Alice, Bob and Charlie at a fair

[karthickgururaj]

One more guessing game!

 

Alice, Bob and Charlie go to a fair. They go to a stall where there is a machine that displays numbers from 1 to 9 at random, one after other, for the sole reason that this puzzle can be framed and posted to brainden.

 

It is not known whether the random process that selects a number in the machine is based on uniform distribution or not.. could be anything.. But it is known that the machine is memory-less, the current number being displayed doesn't influence the next number in anyway.

 

All three stand and watch the numbers for a while. Suddenly Alice turns and tells Bob - "If you can guess the next number within an error margin, I'll give you $100. I won't tell you what the error margin is, but it is not zero". Charlie also offers a similar bet - "If you can guess exactly what the next number is, I'll give you $100. In other words, the bet is same as what Alice is offering, but the error margin in my case is zero".

 

Bob can make two different guesses (if he wants to), one for Alice and other for Charlie.

 

The numbers that they saw before the bet was posed is: { 1, 9, 4, 8, 9, 1, 2, 1, 3, 6, 9, 4, 9, 3, 6, 9, 6}

 

What should Bob do?

[gavinksong]

I would just guess whatever is closest to the mean value for Alice, and the mode for Charlie.

[plasmid]

Does Bob lose anything to Alice or Charlie if he guesses wrong?

[karthickgururaj]

Does Bob lose anything to Alice or Charlie if he guesses wrong?

Bob doesn't get $100..

[karthickgururaj]

I would just guess whatever is closest to the mean value for Alice, and the mode for Charlie.

Let's see if there are any other thoughts

[gavinksong]

I would just guess whatever is closest to the mean value for Alice, and the mode for Charlie.

Let's see if there are any other thoughts

Is there any sort of probability distribution for Alice's margin of error? Can it be anything between 0 and infinity? Edited November 7, 2014 by gavinksong

[karthickgururaj]

 

 

I would just guess whatever is closest to the mean value for Alice, and the mode for Charlie.

Let's see if there are any other thoughts

Is there any sort of probability distribution for Alice's margin of error? Can it be anything between 0 and infinity?

 

No.. nothing of that sort.. It is non-zero, most likely just "1", or at the max "2", if she is feeling generous.

[karthickgururaj]

 

 

 

I would just guess whatever is closest to the mean value for Alice, and the mode for Charlie.

Let's see if there are any other thoughts

Is there any sort of probability distribution for Alice's margin of error? Can it be anything between 0 and infinity?

 

No.. nothing of that sort.. It is non-zero, most likely just "1", or at the max "2", if she is feeling generous.

 

Also, to clarify, your answer is the same as what I thought was right.

[plasmid]

I would agree with gavinksong that...

...the best way to bet with Bob would be to guess the mode, which with the number distribution that was shown would be 9.

For Alice, you would have to look at the actual distribution and, for each number, count out how many times you see a number appear within any given margin.

The distribution is:

1: 3

2: 1

3: 2

4: 2

5: 0

6: 3

7: 0

8: 1

9: 5

The optimal guesses if you knew the error margin would be

Error margin 1: either 8 or 2 (both had 6 hits within +/- 1)

Error margin 2: either 7 or 8 (9 hits within +/- 2)

Error margin 3: 6 (13 hits)

Error margin 4 or greater: 5 (everything)

If you have absolutely no idea what Alice's error margin is, then you can't give a definite answer. If you know the error margin is likely to be either 1 or 2, then go with 8.

[karthickgururaj]

I would agree with gavinksong that...

...the best way to bet with Bob would be to guess the mode, which with the number distribution that was shown would be 9.

For Alice, you would have to look at the actual distribution and, for each number, count out how many times you see a number appear within any given margin.

The distribution is:

1: 3

2: 1

3: 2

4: 2

5: 0

6: 3

7: 0

8: 1

9: 5

The optimal guesses if you knew the error margin would be

Error margin 1: either 8 or 2 (both had 6 hits within +/- 1)

Error margin 2: either 7 or 8 (9 hits within +/- 2)

Error margin 3: 6 (13 hits)

Error margin 4 or greater: 5 (everything)

If you have absolutely no idea what Alice's error margin is, then you can't give a definite answer. If you know the error margin is likely to be either 1 or 2, then go with 8.

Thanks, this is interesting

 

I agree with you on the answer to be given to Bob. This would be 9.

 

I understand your approach.

