Hey, this is Presh Talwalkar. Alice and Bob are on a game show. Each is secretly told a whole positive number. They are told the two numbers are consecutive, but neither knows the other person’s number. For example, if Alice is told 20 she does not know if Bob was told 19 or 21. And if Bob is told 21 he does not know if Alice was told 20 or 22. The point of the game is to guess the other person’s number, but there are rules to the game. Alice and Bob cannot communicate with each other and they are not allowed to plan a strategy either. The two are in a room where a clock rings every minute. After the clock rings either player can call out a guess of the other player’s number or they can both stay silent. Here’s a clock that rings every minute on the minute. The game continues until Alice or Bob makes a guess. After the first guest is made the game ends. Alice and Bob win 1 million dollars if the guess is correct and they lose and get nothing if the guess is incorrect How can Alice and Bob best play this game to win? Each knows the other is perfect at logical reasoning. Can you figure it out? Give this problem a try and when you’re ready keep watching the video for the solution. At first it seems like Alice and Bob can do no better than random chance. If Alice is told 20 for instance there is no way to know if Bob has 19 or 21. But since Alice can limit Bob’s number to two possibilities, she can at least have a 50% chance of guessing correctly. Bob has the same issue: if he was told the number n, then he cannot be sure if Alice was told n minus 1 or n plus 1. If Bob guesses between two possibilities then he also has a 50% chance of guessing correctly. Acting individually Alice and Bob have a 50% chance of guessing the other person’s number and winning the game. Remarkably they can do much better. There is a method they can win 100% of the time if they use logic perfectly. One key detail is the two are given positive consecutive numbers. When Alice gets the number n, she usually has to consider if Bob has n minus 1 or n plus 1. But suppose Alice gets 1. Bob could have 0 or 2 But 0 is not a positive number so Bob must have 2. (Search your feelings Alice, you know it to be 2.) A player that gets 1 will know the other person has 2 and since the player is sure the player will guess on the very first ring on the clock, they will win 1 million dollars for sure. This was from Alice’s perspective But of course if Bob got 1 he would also figure out that Alice has 2 on the first ring of the clock. But what about larger numbers? We can continue the logic. Suppose Alice gets two and the first ring of the clock passes. Bob could have 1 or 3. But if Bob had 1 he would have guessed on the first ring of the clock. So Bob does not guess after one ring then Bob must have 3. Alice realizes this after the first ring and she will guess on the second ring of the clock. In this way Alice and Bob will win 1 million dollars for sure. The logic would be similar if Bob got 2 and the first ring of the clock passes he would know that Alice had 3. So now what about larger numbers? We can continue the logic inductively. Suppose Alice gets n, and the clock rings n minus one times with no guess from Bob. If Bob had n minus one he would have guessed on the n minus one ring of the clock. Since Bob did not guess then Bob must have had the larger number n plus one. Alice therefore guesses that Bob has n plus one after the clock rings exactly n times. They will win the game for sure and get 1 million dollars. Bob would of course reason similarly. So in summary Alice and Bob can always win the game when they’re given two numbers n and n plus 1. The person who gets the smaller number n will always guess after the clock ring exactly n times that the other player has n plus 1. This will be correct, and they will always win the game. Did you figure it out? Thanks for watching this video. Please subscribe to my channel. I make videos on math and game theory. You can catch me on my blog Mind Your Decisions, which you can follow on Facebook, Google+ and Patreon. You can catch me on social media at preshtalwalkar. And If you like this video please check out my books. There are links in the video description.

Without scrolling through the comments or watching the video: When your number comes up (that many minutes have passed), guess the number higher. It's the only way to know for sure that the other number isn't lower.

I knew this one already, and in all these years it has never convinced me, even though I had found the solution with my colleagues when posed the riddle.

I've never been able to find a convincing way to explain why the solution makes sense, and neither, actually, why it should stop working for bigger numbers.

I just discussed this with my father, and he pointed out that this would only work if one were to be absolutely sure that there are no other solutions. Until it's possible that there are other logical ways to solve this, this paradoxically loses its status as a viable solution, because it's based on the assumption that both player are following the same line of reasoning. Right?

