[Guest]

Here is an interesting wager that Paul 'the cheat' is offering you. Without looking into your wallet, guess the amount of money it contains. Let's say your guess is Q. Now, you will be asked to empty your wallet, and the money that comes out of it will be counted. There are three possibilities:

1. Money is P, such that P < Q, in which case you'll have to handover the entire amount to Paul.
   
2. Money is R, such that R > Q, in which case you can retain the amount Q, and shall handover the difference (R-Q) to Paul.
   
3. Money is exactly Q, in which case not only you retain the entire amount, but also get an additional matching amount from Paul.
   

The question here is: Is this a fair deal?

Also, if you know that you normally have an amount in your wallet with mean M and standard deviation S, then what will be your optimal guess?

[Guest]

BTW, am I at fault to distribute the money normally here?

[Guest]

No, this is certainly not a fair deal. The normal distribution implies that the probability that you X=Q is 0. (X is value in wallet, Q is guess) In reality however the probability is greater than 0 because there is not an infinite amount of money values possible (cents is as low as you go, and you cannot have infinite money). This means that the probability it X=/=Q is very close to 1, meaning that you almost certainly lose money. Therefore it is unfair.

[Guest]

Not a fair deal.

You will only earn iff your guess is exactly equal to the amount in the wallet. Since the amount in the wallet is normally distributed (which is a continuous distribution), the probability that your guess will be exactly equal to the amount is zero (the integral of a line = 0).

Hence, the probability that you'll win is zero. And with probability=1 (which means ALL the time), you will lose. This is not a fair deal.

[Guest]

Perhaps I am not your typical person, but i don't see this as a distribution scenario. I see it as simple logic. In 99.999% of cases I know for certain how much money is in my wallet (in the rare cases i am carrying cash at all). Therefore i would most certainly take the bet to double my money.

[Guest]

Perhaps I am not your typical person, but i don't see this as a distribution scenario. I see it as simple logic. In 99.999% of cases I know for certain how much money is in my wallet (in the rare cases i am carrying cash at all). Therefore i would most certainly take the bet to double my money.

Right, but as stated for the majority of the time this is in fact a 'bad bet'. Meaning that if this was a casino game it wouldn't last long because not enough people would win to keep them coming back to bet (not this exact game mind you but one with the same odds of winning).

[Smith]

Anybody gambling with a guy known as 'Paul "the cheat" ' needs intervention!

[Guest]

I would always say -$1M, therefore I would hand over negative money to Paul and be rich, rich I tell you!!!

[Guest]

Not a fair deal.

You will only earn iff your guess is exactly equal to the amount in the wallet. Since the amount in the wallet is normally distributed (which is a continuous distribution), the probability that your guess will be exactly equal to the amount is zero (the integral of a line = 0).

Hence, the probability that you'll win is zero. And with probability=1 (which means ALL the time), you will lose. This is not a fair deal.

Not true, money is descrete, not continous. There is a probability you will win, and one you will lose assuming SD>0

[Guest]

I would always say -$1M, therefore I would hand over negative money to Paul and be rich, rich I tell you!!!

Actually you would not. -$1M would certainly be less that whatever is in your wallet, in which case you would have to hand over all of your money.

[Guest]

Actually you would not. -$1M would certainly be less that whatever is in your wallet, in which case you would have to hand over all of your money.

You are right, man, I will start carrying IOUs then

[Cavenglok]

Actually you would not. -$1M would certainly be less that whatever is in your wallet, in which case you would have to hand over all of your money.

Unless you were in incredible debt.

[Guest]

OK, it seems most people have used facts interchangeably. So, here are the questions again:

1. (Easy) Whether it's a fair deal for anybody in general, irrespective of his/her money holding pattern?
   
2. (Medium) Can I assume the money in my wallet to be normally distributed despite knowing the money to be a discrete quantity?
   
3. (Tough) If I know I usually hold M amount on an average, and the range so far has been M-3S to M+3S, then what should be my optimal guess if I were to take the wager?
   

[Guest]

I was about to say that, i.e. people started to mix up all questions.

To answer each point:

well... that depends :-) on the distribution of the money you've got in your pocket (or at least, an estimation of that distribution). If the amount of money you have is continuously distributed, there is no way this can be fair since the probability you find out the exact amount is zero. Hence you'll keep paying.

