A few years ago I read a great introduction to probability titled Understanding Probability by Professor Henk Tijms. This was not for school or for work. I just read probability textbooks in my spare time for fun.

So anyway, I was thrilled to receive an email from Henk Tijms. He suggested an interesting puzzle from his new book Probability: A Lively Introduction.

People always email me asking how to learn more about probability. So here are a couple of books to get started–do check them out!

Professor Tijms suggested a couple of problems. I am sharing one today, which I have slightly re-worded.

**The puzzle**

Your birthday cake has *n* burning candles, and you need to blow out all the candles before anyone can eat cake.

Each time you try to blow out the candles, the number of candles that remain burning is a random event.

If the cake starts with *k* burning candles, then after your attempt to blow them out, it will end up with a whole number of burning candles between 0 and *k* – 1, and each possibility occurs with equal chance.

(In other words, each time you will blow out at least 1 candle; and for *k* > 1, you blow out a random number of candles between 1 and *k*).

What is the expected number of times you must blow out the candles until all of the *n* candles are blown out, and everyone can eat cake?

Give it a try, and then watch the video for a solution.

**Can You Solve Professor Henk Tijms Birthday Candles Puzzle?**

Or keep reading.

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**Answer To Professor Henk Tijms’s Birthday Candles Puzzle**

We can solve this by considering small cases and then generalizing. Let’s write *e*_{k} to be the expected number when there are *k* candles left on the cake.

When there are no burning candles, we will be done, so *e*_{0} = 0.

When there is 1 candle, we are sure to extinguish it in one blow.

*e*_{1} = 1

If there are 2 candles, we will use 1 attempt for sure, and then there are 2 possibilities with equal chance. We might blow out all candles, and then we are done *e*_{0} = 0. Or we might leave 1 candle, in which case we will still expect *e*_{1} = 1 more blows.

*e*_{2} = 1 + (1/2)(*e*_{0} + *e*_{1})

*e*_{2} = 1 + (1/2)(0 + 1)

*e*_{2} = 3/2

If there are 3 candles, we will use 1 attempt for sure, and there are 3 possibilities with equal chance. We can blow out all candles (we expect no more blows); we can blow out all but 1 candle (we expect *e*_{1} more blows); we can blow out all but 2 candles (we expect *e*_{2} more blows). So we get the equation:

*e*_{3} = 1 + (1/3)(*e*_{0} + *e*_{1} + *e*_{2})

*e*_{3} = 1 + (1/3)(0 + 1 + 1.5)

*e*_{3} = 11/6

We can generalize the pattern. If there are *k* candles, we will use 1 attempt for sure, and then we end up with *k* equally likely events: between 0, 1,…, *k* – 1 candles left, associated with expected attempts between *e*_{0}, *e*_{1}, …, *e*_{k – 1}, respectively. So we have:

*e*_{k} = 1 + (1/*k*)(*e*_{0} + *e*_{1} + … + *e*_{k – 1})

*k**e*_{k} = *k* + *e*_{1} + *e*_{2} + … + *e*_{k – 1}

So how can we solve this recurrence relation? I did not have luck at first, but I did not give up!

I read up on recurrence relations but did not find any easy methods. Still I did find something worth trying: you can sometimes simplify by subtracting consecutive terms. So let’s consider incrementing to *k* + 1 candles. We would then have:

(*k* + 1)*e*_{k+1} = (*k* + 1) + *e*_{1} + *e*_{2} + … + *e*_{k}

Let’s now subtract the *k* candles equation from the *k* + 1 candles equation. Many terms cancel, so we end up with:

(*k* + 1)*e*_{k+1} – *k**e*_{k} = 1 + *e*_{k}

We then simplify:

(*k* + 1)*e*_{k+1} – (*k* + 1)*e*_{k} = 1

(*k* + 1)(*e*_{k+1} – *e*_{k}) = 1

*e*_{k+1} = 1/(*k* + 1) + *e*_{k}

So that’s a simple formula to get the next term! To get the next term *k* + 1, we add the fraction 1/(*k* + 1). And that’s it: this is the harmonic series!

