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St. Petersburg Paradox In search of a stable measure of fair price.

!Arkady Gershteyn September 2014

!1STATISTICS REPORT

St. Petersburg Paradox (1713, 1738) In search of a stable measure of fair price.

The Problem A fair coin is flipped until it lands heads. The number of tosses required is n. The payout is $2n. Question: How much are you willing to pay to play this game once? Daniel Bernoulli (below) publicized this problem. Nicolaus I Bernoulli (below) invented it.

Expected value = Infinity. Question: Are you willing to pay any fixed price? Definitely not.

n Probability, P(n) Money Payout, M(n)

Product

1 1/2 2 1

2 1/4 4 1

3 1/8 8 1

n 2-n 2n 1

!2STATISTICS REPORT

Analogous situations: !* Doubling up the stakes each time when playing at cards or the roulette table. !Edvard Munch, “At the Roulette Table”

!* Doubling up the amount at risk when investing in a stock with stop loss in place. !* Doubling up the lot size in FOREX when taking a position in the same currency pair, under normal conditions (i.e. neither country experiences civil war, natural catastrophe, war, or other force majore). !!!!!

!3STATISTICS REPORT

Problems: 1. High chance you will run out of money before you can reach that win. aka as

“Gambler’s Ruin”. !2. Casinos watch out for this behavior and could ban you from their establishment. !Casino Monte Carlo

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!4STATISTICS REPORT

New Concept: Probability of Realizing Expected Value in fixed n trials. !Other situations where Expected Value fails: !*Lottery, example: Probability that you win $10 million is 1%, would you be willing to pay just under $100k to try your luck? $50K/ticket? !*On other hand, if the above lottery ticket

is offered in plural then YES. Buy out 99% of the tickets at $50k/ticket and almost certainly guaranteed to about double your money.

!*In fact in UK, a group of consumers pooled their resources and bought out 80% of lottery tickets and won, dividing the gains among all members. !Critical feature: one shot deal. !!*Conclusion: Expected value only

useful when can repeat the game/experiment many times so that improbable events tend to realize at least some of the time. !Concept of probability of realization of expected value becomes crucial to estimate risk and reward tradeoff associated with small n. !Sampling without replacement (ex. lottery with only 1 winning ticket) have probability of realization of expected value equal to % lottery tickets purchased.

!5STATISTICS REPORT

Sampling without replacement (ex. lottery with 10 out of 100 tickets winning) and sampling with replacement (toss a 6-sided cube with only 1 sides being winning) require modeling to estimate probability of realization of expected value at specific n. !Computer Simulation of Capped St. Petersburg Problem Abstract

*I conduct a computer simulation experiment of the St. Petersburg Paradox problem and it's variants.

*The computer simulation will be limited to the finite formulation of the St. Petersburg problem where there will be a max number of allowed coin flips.

*If the computer reaches that maximum number of allowed flips then the payoff will be 2^max # of allowed flips regardless of the outcome (Heads/Tails) of the last coin toss.

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!6STATISTICS REPORT

Resolution The paradox is resolved, in my opinion, when you consider that probability theory applies only to large (or at least moderate) number of repetitions of a certain game/experiment, not to a one-off game. Here is an article that resolves the St. Petersburg Paradox: http://mpra.ub.uni-muenchen.de/5233/

!Here is a brief summary, in my words: The attached article solves the St. Petersburg paradox - the key is that Expected value is based on many (or infinite) number of games, and should not be applied to just one game. * When you look at just one game the Expected value of St. Petersburg

paradox is finite! In fact one solution comes from Daniel Bernoulli himself, who has the person who published the paradox in the first place!

http://www.jstor.org/discover/10.2307/1909829?uid=2&uid=4&sid=21104728591283 He proposes to value utility instead of payout, computed as logarithm of payout divided by assets that the individual has before the game commences.

