An infinite coin-flipping game [closed]

Question

Consider a coin-flipping game that costs $1 to play.

• If heads, I win $3.
• If tails, I win nothing.

I start playing this game with $100, and continue playing infinitely. Am I guaranteed to eventually lose all my money?

Answer 1

We can model your winnings as an infinite Discrete-Time Markov Chain, where the state is your current dollar amount. From state $i$, you either move to state $i-1$ or to state $i+2$, each with probability 1/2. Starting from state $i$, we can calculate the probability of ever reaching the state $i-1$, i.e. losing a dollar. Call this probability $x$.

The crucial question is whether $x < 1$. A full proof is beyond the scope of this answer, but note that after a large number of flips $n$, the distribution of the current winnings is approximately a normal distribution with mean $100 + n/2$ and standard deviation $O(\sqrt{n})$. Examining the formula for the tail of a normal distribution, we see that the chance of a normal distribution falling $\Omega(\sqrt{n})$ standard deviations below the mean is $e^{-\Omega(n)}$. This series converges, so it gets increasingly rare to ever fall below \$$100$, if it hasn't happened in the first few steps. So $x < 1$.

Now, let's calculate $x$. There is a $1/2$ chance of immediately moving from $i$ to $i-1$, and a $1/2$ chance of moving to $i+2$, from where we will need to descend three times to reach $i-1$. Each of these descents is independent, so $x$ satisfies the equation:

$$x = 1/2 + x^3/2$$

The unique solution in the interval $(0, 1)$ is $x = (\sqrt{5} - 1)/2 = \phi - 1 \approx 0.618$.

To run out of money, we must lose \$$100$. This requires losing a dollar 100 times, independently. This happens with probability $x^{100} \approx 10^{-21}$.

It is very unlikely that you will ever run out of money, as you play the game infinitely.

Answer 2

Answer:

No

Explanation:

First, let's map the problem to a random walk on the nonnegative numbers, with the initial position $100$ and you move one unit to the left with probability $1/2$ and two units to the right with probability $1/2$. Each walk that eventually lands you at zero involves $k$ right-steps (so $2k$ units to the right) and $100+2k$ left-steps, where $k$ can be any nonnegative integer. For a given $k$, there are several moments in time you could have made the $k$ right-steps. In fact, there are at most $\binom{100+3k}{k}$ moments where you could have made the $k$ right steps (this is an overestimate because we can't take all the left-steps before the right-steps, but we need only an upper bound on the probability that you will reach $0$, so this is okay). Therefore, the number of walks involving $k$ right-steps is at most $\binom{100+3k}{k}$. The probability for such a walk is $2^{-(100+3k)}$ because it involves a total of $100+3k$ steps, so the probability that you will reach $0$ is at most $$\sum_{k=0}^\infty \binom{100+3k}{k}2^{-(100+3k)} < 10^{-20}.$$

Answer 3

Yes, you are guaranteed to eventually lose all your money if you continue playing infinitely. Although you have a positive expected value, each flip is independent, and there is significant variance. There will be long streaks of tails (losing streaks) that can deplete your balance. Since you have a finite starting balance of $100, you are not infinitely resilient to these streaks of losses.

This is due to the concept of gambler’s ruin in probability theory, which states that with infinite play, the probability of going broke approaches certainty unless you have an infinite bankroll.

Answer 4

This is more of a maths questions, but the answer is no, you are not guaranteed to lose all your money. I note that the other answers, at this time, talk about the likelihood of winning, but that is not what you asked.

A guaranteed loss means that there is no path through the game in which you don't lose your money. If we can find just a single path in which you don't lose; then we have proved the point. This is easy to do; in fact there are many. The obvious one is that you always flip heads. It becomes increasingly unlikely as you play more iterations of the game, but however small the chance is, it is never non-zero.

The conditions of the game, such as the starting funds and the amount you win are irrelevant and a distraction. You could start with a single dollar instead of 100 and win 1 cent instead of $3. The answer is the same.