After I saw the tree/cells problem on this channel (which is exactly the same problem btw), the provided solution seemed clearly logic to me. But with this second look at the problem, I think this solution is flawed.

My problem with this solution, is that it seems "logic", but it's not more logic than thinking, as Bob :

"I'll wait until we are at the (n+1)th minute where n is my number, and say n+1 is the other's number"

Because, if one thinks like that, and knows the other thinks the same way, they'll win.

It is obviously "a strategy" of some sort, so it isn't considered as a right answer.

But the provided solution is not better, because Bob KNOWS that Alice CANNOT have 1 or 2 or 3 since the beginning. It is not logical at all to think "if she had 1, she would have guessed my number" because it's just totally impossible. This reasonning assumes to consider clearly impossible situations (Alice and Bob both know that neither can have the number 1,2,3,4,5…) as possibilities, only to use them as a legitimate way to count the minutes and claim it is logic.

This "logical way of thinking" of Bob and Alice is just a strategy like any other, so yes, they'll win if they both think like that.

It is probably not the same thing at all, but it makes me think of the unexpected hanging paradox, the reasonning of the prisonner seems perfectly logic, but it's incorrect anyway.

Finally – A riddle on this channel which I managed to solve.

Yawn

What I don't understand is how this seems impossible? The answer is extremely obvious . This video needed to be 3 seconds long to explain itself.

What if they don't know who has the larger number?

If he has 19, he knows the minimum she could have is 18, and that means the minimum possibility she could think he had would be 17. But he knows that if she thought he had 17, she would think the minimum he thinks she could have would be 16. And if she thought he might think she had 16, then she would think the minimum he thinks she might have would be 15. But he knows she can't actually think he has 15, because he has 19, and that means the lowest she could have would be 18. And so, we come to the problem here. It only works if they have great logic and can conveniently forget the restrictions. They must follow the hypothetical logic all the way, and ignore the practical logic that says the other person cannot think they have a number less than 2 below what they actually have. This means either they must have imperfect logic, or they must both for some reason decide to ignore the information they have, because really they have too much information for this solution to work as described.

Not logical as fuck, dislike.

easy pezy

“Each knows the other is perfect at logical reasoning” is not enough. It needs to be what is called “common knowledge”. Everything of the form of “A knows that B knows that … that A [B] knows that B [A] is perfect at logical reasoning” must be true for the game to work for arbitrarily large numbers.

Love it

they can't always win the game because you said they can't plan strategy so they can't do that you brought this on yourself dued

Can't believe I actually guessed it right 😂😂

Total bollocks. "Perfect logic" is undefinable. There are an infinite number of solutions to this, so the probability of A and B using the same solution is essentially zero.

This puzzle is beyond stupid.

Suppose Alice has 1913453145

She will die before she can make a guess.

Slipped into your secret Brooklyn accent at 5:00 didn't ya?

Assumptions should not be made in logic puzzles. At higher numbers the lower number logic does not hold and they have to "assume" the other player would guess at the Nth number (supposing that player has N). This puzzle is flawed.

This is wrong, it doesn't work for any numbers above 2 & 3.

The concept about the smaller number not being a positive number does not apply on any combination of numbers greater than 1 & 2.. Therefore the question and solution presented is incorrect for all numbers greater than the above mentioned combination.

I considered the case where someone as 1 but I didn't reason up to the solution; I guess I couldn't decide if Alice or Bob could hear each other's guesses. You said they couldn't communicate.

0 is positive

0 is not positive anymore.

lol this is the red face/blue face puzzle worded otherwise

This was surprisingly easy

I fail to realise why your number and the number of the other is dependent on when someone doesn't give an answer. Also, this sounds like something that has to be strategized beforehand.

So they have to figure out this logic in their head in a minute and be able to connect them in game without communicating. Seems impossible. Also if it's any number, like for. eg the authors could give them 329, how can they both know if their counterpart is continuing the logic or not, when they are just passing over 300 times.

Give them seven digit numbers. Then they have to guess.

what if alice has 1 and bob has 2 and both are silent at the 1st ring?

What happens if it was 3 and 4 and he gesses 4 and it gets out 2 lel

This is not logic, it's a strategy. Unless you watch constantly videos like this, a person who doesn't solve this kind of puzzles don't get the idea.