!7STATISTICS REPORT

http://mpra.ub.uni-muenchen.de/5233/

http://www.jstor.org/discover/10.2307/1909829?uid=2&uid=4&sid=21104728591283

Excerpt from Peters, Ole (October 2011b). "Menger 1934 revisited". Journal of Economic Literature. http://arxiv.org/pdf/1110.1578v1.pdf !But Karl Menger then responded in 1934 with the Super Petersburg Paradox that the logarithm does not resolve. !*The series of experiments that I propose is to consider the question: How does mean, median, and 75% percentile (upper quartile) change as a function of # of repetitions of St. Petersburg game, and type of St. Petersburg game. By type of St. Petersburg game, I mean that you can cap the game, at say a max # of coin flips. Thus if the max # of coin flips allowed is 6 then the max possible outcome for one game is $64 (2^6).

!*Furthermore, there might be another measure, besides mean or median then could better speak to the expected value of the St. Petersburg Game played once, 10 times, 100 times, 1000 times, etc.. It would be interesting to search for such a measure.

!*For an article that includes computer simulations of the St. Petersburg Paradox, please take a look at:

http://journal.sjdm.org/9226/jdm9226.html !!!!!!!!!!!!

!8STATISTICS REPORT

http://arxiv.org/pdf/1110.1578v1.pdf

http://journal.sjdm.org/9226/jdm9226.html

Experimental design !2 Factors: Factor A: Variant of St. Petersburg Paradox Problem, max # of coin tosses=3,5, 7, 9 (4 levels) Factor B: # or reps= 10,100, 1000, 10000 (4 levels) !Calculations based on Probability Theory !Variant 1 (3 tosses max): E[x]=(1/2)x2 + (1/4)x4 +(1/8)x8 +(1/8)x16 = 5 Var[x] = E[x^2] – (E[x])^2 = 46-25= 21 Std.Dev = SQRT(21)=4.58 !Variant 2 (5 tosses max): E[x] = 7 Var[x] = E[x^2] – (E[x])^2 = 190-49= 141 Std.Dev = SQRT(141)=11.87 !Variant 3 (7 tosses max): E[x]=9 Var[x] = E[x^2] – (E[x])^2 = 766-81=685 Std.Dev = SQRT(685)=26.17 !Variant 4 (9 tosses max): E[x]=11 Var[x] = E[x^2] – (E[x])^2 = 3070-121=2949 Std.Dev = SQRT(2949)=54.30 !Expected Value by Variant and n

!9STATISTICS REPORT

Variant/n per sample

10 100 1,000 10,000

1 (3 coin throws) 5 5 5 5



4 (9 coin throws) 11 11 11 11

SAS Code, excerpts Variant 1 (n=10; n=100; n=1,000; n=10,000) /* efficient simulation that calls a SAS procedure */ %let N = 10; %let NumSamples = 10000; data Uniform(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else payoff=16; output; end; end; run; proc means data=Uniform noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !/* efficient simulation that calls a SAS procedure */ %let N = 100; %let NumSamples = 10000; data Uniform2(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else payoff=16; output; end; end; run; proc means data=Uniform2 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats2 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run;

!10STATISTICS REPORT

/* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats2; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!/* efficient simulation that calls a SAS procedure */ %let N = 1000; %let NumSamples = 10000; data Uniform3(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else payoff=16; output; end; end; run; proc means data=Uniform3 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats3 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats3; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!!/* efficient simulation that calls a SAS procedure */ %let N = 10000; %let NumSamples = 10000;

data Uniform4(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else payoff=16; output; end; end; run;


proc means data=Uniform4 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats4 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats4; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !Variant 4 !!/* efficient simulation that calls a SAS procedure */ %let N = 10; %let NumSamples = 10000; data Uniform13(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); seed4= RAND('UNIFORM'); seed5= RAND('UNIFORM'); seed6= RAND('UNIFORM'); seed7= RAND('UNIFORM'); seed8= RAND('UNIFORM'); seed9= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else if seed4>0.5 then payoff=16; else if seed5>0.5 then payoff=32; else if seed6>0.5 then payoff=64; else if seed7>0.5 then payoff=128; else if seed8>0.5 then payoff=256; else if seed9>0.5 then payoff=512; else payoff=1024; output; end; end; run; proc means data=Uniform13 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats13 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats13;


histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!!/* efficient simulation that calls a SAS procedure */ %let N = 100;

