You flip four fair coins, if you get 4 heads I will give you $10.00
You pay me $1.00 to play.
Should you play?
I want to have a discussion about the followup from this question, as I believe each play would be independent, but was told no it was conditional...
this means that you can keep playing the game, keeping the number of heads you already got and tossing the remaining coins. Every time you do this, you'll have to pay 1\$ to toss the remaining coins.
On
With the reservations I made in my comment about the question of defining "Should X take this gamble or not" without referencing risk aversion of preferences, below I compute the distribution of this process numerically. I assume the gambler keeps playing until he gets the four heads.
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
def gamble():
'''
The stochastic process
'''
plays = 1
outcome = -1 # Has to pay 1 to join game
toss = np.random.randint(0,2, size = 4) # First toss
heads = np.sum(toss)
if heads == 4: # 4 heads
outcome += 4 # Receives 4
else:
while not heads == 4: # Keeps playing until he gets 4 heads
outcome += -1
plays += 1
toss = np.random.randint(0,2, size = 4-heads)
heads += np.sum(toss)
outcome += 4
return (outcome, plays)
simulations = [gamble() for i in range(1000000)]
Some statistics of the empirical distribution are as follows:
\begin{array}{|c|c|c|c|}
{} & money & plays \\
mean & 0.49 & 3.51 \\
std & 1.75 & 1.75 \\
min & -20.00 & 1.00 \\
25\% & 0.00 & 2.00 \\
50\% & 1.00 & 3.00 \\
75\% & 2.00 & 4.00 \\
max & 3.00 & 24.00 \\
\end{array}
Again, notice that, even though the mean and median are positive, the distribution of gains is left skewed (empirical skewness is around $-1.33$): you have small probabilities of getting very low results.
An agent who is risk averse and loss averse would be specially wary of playing such a gamble. Again, there is no way of answering whether he should play this gamble or not without imposing more structure in the problem (the preferences of the agent).
Answering Rohit Pandey's comments:
Yes, a rich person would be more willing to pay a high amount to play the St. Petersburg lottery. As I said, to analyze this better from an individual point of view you would need to impose more structure. There's a whole area in economics called choice under uncertainty who was first developed by von Neumann and Morgenstern that connects uncertainty to utility. Economists typically assume that individuals are risk averse, and that this risk aversion decreases with wealth (the more wealth someone is, the more risk he is willing to take). This is directly connected to the also common assumption of decreasing marginal utility from wealth: an additional dollar to Elon Musk is of negligible worth, whereas an extra dollar for someone in extreme poverty is worth a lot.
"Also, if you want to keep things real, then any real casino wouldn't have infinite resources either and when the terms are bounded by the money available on Earth, the expectation argument starts making sense again."
The expectation argument does make sense, but for a specific type of risk aversion (risk-neutral utilities). In general, individuals are not risk neutral, but risk averse. This is more of an empirical fact, not a mathematical necessity. Mathematically, risk-neutrality implies an utility that is linear on wealth, while risk-aversion implies an utility concave on wealth. To see why the problem does not disappear by simply considering finite resources, imagine the following gamble: you pay $ \$0$ and win $\$0.01$ with probability $p = \frac{10^6-1}{10^6}$ and lose $\$9,998$ dollars with probability $1-p = \frac{1}{10^6}$. This lottery has positive expected value, but I don't think many people would take it. From a theoretical point of view, this reluctance to accept such a lottery can come, for example, with a concave utility, e.g., $u(w) = \sqrt{w}$. Imagine an individual with an initial wealth $w_0 = 10,000$. The expected utility from taking this lottery $g$ would be:
$$\mathbb{E}(u(g)) = pu(w_0+0.01) + (1-p)u(w_0-9,998) = 99.99 < u(w_0) = 100$$
And such individual would not take that bet.
You shouldn't play since the probability of winning \$10 is $\frac {1} {2^4} = \frac{1}{16}$. So, the expected payout is $\frac{10}{16}<1$ (the cost of playing is 1\$).
EDIT: OP mentioned in the comments that you can actually continue playing this game. So, if you get 2 heads and 2 tails for example, you can pay another 1\$ and toss the two remaining coins and keep repeating this process till you get all 4 heads. This makes the problem slightly more interesting. We can now think of the process as a Markov chain with 5 states. Each state represents the number of heads accumulated so far (0 through 4). It is clear that you won't go backwards. If you have 2 heads for example, you can only go to states 2, 3 or 4 if you play again. The complete Markov matrix for this game becomes (some entries left blank for you to fill in) $$Q= \left[ \begin{matrix} \frac 1 {16} & . & . & . & \frac{1}{16}\\ 0 & \frac{1}{8} & . & . & \frac 1 8\\ 0 & 0 & \frac{1}{4} & \frac 1 2 & \frac 1 4\\ 0 & 0 & 0 & \frac 1 2 & \frac 1 2 \\ 0 & 0 & 0 & 0 & 1 \end{matrix} \right] $$
It is clear that the last state (4 heads) is an absorbing state. If you keep playing, how many times (on average) will you need to pay 1\$ before you reach the absorbing state of 4 heads and collect your 10\$ reward? This is the number of steps within transient states for the Markov matrix. Refer here for the fact that the average number of steps needed before getting to the 4\$ absorbing state is the very first entry of the vector:
$$t = (I-Q)^{-1} \Bbb I$$
Here, $I$ is the identity matrix and $\Bbb I$ is a vector of all ones. If this expected number of steps is less than 10, you should certainly play the game.
Another less precise and more intuitive way to look at this (if someone stops you on the street and offers you the opportunity and you can't invert matrices in your head) is that the number of remaining coins roughly halves on average every time you play. So, you should need on the order of $\log_2(4)$ tosses until all four coins become heads. This is roughly the amount you'll end up paying and much smaller than the 10\$ reward waiting for you.