I understand how it's solved. But this is mere strategy.

Like some others here, I'm not convinced that the proposed solution is purely logical. An assumption is made that a strategy exists, and that it is the

onlystrategy that can be arrived at by reason alone, therefore Alice and Bob will both find it if they reason perfectly.The proposed strategy certainly works, but it's not the only one. Generally the problem is to encode a positive integer N using some positive integer R of rings. The solution R = N is arguably the "simplest" or "most obvious", but those are subjective judgements, not logical conclusions. There are other strategies — for example, R = 2N would also work.

To argue that Alice and Bob are

guaranteedto both arrive at the R = N strategy, you would somehow have to prove that there isno otherline of reasoning that would lead to a different strategy. I'm not sure what such a proof would even look like. Given that infinitely many working strategies obviously exist, it seems similar to proving that a theorem is true but can't be proven. Where's Kurt Gödel when you need him…Lol the clock that rings every minute was ringing like every second in the video.

Really enjoying your channel brother, I discovered it a few days ago & I can't stop trying all the questions. It's been a couple of decades since I did problems like this in school, I prefer cryptic crosswords, but you've got me hooked, where were u 25 years ago? I could have been winning the Nobel peace prize instead of doing deliveries everyday. But seriously, excellent channel & great mental excercise. Thanks

& I got this one right bcause it was more about logical reasoning, rather than the complex algebra & equations on some of your questions, but I'm learning a bit more each question, which is what it's all about. In school I wasn't interested at all, but now I can't get enough of mathematics, I think I've finally gotten old.

Change alice to vegana😂

Ehat about the numbers 99 !

The answer doesn't make sense. You already know your number and that your partner's number is either "n-1" or "n+1." Hence, it is not logical to start with "what if my partner's number is 1" and come up with the inductive reasoning solution. You can't suppose the premise "what if my partner's number is 1" if you already know it's not 1.

Oh my…. More absurd stuff on this site. Your opponent has to be as smart as you are in order for this to work. Pointless exercise in another "doesn't exist and never will exist in the real world" silliness from this website. Skip.

win

Assume the numbers are 0 and 1. On the first turn, the player who has 0 knows that the other has 1 and calls it.

If they do not, then on turn 2, the player who has 1 knows that the other has 2, and calls it.

If they do not, then on turn 3, the player who has 2 knows that the other has 3, and calls it.

…

if they don't, then on turn N, the player who has the number N-1 can guess with certainty that the other number is N.

This is flawed in so many ways!

Even if they were told to think of a way to use the clock as a counter, this "puzzle" still would be unsolvable!

There would still be still many possible ways how this counter could work. Do you start at 0, 1, 2 or 20, 30? Even if it displayed the numbers, there would still be different ways to resolve this. For example when see your number x +1 you say x or, like you do it, when you see x you say x+1.

If they are soo logical they might think of the absolute fastest way to count and to resolve, this gets rid of many possibilities, still infinite open.

I do appreciate your channel, but I am disgusted by your clickbaitey titles. Who could find this "seemingly impossible"?

This is the same riddle as with the cows and others…

Solution:

When the clock ends,

Alice guesses 21

and Bob guesses 20.

This puzzle doesn't work for numbers greater than 3.

Let's suppose Bob has 4 and Maria 5. After one minute without answer, they'll both know, indeed, that the other one doesn't have a 1.

Problem is: they already knew that at the beginning of the game!

Bob has a 4, so he knows that Maria has either 3 or 5. He also knows that if Maria had to guess his number, she would choose either 2, 4, or 6 (if she had 3, she'd pick either 2 or 4, and if she had 5, she'd pick either 4 or 6)

Same goes for Maria, she knows Bob has either a 4 or a 6, and that if he had to guess Maria's number, he could say either 3, 5, or 7.

So, to recap: Bob and Maria both know, that neither of them has a 1, and they also know, that the other one is aware of that fact.

Question: why would they wait one minute to confirm that fact, since they both already know it? They both know that they don't have a one, as they know that the other one knows that to.

So waiting a turn to eliminate the possibility of one of them having a 1 would be meaningless, ergo it would have to be strategicaly decided, wich is against the rules.

