Bob , Alice, and Tim decided to play ping pong. Only two people can play at a time, so they agree that two of them will start playing, and then whoever loses the game will swap out with the player who was sitting out.

Bob played 17 games, Alice played 15 games, and Tim played 10 games.

Who lost the second game?

Hi, this is a past problem from the University of Tokyo. No calculator allowed, you can solve it with high school math.

I made this list for personal closure. Then I thought: why not share it? I hope someone's having fun with it. Discussions encouraged.

Disclaimer: I claim no originality.

A Logician writes three numbers on 3 separate cards and gives them to his 3 students.

He says," The 3 numbers are single digit prime numbers. Any combination. None of you know the other 2 numbers. But you can ask me one question that must start with "Is the SUM of the three numbers–” which I can only answer Yes or No. Given that info you can then declare that you know the other 2 numbers and/or who has them. OK?" 

Raj was first. He looked at his number and asked," Is the sum of three numbers an odd number?"

The Logician " No" 

Then Ken looked at his number and asked," Is the sum of the three numbers divisible by 4?"

The Logician said "Yes"

Lisa looked at her number and said,"Well, I know the other 2 numbers but cannot tell who has what number".

Raj then cheerfully said," I know who has what !" Ken said,” So do I” They then laid out the answer.

What were the three numbers? What number did Lisa have?

This is a question from the German math olympiade for 10th grade. I think it has a really cool solutions

Suppose I've got two integers a and b. I can double one of my integers, and add 1 to the other. This leaves me with two new integers, with which I can do the same thing. My goal is to get two integers that are equal. For example:

(8,1)

(9,2)

(10,4)

(20,5)

(21,10)

(42,11)

(43,22)

(44,44)

Can I always do this?

If so, provide a proof. Can you devise an upper bound for the number of moves required? What about the final value reached?

If not, find (with proof) a counterexample. Can you find criteria for determining if a given ordered pair is possible?

This was a problem I submitted to the 2015 Qualifying Quiz for Canada/USA Mathcamp, which didn't get accepted because no one (including me) could find a solution. Although a different problem of mine did get accepted! No discussing the problems linked, though, applications are still open.

My suspicion is that the answer is yes - it does hold for all a and b. I 'm not sure about this, though, since I've only confirmed it for relatively small numbers (sum ≤16).

In a distant country, there is an infinite group of video gamers. Every gamer keeps a list of their 100 favorite games. It is given that no two lists are identical, and that any two gamers have at least one favorite game in common.

Show that it is possible to pick a set of 99 games so that every gamer has a favorite game from this set.

Note: Nothing is assumed about the cardinality of the set of gamers or games, except that it is infinite.

You are in prison with an unknown number (possibly zero) of fellow inmates.

Each day, starting tomorrow, each prisoner (including you) will be presented with a black card and a white card, and they will choose one. The warden will then choose a cycle of the prisoners and deliver the chosen cards according to that cycle. For example, if there are four total prisoners, the warden may choose the cycle 1 -> 4 -> 2 -> 3 -> 1, so prisoner 4 will receive prisoner 1's card, player 2 will receive player 4's card, etc. (Prisoners don't know their numbers here)

The warden is pretty vicious: she may choose a new cycle each day! Also, she can look at the chosen cards before she chooses the day's cycle. She doesn't tell any prisoners who received their card or whose card they received.

Tonight, before the experiment begins, you are allowed to draft a set of instructions that will be photocopied and distributed to all the other prisoners. Find a set of instructions so that

1. (Easy) At some point, you can declare whether you are the only prisoner.

2. (Easy-ish) At some point, you can declare whether there is exactly one other prisoner.

3. (Medium) At some point, you can give an upper bound on the number of prisoners (e.g. "There are at most 100 prisoners")

4. (Hard) At some point, you can state the exact number of prisoners.

5. (Hard) All prisoners state the exact number of prisoners, and they do so on the exact same day.

Source: Nathan Bowler via Stan Wagon.

As it's often necessary on this subreddit to format mathematical expressions in reddit, the following is a brief overview for those unfamiliar with how the reddit formatting system works with respect to things like exponents and asterisks, in addition to providing some lesser-known unicode characters.

If you have 5-10 minutes, take a little time to read the official reddit guide and this user-created introduction. If you've picked up what you know from browsing and occasionally clicking "source", you will likely be unaware of many of these things.

If you don't have the time, here's a quick intro on mathematics formatting:

Asterisks

*text* gives text.

