Puzzle: probability to guess one number given a second number...

[Speakpigeon]

What solution do you give to this puzzle?

I give three different formulations, if that can help.

The original one is supposed to be the last one here...

In this game there are two people, labelled Dealer and Player. The Dealer writes two different numbers down on two slips of paper, and seals them in envelopes. It doesn't matter what the numbers are. They can be 0, 5, 4081922, -382.393193, pi, anything. They just have to be different numbers. The Dealer then hands the two envelopes to the Player in any order they please. The Player then selects one envelope to open. They then must decide whether the number in the other envelope is greater or less than the number in the envelope that was just opened. Obviously it's easy to win 50% of the time. The challenge of the game is to come up with a strategy which wins more than 50% of the time. Can you think of a strategy?

You are on a game show. The host has chosen two different whole numbers (integers) and has hidden them behind doors A and B. He allows you to open one of the doors, thus revealing one of the numbers. Then, he asks you: is the number behind the other door greater or smaller than the number you have revealed? Your task is to answer this question correctly with probability strictly greater than one half.

Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you’ve seen. Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

I give here my own view on this, hidden to not influence anyone...

Show

We don't have any relevant information as to the relation between the second number and the first number. 50% is the best anyone rational should expect.

EB

 

[Gun Nut]

It's mathematically impossible to beat 50/50 here. There may be a non-mathematical (soft) solution, though.
Most people cannot remember more than 7 digits, and one may suppose that a larger length number would not be chosen to write down by the average person.
therefore, an artificial bound of the selection pool of 9999999 may be guessed with some defensibility.
For the revealed number, if it is larger than 5555555, then one may have greater odds of success by guessing "lower", and the converse.

But, strictly speaking, since the selection pool is theoretically infinite, there is no middle point for which to base your guess.

Can we know the size of the piece of paper the number is written on and the size of the numerals written? Bounds for the size of the number can then be interpreted.
Can we observe the writing of the number? The time taken to print the numerals (and other visual keys) can indicate relative length of the two numbers.

 

[repoman]

It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

Seeing how bluffs and reverse bluffs work would be fun.

On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.

 

[steve_bank]

Put two balls in a box with different numbers on them. Reach in and pick one. What is the probability it is less than or greater than the one left in the box.

I'd say in a series of random trials the probability is 50/50 that you always pick the greater or the lessor.

If I always bet that the pick will be the greater I will be right 50% of the time.

It is an easy experiment to do at the kitchen table.

In the envelope puzzle it depends on how the player makes the choice. If he picks one condition all the time it is 50/50. If he randomly picks one or the other then I'd have to review joint probabilities. I believe the probabilities multiply so he would be right 25% of the time. If the player picks greater a percent of the time and lesser a different percentage it can be calculated, can;t do it off the top of my head.

The experiment.

Put two balls in a box with different numbers.

First bet it is always the greater. Pick a ball from the two in a box 20 times and record how many times it is > or <.

The flip a coin before each pick. Heads bet > tails bet <. Record how many times you would win based on the coin toss.

Call the exercise an intro to probability and statistics for the inexperienced philosopher investigating how math work and how math develops. Games of chance were the germesis of probability and statistics.

 

[steve_bank]

We don't have any relevant information as to the relation between the second number and the first number. 50% is the best anyone rational should expect.

Astounding insight. Does this mean if I randomly pick between two possibilities it will always be 50/50? Are you sure about that?.That would have serious implications.

 

[steve_bank]

Looks like I made a mistake.

I took two coins and put an x on one and a o on the other. Then I put them in a box.

On a third coin I put an x on one side and an o on the other, and put it in a second box.

I shook the boxes and dumped the x/o coin on the table and drew one of the two coins in the other box. If the two coins matched it was a win, mismatch and a loss.

My first run of 20 trials showed 35% win. I repeated with a better job of randomizing and it was close to 50%.

It seems counter intuitive to me but the experiment says otherwise. I saw it as two independent uncorrelated random variables thinking the two probabilities should multiply.

