# Friday Morning Math Problem

BACKGROUND

In the St. Petersburg paradox, Petra von Player and Paul de Bank agree to play a game based on the toss of a coin. If a head is thrown on the first toss (prob = 1/2), Paul de Bank pays Petra von Player a dollar, and the game is over. If the first toss is tails and the second toss is head (prob = 1/4), Paul de Bank pays Petra von Player two dollars, and the game is over. If the first head appears on the Nth toss (prob = 1/2^N), Paul de Bank pays Petra von Player 2^(N-1) dollars, and so on. Each time the winning head does not appear, the payoff doubles.

The paradox comes in the form of the question, "How much should Petra von Player be willing to pay to play this game?" The answer is, she should be willing to pay an infinite amount of money to play this game.

Resolution:

The resolution to the paradox lies in the fact that the expected winnings of Petra von Player are infinite:

EW = 1 * 1/2 + 2 * 1/4 + 4 * 1/8 + ... = 1/2 + 1/2 + 1/2 + ... = infinity

If Petra is to make money off the game, she should be willing to pay a quantity less than the expected payoff to play the game. But if the expected payoff is infinite, Petra should be willing to pay an infinite amount of money to play the game - hence the paradox. (The resolution comes when we recognize that if Petra von Player can play for 10^100 years, there is a good chance that eventually she will win a payoff that makes it worth the wait, and that Paul de Bank will actually have 10^65 dollars or whatever.)

PROBLEM

Suppose Paul de Bank has a trillion (10^12) dollars.

a. How many coin tosses would it take Petra von Player to clean out Paul de Bank?

b. How much should Petra von Player be willing to pay to play?

c. What is the probability Paul de Bank will get cleaned out?

d. What is the probability that Petra von Player will lose money on the game (assuming Petra von Player pays the amount from part b)?

Solution
Part A - number of coin tosses for von Player to clear out de Bank?

We can find the number of times we have to double 1 to reach Paul de Bank's billion dollar fortune:

>>> np.log2(1e12)
39.86

>>> np.power(2,39)
549,755,813,888


But remember that bets start at 1, or $2^0$, so $2^{39}$ takes 40 wins to reach.

If Petra von Player wins 40 times in a row it would leave de Bank with a measely

>>> 1e12 - np.power(2,40-1)
450,244,186,112


half a trillion dollars left to live on. If Petra von Player wins 41 times in a row, it leaves de Bank underwater

>>> 1e12 - np.power(2,41-1)
-99,511,627,776


and de Bank ends up with 99 billion dollars of debt.

An alternative way to think about it is to use the back of the envelope approximation: $2^{10} \sim 10^3$

so it follows that $10^{12} \sim (10^3)^4 \sim (2^{10})^4 \sim 2^{40}$

Our back of the envelope estimate is about 40 times, which matches the more precise calculations above.

Part b - amount to play?

Given our maximum of 40 tosses, and the 50% probaability, a maximum of 40 tosses leads to an expected payoff of

E = 1/2 + 1/2 + ... + 1/2
E = 40*1/2
E = 20


So we should expect to win 20 from the game. Part c - What is probability Paul de Bank will get cleaned out? The probability of getting a heads once is $\dfrac{1}{2}$; twice is $\dfrac{1}{4}$; and so on. The probability of 40 heads in a row is: $\dfrac{1}{2}^n = \dfrac{1}{2}^{40} = \dfrac{1}{2^{40}} = 9.09 \times 10^{-13}$ Alternatively, using our back of the envelope estimate of $2^{10} \sim 10^3$ we get a back of the envelope estimate of \begin{align} \dfrac{1}{2}^{40} & \sim & 2^{-40} \sim (2^{10})^{-4} \\ \dfrac{1}{2}^{40} & \sim & (10^3)^{-4} \\ \dfrac{1}{2}^{40} & \sim & 10^{-12} \end{align} if Paul de Bank paints a single one dollar bill red, and then put all of his dollars into one giant pile, it's the probability of pulling out that one single dollar bill Part d - What is the probability that Petra von Player will lose money on the game? To make back20, we would need to win a power of 2 quantity more than 16 - which is 32

>>> np.log2(20)
4.32

>>> np.power(2,5)
32


To make $32 we have to win 6 times (remember, the first bet is$1, so we are starting at $2^0$ and it takes 6 wins to receive a payoff of at least $2^5$), which has the probability: $\dfrac{1}{2^6} \sim 0.0156$

so probability of Petra von Player losing money on the game is

>>> 1-(1/np.power(2,6))
0.9843