We can now write a simple formula for the series, which starts with *e*_{1} = 1:

*e*_{n} = 1 + 1/2 + 1/3 + … + 1/*n*

Interestingly, since the harmonic series diverges, the expected number also increases indefinitely with age.

**Source: Professor Henk Tijms**

Personal website

http://personal.vu.nl/h.c.tijms/

If you buy from the book links below I may receive a commission for sales. This has no effect on the price for you.

Books by Henk Tijms I recommend:

*Understanding Probability*

http://amzn.to/2HBR9zt

*Probability: A Lively Introduction*

http://amzn.to/2EJaOzH

The puzzle in this post is adapted from *Probability: A Lively Introduction*.

Hi, I am Lucas from Brazil and I love your videos. Here is a problem from a book with math problems that I could not solve and the book doesn’t say the answer, just said “This one we leave to the students!”

I translated the problem for you, and I am also sending you a photo of the original problem and the book link from where I got the problem. Link: http://www.obmep.org.br/bq/bq2009.pdf (page 34 of book, problem 13)

I replied “Thanks for writing (and for translating too)! I also appreciate the original photo.”

I have never seen this false proof before and wanted to share it. Check it out in the video.

**“Prove” 3 = 0. Can You Spot The Mistake?**

Or keep reading.

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**Mistake In False Proof 3 = 0**

Here is the false proof if you could not watch the video.

*text version

Start with *x* as a solution to:

*x*^{2} + *x* + 1 = 0

Since *x* ≠ 0, we can divide both sides by *x* to get:

*x* + 1 + 1/*x* = 0

From the original equation we also have *x* + 1 = –*x*^{2}, so we substitute into the equation above to get:

–*x*^{2} + 1/*x* = 0

1/*x* = *x*^{2}

1 = *x*^{3}

*x* = 1

So we have *x* = 1, so let’s substitute that back into the original equation *x*^{2} + *x* + 1 = 0:

1^{2} + 1 + 1 = 0

3 = 0

But this conclusion is absurd! And yet every step seemed correct. So where was the mistake?

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**The mistake**

The first step is fine:

*x*^{2} + *x* + 1 = 0

Then dividing by *x* is also fine.

*x* + 1 + 1/*x* = 0

Both of the above equations have 2 solutions, which are -1/2 + *i*(√3)/2 and -1/2 – *i*(√3)/2.

The mistake in the false proof originates in the very next step. We then substitute *x* + 1 = *x*^{2} to get:

–*x*^{2} + 1/*x* = 0

Now this equation actually has 3 solutions. It has the same two complex solutions, but it also has another extraneous solution of *x* = 1.

As *x* = 1 is a new solution at this step, we cannot then substitute it back into the original equation.

It’s a pretty sneaky trick in this problem. But the lesson is that not all steps in a proof are reversible, so you have to check for extraneous solutions, or else you could end up with nonsensical results like 3 = 0.

**Sources**

Link to book from Brazil

http://www.obmep.org.br/bq/bq2009.pdf (page 34, problem 13)

Math Forum post

http://mathforum.org/library/drmath/view/69477.html

Hi Presh:

I got this problem from a Chinese math teacher. It was used to identify gifted kids for competition.

ABCD is a parallelogram. In the diagram (see below), the areas of yellow regions are 8, 10, 72, and 79. Find the area of red triangle. The diagram is not to scale.

Math prerequisites:

1. Basic arithmetic.

2. Area of parallelogram and area of triangle

I was told a few Chinese fifth graders solved it in less than one minute. Oh well, it took me almost ten.

Bill from New York

I spent over 20 minutes on this problem and did not make any progress–it stumped me! So I requested the solution and Bill provided it to me. It was amazingly elegant!

Can you figure it out? Watch the video for a solution.

**Can You Solve A 5th Grade Math Problem From China? (To Identify Talented Students)**

Or keep reading.

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**Answer To A 5th Grade Math Problem From China**

It turns out the problem can be solved with a simple calculation. The area of the red triangle is:

79 + 10 – 72 – 8 = 9

Wow! But why does this calculation work? Even though I did not solve the problem, I was very interested in learning the method in order to verify the solution and to improve my problem solving skills.