 

[Speakpigeon]

There may be a non-mathematical (soft) solution, though.
Most people cannot remember more than 7 digits, and one may suppose that a larger length number would not be chosen to write down by the average person.
therefore, an artificial bound of the selection pool of 9999999 may be guessed with some defensibility.
For the revealed number, if it is larger than 5555555, then one may have greater odds of success by guessing "lower", and the converse.

But, strictly speaking, since the selection pool is theoretically infinite, there is no middle point for which to base your guess.

Can we know the size of the piece of paper the number is written on and the size of the numerals written? Bounds for the size of the number can then be interpreted.
Can we observe the writing of the number? The time taken to print the numerals (and other visual keys) can indicate relative length of the two numbers.

We could always modify the premises and a realistic investigation would certainly try to do that but I don't think we could determine the probability that each number be picked out, even if we assumed 0 above some particular number deemed too big to think of or to write down etc.

I agree with your idea about 7 digits but that would only work for most people, as you say.

It's mathematically impossible to beat 50/50 here.

That's what I think. However, one interesting twist is that there is an infinity of numbers to choose from for both the first and the second number so that I don't see how we could make any rationally justified calculation of probability. The probability for any one number to be selected is zero or infinitely small if that's different. So, I would say that the best we can do is just trust our intuition, which says 50/50.

Do you have a calculation to justify 50/50?
EB

 

[Speakpigeon]

It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

Seeing how bluffs and reverse bluffs work would be fun.

And some people seem to believe they have a strategy to beat the 50% mark. They would loose their shirt.

On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.

Any way we could suss the probability of the second number given the first would do. But I think this is what we don't have in this case.
EB

 

[repoman]

It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

Seeing how bluffs and reverse bluffs work would be fun.

And some people seem to believe they have a strategy to beat the 50% mark. They would loose their shirt.

On the topic of method to generate the number, what about the decent chance of having each number generated by the same mathematical relation.

For example if the first number was 64.81 i would guess lower because 36.49 or 25.64 might be the lower. Unless that would be a feint.

Any way we could suss the probability of the second number given the first would do. But I think this is what we don't have in this case.
EB

I misread the scenario. But if A knew what both numbers were and B picked a random envelope and A also saw that number then this would be a good case for a research study. It has probaly already happened.

I wonder how much of a winning spread A could get if skilled at "head games" versus a skilled or nervous opponent B.

 

[Speakpigeon]

I misread the scenario. But if A knew what both numbers were and B picked a random envelope and A also saw that number then this would be a good case for a research study. It has probaly already happened.

I wonder how much of a winning spread A could get if skilled at "head games" versus a skilled or nervous opponent B.

You mean like in the old cartoons of Fritz the Cat, where A could read the figures on the eyes of B as if reading on a jackpot machine the dollars he expects to win?

That would seem a bit far fetched but then, the question is whether there's a strategy to beat the 50% mark. Some dudes think they've demonstrated there is such a strategy to win 50% of the time plus a delta which is infinitesimally small.
EB

 

[repoman]

Let me dispense with numbers to get to my point.

Lets say there are red and black marbles and randomly (coin flip, whatever) that red or black is chosen as the winning marble. Player A knows this and player B does not. If black wins and B picks it (meaning B is left with red) and there is a ante/fold betting system like in poker, wouldn't a good "head game" player B in person with visual cues be able to win?

Even having a 55% chance of winning is a massive "win".

 

[steve_bank]

The 50/50 theoretically comes from the definition of random variables..

A random variable is said to have no memory as a general definition. Flip a coin 100 times and the last HT sequence does not affect the outcome of the next trial which is always 50/50. Random varibles are said to be uncorrelated, oe variable des not affect the ourtsome of the other.

Toss two 6 sided die one and then the other, each die is an independent random variable. The probabilities of each die tossed together are uncorrelated.

Chronic gamblers think they see patterns of dice in craps that can predict the next outcome. In statistics it is called subjectivism. The belief that knowledge of previous random events can predict future random events.

Rolling a die or picking one of six balls in a box is the same statistical experiment.