%let NumSamples = 10000; data Uniform14(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); seed4= RAND('UNIFORM'); seed5= RAND('UNIFORM'); seed6= RAND('UNIFORM'); seed7= RAND('UNIFORM'); seed8= RAND('UNIFORM'); seed9= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else if seed4>0.5 then payoff=16; else if seed5>0.5 then payoff=32; else if seed6>0.5 then payoff=64; else if seed7>0.5 then payoff=128; else if seed8>0.5 then payoff=256; else if seed9>0.5 then payoff=512; else payoff=1024; output; end; end; run; proc means data=Uniform14 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats14 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats14; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!!!/* efficient simulation that calls a SAS procedure */ %let N = 1000;

%let NumSamples = 10000; data Uniform15(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */


seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); seed4= RAND('UNIFORM'); seed5= RAND('UNIFORM'); seed6= RAND('UNIFORM'); seed7= RAND('UNIFORM'); seed8= RAND('UNIFORM'); seed9= RAND('UNIFORM'); if seed1>0.5 then payoff=2; else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else if seed4>0.5 then payoff=16; else if seed5>0.5 then payoff=32; else if seed6>0.5 then payoff=64; else if seed7>0.5 then payoff=128; else if seed8>0.5 then payoff=256; else if seed9>0.5 then payoff=512; else payoff=1024; output; end; end; run; proc means data=Uniform15 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats15 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats15; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!!!!!

/* efficient simulation that calls a SAS procedure */ %let N = 10000; %let NumSamples = 10000; data Uniform16(keep=SampleID payoff); do SampleID = 1 to &NumSamples; /* 1. create many samples */ do i = 1 to &N; /* sample of size &N */ seed1= RAND('UNIFORM'); seed2= RAND('UNIFORM'); seed3= RAND('UNIFORM'); seed4= RAND('UNIFORM'); seed5= RAND('UNIFORM'); seed6= RAND('UNIFORM'); seed7= RAND('UNIFORM'); seed8= RAND('UNIFORM'); seed9= RAND('UNIFORM'); if seed1>0.5 then payoff=2;


else if seed2>0.5 then payoff=4; else if seed3>0.5 then payoff=8; else if seed4>0.5 then payoff=16; else if seed5>0.5 then payoff=32; else if seed6>0.5 then payoff=64; else if seed7>0.5 then payoff=128; else if seed8>0.5 then payoff=256; else if seed9>0.5 then payoff=512; else payoff=1024; output; end; end; run; proc means data=Uniform16 noprint; by SampleID; /* 2. compute many statistics */ var payoff; output out=OutStats16 mean=SampleMean median=SampleMedian q3=Sample_Q3 p90=Sample_p90 p99=Sample_p99 var=Sample_variance; run; /* 3. analyze the sampling distribution of the statistic */ proc univariate data=OutStats16; histogram SampleMean SampleMedian Sample_Q3 Sample_p90 Sample_p99 Sample_variance; run; !!! !!!!!!!!!!!!!!!!!!!!!


Tables & Results !Expected Mean !

Empirical Mean, with S.E. based on 10,000 runs

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Empirical Mean Variance, Std. Dev. and S.E. of Variance, based on 10,000 runs

!Empirical Mean Median based on 10,000 runs


10 100 1,000 10,000

1 (3 coin throws)

5.00 (0.0015) 5.00 (0.0045) 5.00 (0.0014) 5.00 (0.00046)

2 (5 coin throws)

7.00 (0.038) 6.99 (0.012) 7.00 (0.0037) 7.00 (0.0012)

3 (7 coin throws)

8.96 (0.083) 9.00 (0.026) 8.99 (0.0083) 9.00 (0.0026)

4 (9 coin throws)

10.92 (0.17) 10.92 (0.054) 11.00 (0.017) 11.00 (0.0054)


10 100 1,000 10,000

1 (3 coin throws)

21.01;12.51;0.125

21.00; 3.76; 0.038

21.01; 1.20; 0.012

21.00; 0.38; 0.0038

2 (5 coin throws)

141.20; 181.95; 1.82

140.48; 57.01; 0.57

140.89; 18.03; 0.18

141.02; 5.66; 0.057

3 (7 coin throws)

686.44; 1,745.60; 17.46

679.95; 546.06; 5.46

681.93; 174.99; 1.75

685.27; 54.80; 0.55

4 (9 coin throws)

2,810.51; 14,335.77; 143.36

2,876.32; 4,621.42; 46.21

2,945.15; 1,483.87; 14.84

2,946.19; 470.97; 4.71


10 100 1,000 10,000


!!Empirical 75th percentile (Q3) based on 10,000 runs

!Empirical 90th percentile (P90) based on 10,000 runs

!