The answer can be N-1 as well.

But that depends on whether you can see if your partner has chosen to stay silent and

stillbe allowed to speak your guess.Because if Alice has N, she'd think that if Bob has N-1, he would think she has either N-2 or N. In that case, he would have spoken at N-1.

Except he doesn't, therefore he can't be N-1. Thus, he's N+1.

So she says he's that number, after N-1 rings of the bell.

"0 is not positive"

Signed zero says hi.

If they both guess at the same time, would that be a 75% chance of winning? I'd take those odds.

Solved in 1min. eZ …. also mayonaise is an Instrument

Alice's must be a natural number, as is given. If Alice's number were 1, she would instantly know that Bob's number were two – since 0 is not natural, and would call it out. If Alice's number were 2, she would not know if Bob's number was 1 or 3, and thus would not call out either number. If Bob were given 1, he would know that n=2, and would call out 2. If he were given 3, he would know that his options are 2 or 4, and since after one minute Alice still does not know, her number must be 4. This strategy continues, with each player counting the number of beeps.

I think I got that wrong somewhere, but I'm fairly confident the key lies in the numbers being natural.

I feel like the 50% thing is better than this time-consuming confusing thing

same logic as the tree problem

My head hurts now…

I hear people suggesting modulus (60) for really big numbers, which leaves me with the question "Would using modulus (60) work?", and if it did, leave me with a further question "Why use modulus (60) (as you could use modulus (4), modulus(8), modulus (10), or some other smaller modulus instead)?

But you're assuming that bob is as smart as Alice or vice versa and they both can Guess within one minute

Your rules says they cannot plan a strategy. This would involve them to plan a strategy ahead so your solution is crap. sorry my friend

Would not work for 100,000,000 minutes. They would die! 🙂

so easy .. wait until the clock matches the person number and guess the other number is the minutes+1

Why don't they just both say the number above them at the same time?

If either Alice or Bob has as much as a two digit number

that's going to be one really boring game show.

I would have done the right strategy, but I thought, when would I guess?

I thought if everybody guessed on the ring of the minute which is your number, then they can always guess the higher number, since the other one hasn't guessed one minute before. Of course this only works, if the other person has the same thought process, but if he doesn't you still have a 50% chance of winning.

After first minute, if no one answers, it is clear no one has the number 1, since the number 0 isn’t positive so 2 would be the only possibility. This means the number can’t be 1. The next minute, if someone has the number 2 they would guess since the only possibility is 3, but if they don’t they continue knowing no one has 2. After another minute, if someone has the number 3 they would guess it, they would guess 4. If no one guess, it is not 3. Keep going u til you get to the number.

The time this will take is equal to the lowest number anyone has BTW which I think is pretty cool.

If the other person stays silent during the first ring of the bell. that conveys the information that the other person does not have a 1. during the second ring, the other person will know that they don't have a 2 because if they had a 2 they would guess 3 since the other person cannot have a 1 (if they follow perfect logic). and you can continue until they come to the number in which Alice understands that Bob does not have a 19. and will then guess correctly.

HOLA

The logic is false as it presupposes the players are gaining information each time the clock rings and reasoning through bootstrapping. This is clearly not the case.

Let's see what Alice knows at the beginning. First, she knows she has 20. Therefore she also knows that Bob has either 19 or 21. But she also knows that Bob either thinks that she has 18 or 20 (if he has 19) or 20 or 22 (if he has 21). The error is in trying to bootstrap this logic further. In other words, Alice already knows Bob already knows that she doesn't have 1, 2, 3…16, 17. Thus these "silent" minutes gain him no additional information as to what she has.

Now there is a way for this problem to work, but that would entail a possible set of consecutive numbers which is far fewer than all positive consecutive numbers. If, for instance, the problem was that Alice and Bob had two consecutive positive numbers from 1-5 it would work. Let's say Alice had 3 and Bob had 4. Alice knows that Bob has 2 or 4. And Alice knows Bob knows that Alice has either 1, 3, or 3 or 5. If Alice had either 1 or 5, she would answer on the first chance. Since she doesn't, Bob knows she must have 3 and he answers on the second chance.