Put 100 balls 10 red 90 blue in a box. Pick one, put it back, shake the box, and pick again. Given enough samples the percentage of red balls will be 10% of the samples balls. It is called ransom sampling. One sample does not affect the probability of the next sample. The samples are said to be uncorrelated.

In the envelope example if the mixing and sampling process are not random then all bets are off. If the dealer does not randomly shuffle the envelopes it is not a random process. Same as with shuffling a card deck. A skilled player, aka card sharp, can affect the odds by biasing the shuffle.

Put a red and blue ball in a box. If you always bet red when a ball is randomly selected you will be 50/50. If you do not you may have a string of wins or a string of losses other than 50/50. If you have ever been to LV at the craps table you can watch the streaks. Some people wait for streaks and bet on thrower thinking he is 'on a hot streak;.

 

[steve_bank]

Random variables. In a coin toss if the toss is random then the probabilities totaling 1 are split between two states H T, 1/2 and 1/2. Likewise the probability of a die toss is 1/6.

For a balanced coin tossed with a random trajectory, no biased pattern, the total probability of 1 will be divided by the sample spaces 2. The probabilities of the sample space must total 1 or 100%.

https://en.wikipedia.org/wiki/Random_variable

In probability and statistics, a random variable, random quantity, aleatory variable, or stochastic variable is described informally as a variable whose values depend on outcomes of a random phenomenon.[1] The formal mathematical treatment of random variables is a topic in probability theory. In that context, a random variable is understood as a measurable function defined on a probability space whose outcomes are typically real numbers.[2]


This graph shows how random variable is a function from all possible outcomes to numerical quantities and also how it is used for defining probability mass functions.
A random variable's possible values might represent the possible outcomes of a yet-to-be-performed experiment, or the possible outcomes of a past experiment whose already-existing value is uncertain (for example, because of imprecise measurements or quantum uncertainty). They may also conceptually represent either the results of an "objectively" random process (such as rolling a die) or the "subjective" randomness that results from incomplete knowledge of a quantity. The meaning of the probabilities assigned to the potential values of a random variable is not part of probability theory itself but is instead related to philosophical arguments over the interpretation of probability. The mathematics works the same regardless of the particular interpretation in use.

As a function, a random variable is required to be measurable, which allows for probabilities to be assigned to sets of its potential values. It is common that the outcomes depend on some physical variables that are not predictable. For example, when tossing a fair coin, the final outcome of heads or tails depends on the uncertain physical conditions. Which outcome will be observed is not certain. The coin could get caught in a crack in the floor, but such a possibility is excluded from consideration.

The domain of a random variable is a sample space, which is interpreted as the set of possible outcomes of a random phenomenon. For example, in the case of a coin toss, only two possible outcomes are considered, namely heads or tails.

A random variable has a probability distribution, which specifies the probability of its values. Random variables can be discrete, that is, taking any of a specified finite or countable list of values, endowed with a probability mass function characteristic of the random variable's probability distribution; or continuous, taking any numerical value in an interval or collection of intervals, via a probability density function that is characteristic of the random variable's probability distribution; or a mixture of both types.

Two random variables with the same probability distribution can still differ in terms of their associations with, or independence from, other random variables. The realizations of a random variable, that is, the results of randomly choosing values according to the variable's probability distribution function, are called random variates.

https://en.wikipedia.org/wiki/Randomness

Randomness is the lack of pattern or predictability in events.[1] A random sequence of events, symbols or steps has no order and does not follow an intelligible pattern or combination. Individual random events are by definition unpredictable, but in many cases the frequency of different outcomes over numerous events (or "trials") is predictable. For example, when throwing two dice, the outcome of any particular roll is unpredictable, but a sum of 7 will occur twice as often as 4. In this view, randomness is a measure of uncertainty of an outcome, rather than haphazardness, and applies to concepts of chance, probability, and information entropy.