1 (3 coin throws)

3.21 (0.014) 3.00 (0.0096) 2.99 (0.0099) 3.00 (0.010)

2 (5 coin throws)

3.22 (0.014) 2.99 (0.0096) 3.01 (0.0099) 3.00 (0.010)

3 (7 coin throws)

3.21 (0.014) 3.01 (0.0096) 2.99 (0.010) 3.01 (0.010)

4 (9 coin throws)

3.22 (0.013) 3.00 (0.0096) 3.01 (0.010) 2.99 (0.010)


10 100 1,000 10,000

1 (3 coin throws)

6.76 (0.040) 5.97 (0.019) 5.99 (0.020) 6.01 (0.020)

2 (5 coin throws)

7.15 (0.058) 5.95 (0.019) 6.00 (0.020) 6.00 (0.020)

3 (7 coin throws)

7.15 (0.063) 6.01 (0.019) 5.99 (0.010) 6.02 (0.020)

4 (9 coin throws)

7.31 (0.087) 5.99 (0.019) 5.99 (0.020) 6.00 (0.020)


10 100 1,000 10,000

1 (3 coin throws)

11.76 (0.039) 14.16 (0.031) 15.93 (0.0069) 16.00 (0)

2 (5 coin throws)

21.54 (0.16) 15.29 (0.054) 15.94 (0.0064) 16.00 (0)

3 (7 coin throws)

31.38 (0.39) 15.35 (0.055) 15.94 (0.0065) 16.00 (0)

4 (9 coin throws)

41.11 (0.82) 15.24 (0.054) 15.94 (0.0067) 16.00 (0)


Empirical 99th percentile (P99) based on 10,000 runs

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10 100 1,000 10,000

1 (3 coin throws)

13.65 (0.040) 16.00 (0) 16.00 (0) 16.00 (0)

2 (5 coin throws)

29.92 (0.23) 60.34 (0.086) 64.00 (0) 64.00 (0)

3 (7 coin throws)

48.90 (0.67) 146.84 (0.71) 150.18 (0.53) 129.24 (0.12)

4 (9 coin throws)

67.54 (1.54) 242.28 (2.18) 151.72 (0.57) 129.07 (0.11)


Discussion !!!!Expected Mean vs. Empirical Mean !*The expected mean is equal to the number of maximum allowed coin tosses plus 2. The 95 % confidence interval of the empirical mean (mean +/- 1.96 * standard error) for sample size 10 includes the expected mean for all variants #1 thru 4. !*As sample size increases the confidence interval becomes narrower and they all include the expected mean, for all four variants. !*The Central Limit Theory applies to means of sample size greater than or equal to 30, so the fact that the empirical means for a sample size of 10 are very close to the expected means is a bit surprising. *It would be interesting investigating at what sample size the CLT begins to work, and if this depends on what kind of non- normal distribution is used (Poisson, Cauchy, bi-modal, etc.) !!!!!!!!!!!!!!!!!!!!!!!!!!!


Expected Variance vs. Empirical Variance !*For variants 1 thru 3, the empirical variance is within 1.0% of the expected variance for all samples sizes (10 thru 10,000). *For variant 4, the empirical variance is -4.7% smaller for sample size 10 than the expected value. This is the largest deviation of variance from expectation of all the 16 experiments. *As the sample size for variant 4 increases, the deviation of observed variance becomes smaller. For variant 4, sample size 10,000, the difference is only -0.10%. *For distributions in variants 1 thru 4, it is clear that the sample variances (and standard deviation) are stable regardless of sample size (across 10 thru 10,000 sample sizes).