This might be a logical answer but I think that it is unrealistic to believe that this could actually work in practice

but i think it is such a strategy…

Got it. Sweet. Once in a while

I thought 0 was a number? Or is it now dropped from math? But Bob and Alice could hand signal maybe?

Alice, your number is……. googleplex! Muhahahaha!

Liked for star wars reference 😛

Hold on… The problem states the “clock rings every minute and after it rings they can make a guess.” That’s useless information. The clock is only telling them how long they have been in there. A buzzer going off every sixty seconds has nothing to do with the number they are guessing. The premise is not stated correctly. I determined if someone had 1 they could figure it out. I wondered why the information was being given about the clock because it has zero to do with how they guess. So I ignored it as some sort of red herring. Reword the setup and then your solution makes sense. Otherwise the only solution would be if one player had 1.

This riddle would make a lot more sense to me if the game would end and the contestants would lose if the clock bring an equal number of times as the greatest number a contestant holds.

This is a terrible puzzle, it isn't logical and the solution is incorrect. They both know their numbers are consecutive, so anything above 1 and neither person would conclude the other would use clock ticks to scale their guesses, it makes no actual sense.

I still do not get the answer. The solution given is one possible strategy. But the two are not allowed to plan a strategy together. In such case, why would one think that since N minutes have passed, the opponent number would be N+1?

I didn't get it but I'm disappointed it was a semantics answer.

The clock rings every minute, how long can a human stay awake?

10000 minutes is enough for any human to give up and take the 50% chance.

What about 9,999,998 and 9,999,999

Alice and Bob must be some crazy mathematician aliens

A simpler “explanation” of the solution: they both count the ticks of the clock. Whoever has the smaller number stands up at the appropriate tick. The other then declares the two numbers.

Say Alice was given the number 20 and Bob 21. Alice and Bob count the ticks of the clock. Alice stands up at the 20th tick. At which time Bob shouts “20 and 21”.

But this explanation breaks the rule of “no communication” between the 2 players! To abide by this rule, Alice simply shouts “We reached 20 ticks and he remained silent. I know I have 20 so he must have 21”.

I really enjoy your good work.

It’s easy: Alice and Bob just plan a strategy together. Sure, the rules say they can’t, but as they have perfect reasoning they’ll notice that breaking the rules has no repercussions. Hence the rule cannot be reinforced. They do not lose the game if they break it. Alice and Bob use this design flaw in the game to win $ 1000000.

How abot this: Alice has 20. She guesses 19, if it's right, they win. If it's wrong, then Bob (who had 21) will know, that alice has 20, and wins

Want an impossible question?

Bob is driving a car at 60 km/h to his work office and his son is playing football(soccer). If his son shoots the ball with all his power what is the temperature of his fridge? 10/10 Einstein can't do this

I don’t get it. Why is how much time passes related to their number? Why is that a thing?

There is no way the clock correlates to the numbers they have. If Alice has 21, how does Bob know that at 21 clock ticks? Absurd.

I have seen a lot of puzzles that use the same kind of solution, and I can tell you they are flawed and wont work. In the problem, each person is given a number and told the other person has one more or one less. In the solution, you state that the players come to a conclusion based on the possibility that they could have ANY number. You cannot come to a logical conclusion starting off with the statement "if bob/alice had 1" if it is an impossibility for them to have 1. In the example above, Alice had 20 and she is told that Bob has 19 or 21, therefore she would NEVER logically start with the assumption that bob has 1 and start the count from there. There are only 6 possible numbers for each person to wait on. From Alice's point of view, she knows that bob HAS to have either 19 or 21, so she can logically draw the conclusion that if Bob has 19, he knows alice has 18 or 20 and if bob has 21, alice either has 20 or 22. From Bob's point of view, he knows Alice HAS to have either 20 or 22, so he can logically assume that if Alice has 20, she knows Bob has 19 or 21 and if Alice has 22, Bob has either 21 or 23. This means that between the two of them, the lowest number that would be possible between them is 18, so it is a logical impossibility to include the number 1 or 2 or so on.

Sooooo…… I guess I got it right?

Worst gameshow ever…

4:22 Things kind of break down here. If you get 3, why would you expect an answer after the first ring? There is no way there COULD be an answer, so logically it should be disregarded.

