 50%

Posted by: BradC | Apr 1, 2005 9:04:52 PM

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