 

But the integer closest to mean (or the "expected" value) turns out to be 5, which the last choice in your argument.

 

(Actually, I constructed the list of samples with the intent of keeping the expected value to be exactly 5, with the number of samples as 18, totaling to 90. But I see now that I made some mistake and hence the number of samples have become 17, yielding expected value as ~ 5.3. Fortunately, doesn't affect the results drastically - the closest integer is still 5 at least).

 

The S.D for the samples turns out to be ~ 3.1378, yielding SEM to be: 3.1378/sqrt(17) = 0.761. 

 

Question is: Should mean be used or should Charlie go with your analysis..?

 

[k-man]

 

I would agree with gavinksong that...

...the best way to bet with Bob would be to guess the mode, which with the number distribution that was shown would be 9.

For Alice, you would have to look at the actual distribution and, for each number, count out how many times you see a number appear within any given margin.

The distribution is:

1: 3

2: 1

3: 2

4: 2

5: 0

6: 3

7: 0

8: 1

9: 5

The optimal guesses if you knew the error margin would be

Error margin 1: either 8 or 2 (both had 6 hits within +/- 1)

Error margin 2: either 7 or 8 (9 hits within +/- 2)

Error margin 3: 6 (13 hits)

Error margin 4 or greater: 5 (everything)

If you have absolutely no idea what Alice's error margin is, then you can't give a definite answer. If you know the error margin is likely to be either 1 or 2, then go with 8.

Thanks, this is interesting

 

I agree with you on the answer to be given to Bob. This would be 9.

 

I understand your approach.

 

But the integer closest to mean (or the "expected" value) turns out to be 5, which the last choice in your argument.

 

(Actually, I constructed the list of samples with the intent of keeping the expected value to be exactly 5, with the number of samples as 18, totaling to 90. But I see now that I made some mistake and hence the number of samples have become 17, yielding expected value as ~ 5.3. Fortunately, doesn't affect the results drastically - the closest integer is still 5 at least).

 

The S.D for the samples turns out to be ~ 3.1378, yielding SEM to be: 3.1378/sqrt(17) = 0.761. 

 

Question is: Should mean be used or should Charlie go with your analysis..?

 

 

 

I'm going to agree with plasmid here.

 

The expected value of 5 makes sense if you assume a uniform probability distribution. Not knowing the distribution one cannot calculate the expected value. For all we know, the machine may be programmed to never (or extremely rarely) produce a 5. Deriving the possible distribution from the sample is your best bet, which is exactly what plasmid did.

[karthickgururaj]

 

 

I would agree with gavinksong that...

...the best way to bet with Bob would be to guess the mode, which with the number distribution that was shown would be 9.

For Alice, you would have to look at the actual distribution and, for each number, count out how many times you see a number appear within any given margin.

The distribution is:

1: 3

2: 1

3: 2

4: 2

5: 0

6: 3

7: 0

8: 1

9: 5

The optimal guesses if you knew the error margin would be

Error margin 1: either 8 or 2 (both had 6 hits within +/- 1)

Error margin 2: either 7 or 8 (9 hits within +/- 2)

Error margin 3: 6 (13 hits)

Error margin 4 or greater: 5 (everything)

If you have absolutely no idea what Alice's error margin is, then you can't give a definite answer. If you know the error margin is likely to be either 1 or 2, then go with 8.

Thanks, this is interesting

 

I agree with you on the answer to be given to Bob. This would be 9.

 

I understand your approach.

 

But the integer closest to mean (or the "expected" value) turns out to be 5, which the last choice in your argument.

 

(Actually, I constructed the list of samples with the intent of keeping the expected value to be exactly 5, with the number of samples as 18, totaling to 90. But I see now that I made some mistake and hence the number of samples have become 17, yielding expected value as ~ 5.3. Fortunately, doesn't affect the results drastically - the closest integer is still 5 at least).

 

The S.D for the samples turns out to be ~ 3.1378, yielding SEM to be: 3.1378/sqrt(17) = 0.761. 

 

Question is: Should mean be used or should Charlie go with your analysis..?

 

 

 

I'm going to agree with plasmid here.

 

The expected value of 5 makes sense if you assume a uniform probability distribution. Not knowing the distribution one cannot calculate the expected value. For all we know, the machine may be programmed to never (or extremely rarely) produce a 5. Deriving the possible distribution from the sample is your best bet, which is exactly what plasmid did.