If you assume a slightly more realistic distribution, for example, discrete amounts being either 1,2,3... up to N, each amount having the same probability to occur then it may be fair. I did the math and you can show that if N is either 1 or 2, the game is unfair but in your favor . If N =3, then the bet is fair only if you bet q = 2 or 3. If N>3, the game is always unfair (that is, in Pauls'favor).

As an approximation, why not. But there are several drawbacks. First, as stated previously, you can can basically NEVER win :-) (while in a more realistic discrete finite setting, you do stand a chance). Second, the distribution is not bounded, hence you can in theory have a negative amount of money in your pocket :-). However if the mean is much larger than the standard deviation, then the likelihood of this can be negligible.

For a continuous distribution (whatever it is), you can work out an equation for the optimal solution. I came up with something like -> (1-N(Q)) - Q * f(Q) = 0, where N(.) is the cumulative distribution of the money in your pocket and f(.) the density. Q is your optimal bet (I need to double check this equation though). Optimal here means that you would lose in average as little as possible. There is no easy solution to this equation, unless one assume simple distribution such as uniform. I did a few quick numerical test in excel, but you can quickly guess where the optimal solution should roughly be --> if you bet above the average M, even slightly, then in average you'll have to pay M. If you bet slightly below average, say Q = M - epsilon, then in average you'll have to pay epsilon.... so ideally you want to always bet slightly below average M (my intuition says somewhere between M-s and M, where s is the standard deviation of the distribution)

[Guest]

I was about to say that, i.e. people started to mix up all questions.

To answer each point:

well... that depends :-) on the distribution of the money you've got in your pocket (or at least, an estimation of that distribution). If the amount of money you have is continuously distributed, there is no way this can be fair since the probability you find out the exact amount is zero. Hence you'll keep paying.

If you assume a slightly more realistic distribution, for example, discrete amounts being either 1,2,3... up to N, each amount having the same probability to occur then it may be fair. I did the math and you can show that if N is either 1 or 2, the game is unfair but in your favor . If N =3, then the bet is fair only if you bet q = 2 or 3. If N>3, the game is always unfair (that is, in Pauls'favor).

As an approximation, why not. But there are several drawbacks. First, as stated previously, you can can basically NEVER win :-) (while in a more realistic discrete finite setting, you do stand a chance). Second, the distribution is not bounded, hence you can in theory have a negative amount of money in your pocket :-). However if the mean is much larger than the standard deviation, then the likelihood of this can be negligible.

For a continuous distribution (whatever it is), you can work out an equation for the optimal solution. I came up with something like -> (1-N(Q)) - Q * f(Q) = 0, where N(.) is the cumulative distribution of the money in your pocket and f(.) the density. Q is your optimal bet (I need to double check this equation though). Optimal here means that you would lose in average as little as possible. There is no easy solution to this equation, unless one assume simple distribution such as uniform. I did a few quick numerical test in excel, but you can quickly guess where the optimal solution should roughly be --> if you bet above the average M, even slightly, then in average you'll have to pay M. If you bet slightly below average, say Q = M - epsilon, then in average you'll have to pay epsilon.... so ideally you want to always bet slightly below average M (my intuition says somewhere between M-s and M, where s is the standard deviation of the distribution)

First of all, it is nice to see someone actually comprehending the problem and putting structured thoughts to arrive at the solution. I agree with all your points except the values of N.

Expected gain out of this wager are = Expected gain when amount is Q - Expected loss when amount is P - Expected loss when amount is R

The wager will be a fair deal only when the expected gain is 0. If it is positive then it is unfair in you favor, and if it is negative then it is unfair in Paul's favor.

Now, let's assume the amount to be uniformly distributed over discrete values between M-3S and M+3S with Mode=Median=Mean=M.

So E(G) = Q x P(G=Q) - {Sum(i x P(i)): M-3S <= i < Q} - {Sum((i-Q) x P(i)): Q < i <= M+3S}

Now, since it is uniformly distributed, hence all P(i) are equal. Thus, for a fair deal.