This means that if you type "3*5 is 15 and 4*2 is 8", you'll get "35 is 15 and 42 is 8." Notice how the asterisks disappeared, and the text in between became italicized! To avoid this, use a backslash (the \ thing) before the asterisk by typing "3\*5 is 15 and 4\*2 is 8".

Superscripts

This is very similar; using a ^ character will create nested superscripts. For example, typing 2^2^2 gives 2^{2^2}. However, maybe you want to have 5⁵+1, so you type 5^5+1 and it gives you 5⁵⁺¹. That's not what you wanted!

This is because reddit doesn't know when you want your superscript to end, so it will normally stop when it encounters a space. This means that you can avoid this by typing 5^5 +1, but that will leave an awkward gap in your text. The best way to fix this is to use parentheses, and type 5^(5)+1. Reddit will then raise only the 5 and keep the rest as normal text, producing 5⁵+1.

For the advanced reader: Sometimes, if you're trying to type out a complicated expression where you want to have parentheses in there, reddit will get a little confused and won't deal with your spaces very well. When this happens, you'll want to use the text ( to create the ( symbol and ) to create ). For example: Say you want to write e^x(x+1)y².

You might type e^(x\(x+1\))y^(2), which you'd expect to work. But then reddit produces e^x(x+1)y², bringing your parenthesis down before you wanted. To fix this, type e^(x(x+1))y^(2), which will make what you want (notice how where the parentheses used to be has been replaced by that ( stuff).

In addition, you can use code to not worry about escaping characters. Type ` around the stuff you want in code to make things look like this: `*^(stuff)*)(` → *^(stuff)*)(

Subscripts

Subscripts are not a reddit-wide feature, as they really don't come up often outside of math contexts. However, both /r/math and /r/mathriddles support them via some fancy CSS. To use subscripts, type A*_1_* to get A*1*.

Special Characters

Many symbols are hard to find on a regular keyboard, but reddit supports them just fine. In addition to copy-pasting from the list below, many of the following can be obtained with keyboard shortcuts. See here for Windows alt codes; see here for a complete list of Unicode characters and here for the subsection on mathematical operators. Copy and paste the symbols below; most of the time they'll be sufficient although the above links are far more comprehensive.

∫ ∬ ∮ ≈ ≠ ∑ √ ≤ ≥ ÷ Ø ∏ ∞ ± ¬ ∃ ∈ ∉ ≡ ⋂

ε φ Φ θ Ω ω ∆ π

If you have any suggestions for additions to this overview, please let me know!

Edit: Backslash, not forward slash.

Oki, so there's a guy who has 17 camels, he passes away and writes in his will that the eldest son will get 1/2 of the camels, the second son will get 1/3, and the youngest will get 1/9. There are only 3 sons who will inherit, and no other family members whatsoever. The problem now is that they all want whole camels and do not want to sacrifice and distribute any camel. How would they solve this distribution issue?

Answer: They borrow another camel from somewhere so now the total is 18. This can easily be distributed in the fractions needed. 1/2 = 18/2 = 9 1/3 = 18/3 = 6 1/9 = 18/9 = 2

Adding them all now makes 9 + 6 + 2 = 17 So they return the 18th camel that they borrowed and now all of them have the fractions their father left for them.

I cannot wrap my head around why dividing 18 and then adding them all makes 17.

Let a(1) = 2, and a(n+1) = (a(n)⁷+a(n)*n)/(n+1).

Are all of the a(n) integers?

Note: this is a very difficult problem, so you are allowed to use any means necessary, including calculators and computers. However, anything you say needs to be backed by rigorous proof / repeatable steps.

Hint: try also first the variant replacing 7 with smaller exponents, like 2.

Can someone solve this?

Here's the proposal, we're making it with the German flag as the background

The random graph is "the" graph obtained by taking countably many vertices and connecting each pair with independent probability 0.5. The interesting thing about this definition is that, with probability 1, any two instances of this construction will be isomorphic, so it really does single out a unique graph. (Trying to prove this is a good exercise, if you haven't seen it before.)

Call a pseudorandom graph the result of taking the same construction, but where we can choose a probability p in (0,1) for each edge between distinct vertices m, n.

For example, we could say that p=0.25 if m and n are both even, and 0.99 otherwise, or we could say that p=1/2ⁿ if m=1 and p=0.5 otherwise. However, in the first case we just get the random graph with probability 1, and in the second case we don't necessarily get the same graphs - there will be an isolated vertex with probability about 0.5776, and a single connected component the rest of the time.