 

While I'm not discounting plasmid's analysis (since I do not know the answer to this one), one point what you said above seems wrong to me:

 

The "expected value" or mean does not assume uniform probability distribution.

The equation,

   mean = 1/n * (sum of all samples)

works for any distribution.

 

(Actually, we can calculating the expected value of the mean above, since the mean can only be calculated if the exact probability distribution is known up-front).

 

And indeed, even if the machine may be programmed to never produce a 5, the expected value can be 5.

 

[k-man]

 

 

 

I would agree with gavinksong that...

...the best way to bet with Bob would be to guess the mode, which with the number distribution that was shown would be 9.

For Alice, you would have to look at the actual distribution and, for each number, count out how many times you see a number appear within any given margin.

The distribution is:

1: 3

2: 1

3: 2

4: 2

5: 0

6: 3

7: 0

8: 1

9: 5

The optimal guesses if you knew the error margin would be

Error margin 1: either 8 or 2 (both had 6 hits within +/- 1)

Error margin 2: either 7 or 8 (9 hits within +/- 2)

Error margin 3: 6 (13 hits)

Error margin 4 or greater: 5 (everything)

If you have absolutely no idea what Alice's error margin is, then you can't give a definite answer. If you know the error margin is likely to be either 1 or 2, then go with 8.

Thanks, this is interesting

 

I agree with you on the answer to be given to Bob. This would be 9.

 

I understand your approach.

 

But the integer closest to mean (or the "expected" value) turns out to be 5, which the last choice in your argument.

 

(Actually, I constructed the list of samples with the intent of keeping the expected value to be exactly 5, with the number of samples as 18, totaling to 90. But I see now that I made some mistake and hence the number of samples have become 17, yielding expected value as ~ 5.3. Fortunately, doesn't affect the results drastically - the closest integer is still 5 at least).

 

The S.D for the samples turns out to be ~ 3.1378, yielding SEM to be: 3.1378/sqrt(17) = 0.761. 

 

Question is: Should mean be used or should Charlie go with your analysis..?

 

 

 

I'm going to agree with plasmid here.

 

The expected value of 5 makes sense if you assume a uniform probability distribution. Not knowing the distribution one cannot calculate the expected value. For all we know, the machine may be programmed to never (or extremely rarely) produce a 5. Deriving the possible distribution from the sample is your best bet, which is exactly what plasmid did.

 

While I'm not discounting plasmid's analysis (since I do not know the answer to this one), one point what you said above seems wrong to me:

 

The "expected value" or mean does not assume uniform probability distribution.

The equation,

   mean = 1/n * (sum of all samples)

works for any distribution.

 

(Actually, we can calculating the expected value of the mean above, since the mean can only be calculated if the exact probability distribution is known up-front).

 

And indeed, even if the machine may be programmed to never produce a 5, the expected value can be 5.

 

 

Expected value is a probability-weighted average of possible outcomes. If probability distribution is uniform, then it turns into a simple average. If the probability distribution is unknown then you cannot calculate it this way. Of course, you can always calculate an average (or mean) of any set of numbers and you can use it as an estimation for the expected value, but with such a small sample set the estimate is not that good.

[DejMar]

If we assume a uniform distribution (each digit having an equal probability of being displayed), a bell-curve forms as the error of margin increases from 0 to 5, with the number 5 being the digit most likely to be included within the margin of error. (The curve flattens out again as the error of margin increases beyond 5. Yet, the number 5 holds the greatest probability of appearing with a margin of error within a uniform distribution of error of margin.) Thus, for a uniform distribution, the best guess for Alice's posed wager is 5, and for Bob's -- any number is as likely as the next.

For a non-uniform distribution, if we calculate the standard deviation based on the sample set of 17 observations, subtracting 1 from the number of observations to get a more accurate deviation for the unlimited set, we find σ ≈ 3.137, and the nonparametric skew, γ ≈ 0.319. The positive skew indicates the distribution is likely to be skewed to the right. As the mean of the sample space is 5.29, the standard deviation from the mean of the 17 observations is ≈ 0.761. For a non-uniform distribution based on the sample space, the tip of the skewed bell curve lies approximately at 5.761, which is closer to 6.

Given the sample space, a non-uniform distribution appears likely, and due to the deviation and skew (and the fact that 6 appeared thrice in the 17 observations), 6 may be the best guess posed for Alice's generous offered wager. Given the mode is equal to 9, the choice of number to be guessed for Bob's generously offered bet would be 9.

Edited November 20, 2014 by DejMar