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A Stable Measure !*After searching for a measure that is stable regardless of variant or sample

size, the following measure came out as the best: Median and Q3(75th

percentile). !*I would explain this by the structure of the distribution, namely the fact that 50% probability that the payoff is 2 makes the median 2, or 4 and sometimes (rarely) their average – 3. The bimodal distribution of with a small bump at 3 is visible from the histograms of the median. !*The reason Q3 is stable is that it is either 4, or 8, and sometime (rarely) their average-6. The 75% cumulative probability falls at 4 or less. !*So basically these “stable measures” just reinforce that some features characteristic of the underlying distribution are constant across all the variants, and appear even at small n (starting at sample size of 10). !!!!!!!!!!!!!!!!!!!!!!!!!!!!


P90 and P99 !*For any given variant, P90 approaches and reaches 16 as sample size increases to 10,000 from 10. All the variants have 16 correspond to a cumulative probability of 93.75% and the cumulative probability of 8 is 87.5%,

so for large n the 90th

percentile falls on the value 16. !*The asymptote of P99 is different for variant #1 (16) vs. #2(64) vs. #3 and #4 (both close to 128). !*This makes sense because: for variant 1 the top 6.25% should have value 16 (maximum payoff) and for variant 2, the top 3.125% should have value 64. !*For variant 3 the 99.22 percentile has value 128, and 98.44 percentile should have value 64. This explains that the percentile in between – P99, yields a value of nearly 128 (exactly 129.24 when sample size equals 10,000). !*The same reasoning applies to variant 4 with almost exactly the same result, 129.07, also very close to 128. !


!!!Fair value of the game? !*For a single game of the St. Petersburg problem, with the original payoff distribution, I am convinced that expected value of the pay-off provides no relevant information. !The expected value (mean) comes from many repetitions of the same game, and is not applicable as a predictor of a single shot. The reason expected value (mean) is useful as a measure of center when you have the possibility to play a game many times, is that the mean is simply an extension of the sum (i.e. it’s just the sum divided by the number of games). !For a one-time game of the St. Petersburg problem paying less than $2 to play means you are guaranteed to win money. So this is clearly not a fair price. !On the other hand, if you pay between $2 and $4 these would seem to be the general ballpark for an estimate of the “fair price”. For instance, let’s say you pay $3 then you have a 50% probability of losing $1 versus a 50% probability of winning $1 or more. !So $3 is below the “fair price”, since though winning versus losing is equally probable, you can win more (ex: if the payoff is 8, 16, 32, etc.). !!!!!!!!!


Fair value of the game? continued !Jean Le Rond d’Ambert’s solution was that small probabilities should be ignored. This solution is unpopular today, yet I subscribe to this approach, when the game is played only once. When we do hypothesis testing we reject a null hypothesis, typically at the 5% significance level, which means that there is a 5% chance that the data could have come from the null hypothesis. Sometimes people use 10% significance level instead.

!!!I think that the 10% is a good cut-off, though arbitrary for a one-time game. So probabilities less than 0.1 should be ignored for one time games. !Furthermore, I would not use expected value (mean) as a measure of center. !The median is a much better measure of the

center. It is symmetrical by definition, there is a 50% chance that the value will be above it, and a 50% that the value will be less than it. So I would stick to the median. !!!!!!!!!!!


Fair value of the game? continued 2 !For the one time St. Petersburg game I would claim that a “fair” price is a little above $3. I would ignore the possibilities of getting payoffs of 16 or more. Then the p.m.f becomes: p(x)=0.5; x=2 p(x)=1/4; x=4 p(x)=1/8; x=8!and this poses the problem that it does not sum to 1. The sum of the probabilities is 7/8. !So I would multiply the individual probabilities by 8/7 each to make sure that the p.m.f. does sum to 1. The expected value becomes 24/7 (=3*8/7) = 3.43. !However, I would not use this number as a measure of center, nor of fair price because I prefer the median. !My best estimate of the “fair” price of the St. Petersburg paradox, one-time game, comes from the Variant 1, sample size 10, median value after 10,000 runs, which is 3.21 (+/- 1.96*0.0014) = (3.207, 3.213). !It is worth noting that even as variant changes, and you allow more coin tosses, the mean median of the sample size 10 remains almost exactly the same, it’s either 3.21 or 3.22 (see table). The standard error is also almost exactly the same, either 0.0014 or 0.0013. !My Answer to the fair price of a one-time St. Petersburg game !$3.21. !!!!!!!!!