Q = {Sum(i): M-3S <= i < Q} + {Sum(i-Q): Q < i <= M+3S}

=> Q = {3S (2(M-3S) + 3S - 1) / 2} + {3S (2 (M+1) + 3S -1 - Q) / 2}

=> Q = 3S (4M - Q) / 2

=> Q = 12 MS / (2 + 3S)

[Guest]

Hi,

I'll stand by my statement when money is distributed discretely over $1, $2, $3,..., $N. But the results are highly dependent of the setting, see below

If you include the possibility of having $0 in your pocket (discrete value going from 0, 1, 2,.., N), then my results are not valid anymore.

In your example: I haven't checked the calculation (they seem sensible and I'll trust you), but even if you get a value for Q (optimal bet), have you checked that this value makes sense, i.e. that Q is in [M-3S, M+3S], and that the value of Q is a discrete number itself (in both cases, you'd be in a case where again you can't win.. and your calculations are somewhat faulty since you start by including the case where we guess the right amount, i.e. the Q x P(G=Q) term).

In your results Q = 12 MS / (2 + 3S), pick a simple case M = 10 and S = 3 (money is distributed between 1 and 19), you get Q = 32.7 which is not a valid bet in this case. In fact, there doesn't seen to be any trivial case (S>0), where Q is a discrete number within [M-3S, M+3S].

[Guest]

Hi,

I'll stand by my statement when money is distributed discretely over $1, $2, $3,..., $N. But the results are highly dependent of the setting, see below

If you include the possibility of having $0 in your pocket (discrete value going from 0, 1, 2,.., N), then my results are not valid anymore.

In your example: I haven't checked the calculation (they seem sensible and I'll trust you), but even if you get a value for Q (optimal bet), have you checked that this value makes sense, i.e. that Q is in [M-3S, M+3S], and that the value of Q is a discrete number itself (in both cases, you'd be in a case where again you can't win.. and your calculations are somewhat faulty since you start by including the case where we guess the right amount, i.e. the Q x P(G=Q) term).

In your results Q = 12 MS / (2 + 3S), pick a simple case M = 10 and S = 3 (money is distributed between 1 and 19), you get Q = 32.7 which is not a valid bet in this case. In fact, there doesn't seen to be any trivial case (S>0), where Q is a discrete number within [M-3S, M+3S].

first off, my calculations were wrong

And yeah, I've come to realize that the problem is heavily dependent on the distribution. Getting Q in the range requires tweaking with S. So, it all depends on how wide is the interval. If interval is very wide, there may not be any optimum guess.

[Guest]

Well in fact, assuming a discrete finite set of value I = {a, a+1, a+2, ..., a+N-1}, that is N values with a uniform probability, you can check whether there is a fair solution or not (there is always an optimal solution, if we say optimal means the solution were the loss is minimized, but not necessarily zero).

the expected gain in that setting given that we bet Q is

g(Q) = E(gain | bet = Q) = -1/N * (sum_{i=0}^{Q-1} (i+a)) - 1/N * (sum_{i=Q+1}^{N-1}(i+a-Q)) + Q/N

which leads to

g(Q) = -Q^2 / N + Q(N+1)/N - (N-1) (1+a/N)

you get a quadratic equation in Q. The game could be fair is there exists values of Q where g(Q) = 0 and given that Q is in an element of I (the problem reduces to solving a second order equation). More generally, we get three cases,

1) if there are two solutions to the equations g(Q) = 0, then the game is in your favor for any Q in between these two solutions, and in Paul's favor otherwise

2) if there is only one solution, the game may be fair for only one value (only if it is in I) and in Paul's favor otherwise.

3) if there are no solution, then it is always in Paul's favor.

The last case occurs when ((N+1)/N)^2 -4*(N-1)/N*(1+a/N) < 0, that is a > (N+1)^2/(4*(N-1)) - N. This says that as soon as N>=3, the r.h.s is negative and hence the condition always satisfied. I may need to check my calculation, but that would imply that the game is always in Paul's favor is there are 3 or more outcomes.

[Guest]

Assume the set to be {100, 101, 102, 103, 104} and let your guess be 100; so your corresponding gains will be {100, -1, -2, -3, -4}.

Hence, expected gains are 90/5 = 18

Thus, the game could still be in your favor if you take advantage of the distribution (probability) and the range.