St. Petersburg Paradox

I don't have ten large in the bank right now. I'd beg, borrow, and steal if the chance of winning wasn't 1/1024.

I'd do it for $100, though, to win $10 million.

Am I a "typical individual?" Or do I miss the point entirely?

Perhaps the resolution comes from a mental budget that caps how much people will devote to gambling. Obviously people are happy to lose $1 for a much smaller chance of winning millions.

In theory a casino could simply offer this 'game' for a $100 bet. They would expect to raise $1,025,000 for every person who won the ten coin flips so they could profit by making the payout an even million. I think people would take it.

Seems to me that Arnold is pointing at a nice (probably not original) solution to the Paradox. In the real world, it's ridiculous to talk about 1000 trillion dollar payouts -- not just hard to believe someone is going to do it, it's completely impossible. And as soon as you cap the payout, the infinite series goes away, and the expected return of the bet is something like $25. (2^50 is 1,100 trillion or so, and each term of the series is worth fifty cents.)

If I could bet $10,000 for 10 flips infinitely, of course I would do it. One out of 100,000 times the coin is bound to come up heads 10 times in a row. But if I am given a one-time only opportunity for 10 flips, I would turn it down. Also, if I have to pay out $10,000 in cash each time I lose, I couldn't afford to do it enough times to put the odds in my favor. In other words, I think there is an unstated assumption that you can make the bet as many times as you want, which changes the parameters of the bet.

I don't think people calculate the expected payoff the way one should - statistically. I think most people simply figure out if the odds are that they win or lose, and most times you will lose this game.

Wouldn't a person's income or accumulated wealth factor into the decision to enter into certain bets?

I can imagine that a person with $10 million in the bank would be far more willing, in the event of a loss, to give up $10,000 than a person with with $10 in the bank.

Are poor people more likely to bet in a lottery, wheere the total potential loss is just the price of the ticket, than people with relatively large amounts of money?

Is the question of income or wealth even relevant?

Interesting explanation. What is your view on the Allais paradox? That's a psychological one two, and perhaps more important from an economic/preference-studies perspective:

http://blogofpandora.blogspot.com/2003_11_01_blogofpandora_archive.html#106802749125727892

People miss the second function here. It is not the simple odds which apply, but the number of times an Individual has to attempt the bet. The Individual is not betting the odds will work in his favor, but the odds they will work in his favor before he loses the ability to bet. lgl

As previous posters have alluded to, it's not really correct to say that "on average, you stand to win by taking the bet."

Just because the expected value is very high doesn't mean that "on average" the participant will come out ahead, due to the extremely large variance.

It will, in fact, be extremely rare for the participant to come out ahead, and it is therefore a rational decision not to take the bet given any level of risk aversion whatsoever.

I have argued in a recent article 'Solving the St Petersburg Paradox-the paradox which is not and never was' SAJEMS 6 (2) 2003 that the St Petersburg game does not lead to a paradox at all. The source of the confusion lies in the traditional solution to the game which holds that the expected value is infinite. This is a special case - where the game is played an infinite times. For any finite number of times the game is played it finite and modest, of the same order that people will offer to play the game.

What most comments are implying is, in most cases, theoretically incorrect in all situations. St Petersburg's Paradox is purely based upon mathematics, which in the end simply is a reoccurrence of what has been already preceded. You see when n-1, k=1 and 2^k then the answer must be simply 2^n-2 dollars. In this case the player has won $2 (which is 2^n dollars). Taking into account the probability of winning the nth toss is (1/2)^n, then p=1/2 is in all cases.