Bill explained the key is to identify triangles whose area is half of the parallelogram.

The area of a triangle is (base × height)/2, and the area of a parallelogram is (base × height).

A triangle whose base equals one side of the parallelogram, and whose height reaches the opposite side of the parallelogram, has exactly half the area of a parallelogram. The same is true for a pair of triangles, if the pair of triangles span one side and if their heights reach the opposite side.

Let’s first label the unknown areas with letters *a*, *b*, *c*, *d*, *e*, and *f*. And let’s label the area we want with the letter *x*.

First consider the two triangles below whose sides together span the top side (length) of the parallelogram. As both of these triangles touch the bottom side of the parallelogram, each triangle has the same vertical height as the parallelogram does between its bottom and top sides.

Therefore, these two triangles have a combined area equal to 1/2 the area of the parallelogram. Their area is:

Upper triangles = (*x* + *a*) + (72 + *b* + 8) = (area of parallelogram)/2

Now consider the triangle with a base of the left side (width) of the parallelogram that touches the right width of the parallelogram. As this triangle touches two opposite sides of the parallelogram, the triangle has the same horizontal distance the parallelogram does between its left and right sides.

Therefore, this triangle also has an area equal 1/2 the area of the parallelogram, and its area equals:

Left triangle = *a* + 79 + *b* + 10 = (area of parallelogram)/2

As both equations equal half the area of the parallelogram, we can set these areas equal to each other.

(*x* + *a*) + (72 + *b* + 8) = *a* + 79 + *b* + 10

We can cancel the terms *a* and *b* on both sides and then solve for *x*.

*x* + 72 + 8 = 79 + 10

*x*= 79 + 10 – 72 – 8 = 9

And like magic we have found the solution!

**Equivalent method: bottom triangles and right triangle**

There’s another way to divide the shapes that works out just as conveniently.

First consider the two triangles whose sides together span the bottom side (length) of the parallelogram. As both of these triangles touch the top side of the parallelogram, each triangle has the same vertical height as the parallelogram does between its bottom and top sides.

Therefore, these two triangles have a combined area equal to 1/2 the area of the parallelogram. Their area is:

Lower triangles = (*c* + 79 + *e*) + (*d* + 10 + *f*) = (area of parallelogram)/2

Now consider the triangles whose sides span the right side of the parallelogram and touch the left side of the parallelogram. As these triangles touch two opposite sides of the parallelogram, each triangle has the same horizontal distance the parallelogram does between its left and right sides.

Therefore, these triangles also have an area equal 1/2 the area of the parallelogram, and their combined areas equal:

Right triangles = (*x* + *c* + 72 + *d*) + (*e* + 8 + *f*) = (area of parallelogram)/2

As both equations equal half the area of the parallelogram, we can set these areas equal to each other.

(*c* + 79 + *e*) + (*d* + 10 + *f*) = (*x* + *c* + 72 + *d*) + (*e* + 8 + *f*)

We can cancel the terms *c*, *e*, *d*, and *f* on both sides and then solve for *x*.

79 + 10 = *x* + 72 + 8

*x* = 79 + 10 – 72 – 8 = 9

And again like magic we have found the red triangle’s area!

]]>You play a dice rolling game where you can win money.

You roll a fair six-sided dice. You can stop and earn the value of the roll (in dollars), or you can roll again.

If you roll again, the same rules apply: you can stop and earn the value of the roll, or you can roll again.

If you choose to roll a third time, you will earn the value of the roll and the game ends.

How should you play this game to earn the most money on average, and how much money can you earn? In other words, what is the expected value of this game?

Watch the video for a solution.

**Can You Solve The Roll, Roll, Roll A Dice Game Puzzle?**

Or keep reading.

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**Answer To Roll A Dice Up To 3 Times Puzzle**

One way to solve this game is to think ahead and reason backwards.

Let *v*_{i} = value of the game before roll *i*.