According to Ramsey theory ideal randomness is impossible especially for large structures, for instance professor Theodore Motzkin pointed out that "while disorder is more probable in general, complete disorder is impossible".[2] Misunderstanding of this leads to numerous conspiracy theories.[3]

The fields of mathematics, probability, and statistics use formal definitions of randomness. In statistics, a random variable is an assignment of a numerical value to each possible outcome of an event space. This association facilitates the identification and the calculation of probabilities of the events. Random variables can appear in random sequences. A random process is a sequence of random variables whose outcomes do not follow a deterministic pattern, but follow an evolution described by probability distributions. These and other constructs are extremely useful in probability theory and the various applications of randomness.

Randomness is most often used in statistics to signify well-defined statistical properties. Monte Carlo methods, which rely on random input (such as from random number generators or pseudorandom number generators), are important techniques in science, as, for instance, in computational science.[4] By analogy, quasi-Monte Carlo methods use quasirandom number generators.

Random selection, when narrowly associated with a simple random sample, is a method of selecting items (often called units) from a population where the probability of choosing a specific item is the proportion of those items in the population. For example, with a bowl containing just 10 red marbles and 90 blue marbles, a random selection mechanism would choose a red marble with probability 1/10. Note that a random selection mechanism that selected 10 marbles from this bowl would not necessarily result in 1 red and 9 blue. In situations where a population consists of items that are distinguishable, a random selection mechanism requires equal probabilities for any item to be chosen. That is, if the selection process is such that each member of a population, of say research subjects, has the same probability of being chosen then we can say the selection process is random.

 

[Bomb#20]

It would be interesting to have world class and terrible poker players as either or both of the participants of this study. Then have raising antes or folds.

Seeing how bluffs and reverse bluffs work would be fun.

And some people seem to believe they have a strategy to beat the 50% mark. They would loose their shirt.
EB

Yes and no. Your three formulations are three different problems. There is a strategy to beat formulation two — watch the game show beforehand and see what the host’s pattern is. Humans are terrible random number generators. In formulation three you lose your shirt if you don’t pick randomly, because there’s always some strategy to beat whatever nonrandom strategy you pick, even if you pick a good pseudorandom number generator. Formulation one is essentially the Man In Black game. “It’s so simple. All I have to do is divine from what I know of you, whether you’re the sort of man who would put the largest number in the envelope he hands his enemy first or second.” Because the dealer is in complete control, there’s no need to even open the envelope and look at the number. The numbers are irrelevant. Regardless of the number in the envelope you look in, the dealer was free to choose a bigger or smaller number for the other nonindependently. So this is not a math problem at all, but a psychology problem. Repoman is correct — the better poker player will win if neither plays randomly.

If you want a puzzle where the player really needs to think about the number he sees, you’ll need to think up a fourth formulation.

 

[Gun Nut]

There may be a non-mathematical (soft) solution, though.
Most people cannot remember more than 7 digits, and one may suppose that a larger length number would not be chosen to write down by the average person.
therefore, an artificial bound of the selection pool of 9999999 may be guessed with some defensibility.
For the revealed number, if it is larger than 5555555, then one may have greater odds of success by guessing "lower", and the converse.

But, strictly speaking, since the selection pool is theoretically infinite, there is no middle point for which to base your guess.

Can we know the size of the piece of paper the number is written on and the size of the numerals written? Bounds for the size of the number can then be interpreted.
Can we observe the writing of the number? The time taken to print the numerals (and other visual keys) can indicate relative length of the two numbers.

We could always modify the premises and a realistic investigation would certainly try to do that but I don't think we could determine the probability that each number be picked out, even if we assumed 0 above some particular number deemed too big to think of or to write down etc.

I agree with your idea about 7 digits but that would only work for most people, as you say.

It's mathematically impossible to beat 50/50 here.

That's what I think. However, one interesting twist is that there is an infinity of numbers to choose from for both the first and the second number so that I don't see how we could make any rationally justified calculation of probability. The probability for any one number to be selected is zero or infinitely small if that's different. So, I would say that the best we can do is just trust our intuition, which says 50/50.