Lets put this another way.

You have X.

For each strike of the clock N, you wait until X=N, Once that occurs, you call out X+1. Your partner follows the same strategy and you are guaranteed to win.

If Bob have 10 and Alice has 9, then on the 9th strike, Alice calls out that Bob has 10.

if Bob has 10 and Alice has 11, then on the 10th strike, Bob calls out that Alice has 11.

What if their nunbers are like 100000 lmao

They should both guess n+1 and 1 of them will always win

I don’t think it was said that they

bothhave to get it rightEdit: they have to say it at the exact same time

Edit 2: This was meant as a loophole kinda joke thing

This is an

incrediblyinformal way of saying but I understand it nowLet’s say that

A has 4

B has 5

So

A thinks “B is either 3 or 5”

B thinks “A is either 4 or 6”

A then thinks that if B is 3, then B will think that A is either 2 or 4

If B thinks that A is 2, then this hypothetical A will deduce that B is 1 or 3

If the most recently mentioned hypothetical B says nothing after 1 minute, the hypothetical A concludes that that hypothetical B is 3

Since this guessed A now should know that B is 3 if A doesn’t guess in the second minute, B knows that A is not 2

B then knows that A is 4 (this is when A guesses that B is 3)

If B doesn’t guess that A is 4 in the third minute, A then knows that B is 5 (this is for sure)

A guesses that B is 5 on the fourth minute

(If A didn’t guess this, B would know that A is 6

If B didn’t say *this*, A would say that B is 7

etc.)

This method doesn't work for 3 or higher. If you have 19, you know the other person has 20 or 18. But when the clock rings, neither of you have narrowed down any options. That person still knows the options are either 20 or 18. Nobody who's perfectly logical would think on the first turn "OK, I ruled out her having a 1" when that was never an option. They certainly wouldn't then count each round assuming the other person has the exact same thought process and go "OK this round I ruled out her having a 2". If you can plan a strategy ahead of time to call out your number +1 when the number of rings reaches your number, then it would work, but you have to agree to that strategy ahead of time.

So in the example that Alice has 20 and bob has 21, if Alice guesses 21, then they win. However, if Alice guesses 19, Bob now knows that Alice's number must be 20, because the two of the have consecutive numbers, so why would Alice guess 19 if she had 22. It wouldn't make sense.

And, the riddle never specified that if one guessed wronng, the game would end. It just said one of them had to guess right.

the first examples were taking advantge that 0 is not positive , but when both numbers are above 3 there will be no advantges ; just because it worked for the small numbers (because 0 is not positive ) you can not carry the same logic when all of the numbers are positive ;if the logic in this solution was right fermats last therome would have been a pice of cake.

I hate how surreal the solution to these kinds of problem seems. They really count on how the 2 of them wont panic and can calmly think about things logically in stressful situation.

I can't believe it . I've actually figured it out . I was thinking then suddenly the answer came to my mind . This video has given me confidence . 😁😁

Wait until "your number – 1" no. of minutes have passed. For example if your number is 18, wait till 18 minutes have passed. if the other person has not called out, your number is higher. Otherwise your number is the lower of the two.

This is because: say Alice got 1, she would immediately know that Bob got two. as for Bob, he would not know if Alice got 1 or 3, but if she doesn't call in the first minute, he will know that she got 3. Now what if Alice got 3? She will not know if Bob got 2 or 4. But if Bob got 2, he would know after the first minute so he would have said it in the second minute.(Because he would just be checking if Alice calls out "1" in the first minute) But if Bob doesn't say it, it means he got 4. And so on for every number after that. Now to watch the video for the solution! 😁

I guess the real answer is n – 3.. That works at numbers higher than 3

If Alice is 20, She would figure that if Bob is 19 he would knew her number at the minute 18 (acording to your logic n – 1) but she saw him keeps quite then she would know that he isn't 19

So till now it's n – 2

Then she would also figure that He would use the same logic, So if he was 19 he would know at the 17th ring, When she saw him keeps quite at the 17th ring then she will know that he isn't 19, Then she will know his number is 21 for sue at the 17th ring

Do i miss something?