First we calculate *v*_{3}. Imagine you have already rolled twice. What is your expected payoff if you continue to third time? At that point the game ends, so you will earn the average value. As a fair dice shows each number from 1 to 6 each with equal chance, you can expect to earn:

*v*_{3} = (1/6)(1 + 2 + 3 + 4 + 5 + 6) = 3.5

Now let’s work backwards. Knowing the third roll returns 3.5, how should you play the second roll?

If you get a roll larger than or equal to 3.5, it would make sense to stop and take the payout. But if you roll less than 3.5, you should not stop, as you can earn an average of 3.5 on the third roll.

Consequently, you should stop for rolls of 4, 5, or 6 (in which case you earn an average of (4 + 5 + 6)/3 = 5); and you should roll again for rolls of 1, 2, or 3. This is an example of an optimal stopping point game in probability.

We can write the expected value from the second roll as follows:

*v*_{2} = Pr(stop 2nd roll)E(*v*_{2} | stop 2nd roll) + Pr(continue to 3rd roll)E(*v*_{2} | continue to 3rd roll)

*v*_{2} = Pr(roll 4, 5, 6)E(roll 4, 5, 6) + Pr(roll 1, 2, 3)E(*v*_{3})

*v*_{2} = (1/2)(5) + (1/2)(3.5) = 4.25

Now let’s work one more step backwards. Knowing the second returns 4.25 on average, how should you play the first roll?

Just like we reasoned before, you should stop if you get a roll larger than or equal to 4.25, and you should continue if you get a roll smaller than 4.25.

Consequently, you should stop for rolls of 5 or 6 (in which case you would earn an average of (5 + 6)/2 = 5.5); and you should roll again for rolls of 1, 2, 3, or 4.

Once again, we can calculate the expected value:

*v*_{1} = Pr(stop 1st roll)E(*v*_{1} | stop 1st roll) + Pr(continue to 2nd roll)E(*v*_{1} | continue to 2nd roll)

*v*_{1} = Pr(roll 5, 6)E(roll 5, 6) + Pr(roll 1, 2, 3)E(*v*_{2})

*v*_{1} = (1/2)(5.5) + (1/2)(4.25) = 14/3 = 4.666…

So we can earn an average of 14/3 dollars in this game, if we optimally stop and roll as follows:

Roll 1: stop for rolls 5, 6; else roll again

Roll 2: stop for rolls 4, 5, 6; else roll again

By rolling carefully, you can increase your payout from an average of 3.5 on a single roll to an average of 14/3, which is an increase in earnings of about 33.3%!

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Source**

Thanks to James M. for the suggestion!

This was a problem in a book of Olympiad style questions

https://math.stackexchange.com/questions/179534/the-expected-payoff-of-a-dice-game

I admit that I could not solve it! This is a good old-fashioned test of observational skills. Can you count all of the triangles?

Give it a try, and then watch the video for a solution.

**How Many Triangles Do You See? Challenging Puzzle**

Or keep reading.

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**How Many Triangles Do You See? A Challenging Puzzle**

The only trick is you might notice the diagram is symmetric vertically, so you could count just the lower half of the diagram and then double it. But even that is not easy.

The answer is there are 26 triangles. Here is a numbering of them–the blue numbers refer to the large six triangles (3 pointed up, 3 pointed down).

It is hard to label all of the triangles. So if the diagram does not make sense, watch the video where I highlight each triangle by constructing the figure step by step, around 1:13 in the video.

**How Many Triangles Do You See? Challenging Puzzle**

**Source**

Saint Louis Science Center Twitter

Problem:

It's Math Monday! How many triangles can you find in the picture below? Visit our free exhibition #MathAlive! opening this Friday, Jan. 27. pic.twitter.com/VsEn94lfNV

— STL Science Center (@STLScienceCtr) January 23, 2017

Answer:

]]>Great guesses. This picture actually has 26 triangles. Check back next week for another math problem and visit #MathAlive! starting Friday. pic.twitter.com/7GJEyDaxHN

— STL Science Center (@STLScienceCtr) January 24, 2017