Do you have a calculation to justify 50/50?
EB

Not a mathematical proof, per say.... that form is above me.
But, while infinity is not rational, it is defensible to say that any point is the "mid-point' within an infinite set. So, given any arbitrary (random) value, the set of values both lower and higher than that value are equal in length (infinite).
There is, however, an edge case that throws the "calculation" off... if the second number is equal to the first. Taking that case into consideration changes the calculation from a "perfect" 50/50 to 0.5 - (1/infinty ).. or something like that.
I don't possess the tools to express that correctly.

 

[Jokodo]

What solution do you give to this puzzle?

I give three different formulations, if that can help.

The original one is supposed to be the last one here...

Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you’ve seen. Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

I give here my own view on this, hidden to not influence anyone...

Show

We don't have any relevant information as to the relation between the second number and the first number. 50% is the best anyone rational should expect.

EB

This is a psychology question, not a math question. If it were a math question, the answer would be: "except in very rare occasions, it is impossible to write a random number between -inf and inf on a piece of paper - no piece of paper it's big enough

 

[Jokodo]

Looks like I made a mistake.

I took two coins and put an x on one and a o on the other. Then I put them in a box.

On a third coin I put an x on one side and an o on the other, and put it in a second box.

I shook the boxes and dumped the x/o coin on the table and drew one of the two coins in the other box. If the two coins matched it was a win, mismatch and a loss.

My first run of 20 trials showed 35% win. I repeated with a better job of randomizing and it was close to 50%.

It seems counter intuitive to me but the experiment says otherwise. I saw it as two independent uncorrelated random variables thinking the two probabilities should multiply.

Oh, they do multiply, the probability for 2 o's is 1/4, as is the probability of two x's. The probability of a match however is the sum if those values. A MATCH is actually a cover term for two distinct events.

 

[steve_bank]

The 50/50 proof is in the demonstration of a coin toss or any discrete random variable like todding a die. I do not think there is a formal proof. From the definition of random for a discrete random process like a coin toss the probabilities are divided equally between the possibilities.

If I toss a well baled die many times and the results are close to 1/6 per side the we say the process is random.

The probability is discrete, all possibilities are quantified, all possible outcomes.

It is true that in a continuous random distribution like Gaussian or normal the probability of any finite probability is zero. Mathematical you end up taking the astragal of the distribution between the same number which alwys results in zero.

For a continuous random variable as dx ->0 for the integral of the PDF between +dx and -dx the integral approaches the finite probability as a limit.

In the envelop experiment the actual numbers are irrelevant, it is required they be unequal. That makes it a binary choice as to which one is > or <, If the choice us random and unbiased like flipping a coin the probability is 1/2. .

 

[skepticalbip]

Las Vegas casinos love people that don't understand randomness and show up with their 'system' to beat the house.

 

[beero1000]

Suppose the numbers are x and y, with x < y. Choose your favorite probability distribution supported on the whole real line, let's say N(0,1), and sample a random value z from it. If the number you see is greater than or equal to z, guess it is the larger of the two.

Using the law of total probability: P(you correctly identify y) = P(you see y)*P(you don't switch | you see y) + P(you see x)*P(you do switch | you see x)

Since we see x or y randomly, P(you see y) = P(you see x) = 1/2. By the strategy, P(you don't switch | you see y) = P(z ⩽ y) and P(you do switch | you see x) = P(x < z).

Therefore, P(you correctly identify y) = 1/2*P(z ⩽ y) + 1/2*P(x < z) = 1/2(P(z ⩽ y) + P(x < z))

We can break up P(x < z) into P(x < z ⩽ y) + P(y < z), and notice that P(z ⩽ y) + P(y < z) = 1.

Therefore, P(you correctly identify y) = 1/2(P(z ⩽ y) + P(x < z ⩽ y) + P(y < z)) = 1/2*(1 + P(x < z ⩽ y) = 1/2 + 1/2*P(x < z ⩽ y).

Since P(x < z ⩽ y) > 0 for any possible choice of x and y, P(you correctly identify y) > 1/2, and the strategy does strictly better than 50%.