Future directions and conclusions !The mean is very sensitive to extreme values, which is not true of the median. The method of quintile regression is gaining momentum in economics and statistics community because it uses rank statistics as opposed to the mean. The key question of whether a measure is stable from dataset to dataset, robust to deviations from normality, and in-sensitive to extreme rare values, all motivate the ascent of the median over the mean. Though probability theory likes to deal with a large number (or even infinite) of repeating the same game, there is growing need to develop metrics that can accurately predict the outcome of a game that can only be played a small number of times. There are plenty of examples of these games, ranging medical procedures where a person is limited to only trying one or two treatments before their time runs out (in case the treatments fail), military operations where only a few battles are allowed until the war is won or lost, individual investment choices where the buy-ins are large!!!!!!!!!!!!!!!!!!!


Appendix Histograms: Distribution of Sample Mean and Sample Median: Variant 1, sample size 10

!

! !Variant 1, sample size 100 !!!!!!!!!!!!!


!!!Variant 1, sample size 1000

! !Variant 1, sample size 1000

! !!!!!!!!!!!!!28STATISTICS REPORT

!!Histogram that displays distribution of Sample Median Variant 1, sample size 10

! !!Variant 1, sample size 100

! !!!!!29STATISTICS REPORT

!!!Variant 1, sample size 1000

! !!!Variant 1, sample size 10,000

! !!!!!!!30STATISTICS REPORT

!!!!Histogram that displays distribution of Sample Mean !Variant 4, sample size 10

! !!Variant 4, sample size 100

!


!Variant 4, sample size 1,000

! !!Variant 4, sample size 10,000

! !!!!


!!Histogram that displays distribution of Sample Median !Variant 4, sample size 10

! !!!Variant 4, sample size 100

! !


!Variant 4, sample size 1,000

! !!Variant 4, sample size 10,000

! !!


!Footnotes: !Photo of St. Petersburg: Attribution: By Victorgrigas (Own work) [CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commonsense Page url: http://commons.wikimedia.org/wiki/File%3ASt.Petersburg_Russia_White_Night_Pepsi_Sign.jpg !Photo of Daniel Bernoulli: http://en.wikipedia.org/wiki/Daniel_Bernoulli#mediaviewer/File:Daniel_Bernoulli_001.jpg !Photo of Nicolas Bernoulli: http://www.ecured.cu/images/thumb/a/af/Nicolas_Bernoulli.jpg/260px-Nicolas_Bernoulli.jpg !Edvard Munch, “At the Roulette Table” http://upload.wikimedia.org/wikipedia/en/1/1f/Edvard_Munch_-_At_the_Roulette_Table_in_Monte_Carlo_-_Google_Art_Project.jpg !Casino Monte Carlo http://upload.wikimedia.org/wikipedia/commons/4/47/Casino_Monte_Carlo.JPG !Pastel painting of Jean LeRond d’Alembert http://upload.wikimedia.org/wikipedia/commons/d/df/Alembert.jpg !Reference: Arkady Gershteyn “St. Petersburg's Paradox: Average Median Estimate” http://figshare.com/articles/St_Petersburg_s_Paradox_Simulation_and_using_Mean_Median_as_valid_estimate/693752


http://commons.wikimedia.org/wiki/File%3ASt.Petersburg_Russia_White_Night_Pepsi_Sign.jpg

http://en.wikipedia.org/wiki/Daniel_Bernoulli#mediaviewer/File:Daniel_Bernoulli_001.jpg

http://www.ecured.cu/images/thumb/a/af/Nicolas_Bernoulli.jpg/260px-Nicolas_Bernoulli.jpg

http://upload.wikimedia.org/wikipedia/en/1/1f/Edvard_Munch_-_At_the_Roulette_Table_in_Monte_Carlo_-_Google_Art_Project.jpg

http://upload.wikimedia.org/wikipedia/commons/4/47/Casino_Monte_Carlo.JPG

http://upload.wikimedia.org/wikipedia/commons/d/df/Alembert.jpg

http://figshare.com/articles/St_Petersburg_s_Paradox_Simulation_and_using_Mean_Median_as_valid_estimate/693752

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