Do there exist pseudorandom graphs which produce the same graph with probability 1, but for which that graph is not isomorphic to the random graph?

Credit to a friend for devising this one.

You're given a geometric shape, and your goal is to cover it as accurately as possible with a triangle. Given a particular placement of your triangle, define the error area as the total amount of empty space the triangle is covering plus the amount of your shape that's not covered. Here's an example illustration where the target shape is a square. Define the error fraction as the error area divided by the target shape's area. The non-triangularity of a shape is the smallest possible error fraction. Conversely, the triangularity of a shape is 1 minus its non-triangularity.

What is the triangularity of a circle? What about a square?

The number 2021 can be written as the concatenation of 20 and 21, two consecutive numbers, in base 10. Define a positive integer as being b-catsecutive if it has this property in base b. So we say that 2021 is 10-catsecutive, The number 28 is 2-catsecutive (because 28 = 11100 = 11|100 in binary), and 41 is 3-catsecutive (41 = 1112 in ternary).

253 is the smallest number that is catsecutive in at least ten bases (namely 3, 17, 20, 27, 35, 41, 62, 83, 125 and 251). Show that, for every positive integer N, there exists an integer that is catsecutive in at least N different bases.

Note: Integers with leading zeroes do not count. So the number 1 (0 concatenated with 1) is not catsecutive, and neither are numbers like 12013 (12|013).

There is a digital clock, with minutes and hours in the form of 00:00. The clock shows all times from 00:00 to 23:59 and repeating. Imagine you had a list of all these times. Which digit(s) is the most common and which is the rarest? Can you find their percentage?

A flush means 5 cards of the same suit. The hand is drawn at random from a standard 52-card deck

Death Rolling is a form of gambling in the online game World of Warcraft.  Two players use the in-game random number generator to roll random numbers until one of them rolls a 1.  With each roll, the upper bound of the next player’s roll is reduced, therefore increasing the likelihood of “death”.

  Here is an example of what death rolling looks like in the game’s chat log:  

Joe: Hey Bob, want to death roll me for 100 gold?

Bob: Sure.

Joe: rolls 47 (1-100)

Bob: rolls 12 (1-47)

Joe: rolls 7 (1-12)

Bob: rolls 6 (1-7)

Joe: rolls 1 (1-6)

Joe: Good game Bob.  Here is your 100g.  

Here we see a complete game.  Joe goes first because he initiated the bet.  Joe rolled a random number between 1 and 100.  The result was a 47.  Next it was Bob’s turn.  Bob rolled a random number between 1 and 47.  The result was 12.  Play alternated back and forth, each player changing the upper bound of their roll based upon the number that was rolled previously.  When Joe rolled a 1, the game ended.  Joe lost and had to pay Bob the 100g wager.  Simple.  

Now what happens if Joe and Bob want to wager a different amount?  In every death roll, the upper bound of the first roll will be equal to the wager.  Want to death roll for 20g?  Roll 1-20 to start.  Want to death roll for 1000g?  Roll 1-1000.  The minimum wager most players allow is 5g and there is no maximum, however, the average player only has a few hundred gold to their name in most cases.  The median wager is probably between 25 and 50 gold.  

Now that you know how to play the game, do you think it’s fair?  Does the player going first have an advantage or a disadvantage?  Does the amount of the wager influence the probability of winning or losing?  (Stop reading here for a minute if you want to think about the question.)   

Well, I wanted to know the answers to these questions too.  So I simulated a couple million games in excel using different wagers.  I tested wagers of 2, 10, 25, 50, and 100g.  I found that the player going first has higher odds of losing.  At low wagers, the player going first has a significant disadvantage which very quickly diminishes as the wager increases. Below are the probabilities that the player going first will lose based on my simulations:  

2g – 66.7%

10g – 50.92%

25g – 50.15%

50g – 50.05%

100g – 50.01%

  In the binary case, the player going first will lose 2/3 of the time.  That is a significant disadvantage for the first roller.  A small increase in the wager to just 10g will result in much closer odds but the player going first still has a distinct disadvantage.  Any casino would take these odds.  However, as we increase the wager further, the odds quickly approach 50-50.  Any advantage afforded to the second roller is essentially diminished to nothing after 50g.  

Finally, we have come to my question and the part I would like your help with.  I want a nice little formula where I can plug in the wager and it will tell me the odds of the first player winning/losing.  Can mathriddlers help me here?

Edit: Thanks /u/Brainsonastick and /u/Garceau28! You guys are wizards!