What would you pay to play (or avoid) a finite St Petersburg game?

[BWV]

The St Petersburg Paradox is a game with an infinite expected value, but the 'paradox' is that a rational person would pay a relatively small sum (maybe $20) to play it (see https://en.wikipedia.org/wiki/St._Petersburg_paradox )

So say the government offered you the game with a tax-free payout up to $2^40 (so it would potentially make you the first trillionaire), how much would you pay to play? (the expectation value is $40)

Worth noting that people regularly pay 2-4x (or more) expected A/T PV value for lottery tickets with big payouts

secondly, supposing the government decided to fine you and gave you the option of a fixed fine or paying the results of a game up to your net worth, what multiple of the expectation value would you pay to not take the risk?

Finally - the gov gives you the option of dividing the payout across up to 2^40 simultaneous games, so for example you could play 100 simultaneous games with payouts beginning at $0.02 each

 

[PeroK]

First, I would need to know how long I will be allowed to play - in terms of number of games. Then I could model a distribution of winnings.

Second, it would depend on my criteria for winning. On the one hand I may not want to risk very much money and on the other I may not view winning a lot of money as being particulary important.

For example, if you offered me a one-off 50-50 bet of my flat and life savings against $10 million dollars, I wouldn't take it. If I lost, then I'm ruined financially; and, if I win, well I don't actually need $10 million. Although I could do some good with it. And even if you made it $100 million, it wouldn't make any difference. I wouldn't risk financial ruin for something I don't need.

 

[BWV]

First, I would need to know how long I will be allowed to play - in terms of number of games. Then I could model a distribution of winnings.

Fair enough, added an option to divide the payouts across n simultaneous games

 

[PeroK]

Fair enough, added an option to divide the payouts across n simultaneous games

I did a 40 simulations of 1024 games, assuming you win $1 if it's tails first time, $2 if it's tails the second time etc. The winnings were almost always between $4,000 and $10,000, with four outliers,all about $14,000.

That fits in with a basic model that for 1024 games you have 512 wins of £1, 256 wins of $2 ... and 1 win of $512 dollars, which amounts to $5,120 total. That would be an alternative way to calculate "expected" winnings. In this case it's $5 per game - but with a small chance to win more or even much more.

If we double the number of games to 2048, then the "expected" winnings are in the range $9,000 - $20,000. For 40 simulations I got them all in this range, except for a $31,000+, a $43,000 and a £1.3 million (!)

For 2048 games you are almost certain to win $9,000, but unlikely to win more than $20,000.

In practical terms, therefore, it critically depends on how many games you are able to play. My answer is:

If I'm going to play 1024 games I would pay $5000, perhaps a bit more; if I'm going to play 2048 games I would pay $10,000, perhaps a bit more etc.

The general formula I would use is $n \times 2^{n -1}$ if we are playing $2^n$ games.

PS for n = 12, i.e. 4,096 games, that would be $25,000. I ran 40 simulations and the winnings were all between $20,000 and $54,000.

 

[pbuk]

So say the government offered you the game with a tax-free payout up to $2^40 (so it would potentially make you the first trillionaire), how much would you pay to play? (the expectation value is $40)

I would pay no more than the amount of change I have in my pocket.

Worth noting that people regularly pay 2-4x (or more) expected A/T PV value for lottery tickets with big payouts

Their decisions are not rational taking only financial considerations into account.

secondly, supposing the government decided to fine you and gave you the option of a fixed fine or paying the results of a game up to your net worth, what multiple of the expectation value would you pay to not take the risk?

This is similar to the decision that you take when deciding on purchasing (optional) insurance. The decision is not based purely on expected value.

Finally - the gov gives you the option of dividing the payout across up to 2^40 simultaneous games, so for example you could play 100 simultaneous games with payouts beginning at $0.02 each

I depends on the distribution of the payouts.

 

[BWV]

I did a 40 simulations of 1024 games, assuming you win $1 if it's tails first time, $2 if it's tails the second time etc. The winnings were almost always between $4,000 and $10,000, with four outliers,all about $14,000.

That fits in with a basic model that for 1024 games you have 512 wins of £1, 256 wins of $2 ... and 1 win of $512 dollars, which amounts to $5,120 total. That would be an alternative way to calculate "expected" winnings. In this case it's $5 per game - but with a small chance to win more or even much more.

If we double the number of games to 2048, then the "expected" winnings are in the range $9,000 - $20,000. For 40 simulations I got them all in this range, except for a $31,000+, a $43,000 and a £1.3 million (!)

For 2048 games you are almost certain to win $9,000, but unlikely to win more than $20,000.

In practical terms, therefore, it critically depends on how many games you are able to play. My answer is:

If I'm going to play 1024 games I would pay $5000, perhaps a bit more; if I'm going to play 2048 games I would pay $10,000, perhaps a bit more etc.

The general formula I would use is $n \times 2^{n -1}$ if we are playing $2^n$ games.

PS for n = 12, i.e. 4,096 games, that would be $25,000. I ran 40 simulations and the winnings were all between $20,000 and $54,000.

the payouts are 2,4,8,16 etc - so that the expectation with a finite max payout of 2^n is n
I ran 20 runs of 10^8 games and the mean payout was $33.5, so less than expectation with a median of $31.5, a max of $73.4 and min of $24.6. The largest payout across all 2B runs was 2^30, $1.07B
so this is a lottery that could be offered at expectation value where the sponsor would most likely make money

 

[BWV]

this is my MATLAB code if anyone is interested:

clear; close all; clc
results=zeros(20,1);
maxp=zeros(20,1);
tic
for i=1:20
m=rand(10^8,1);
test=ceil(-log2(m));
pay=2.^test;
results(i,1)=mean(pay);
maxp(i,1)=max(pay);
end
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