The game:
You are player A, and your opponent is player B. We have a fair coin. The score, $s$ starts at $0$ and the wager, $w$ at \$$1$
We toss the coin and if heads the score adds $1$ and if tails we takeaway $1$. For example if we tossed Heads Tails Tails Heads Heads Heads the score would move as follows: $0 \to 1 \to 0 \to -1 \to 0 \to 1 \to 2$
The game ends when the score is $\pm 2$ If the score is $+2$ then player A wins, and if $-2$ then player B wins. The looser pays the winner the wager.
Here is where your advantage and the strategy comes in. Before every toss of the coin player A gets to decide if they want to double the wager or not.
Example:
First toss: Player A decides to double wager, $w = \$2$. A head is tossed, $s = 1$
Second toss: Player A decides to double the wager again, $w = \$4$. A tail is tossed, $s=0$
Third toss: Player A decides to not double the wager, $w = \$4$. A tail is tossed, $s=-1$
Fourth toss: Player A decides to not double the wager, $w = \$4$. A tail is tossed, $s=-2$
Player A now must pay Player B $\$4$
What is the optimal Strategy, with respect to maximising EV? I have found a way to make the expected value +2 but I believe there is a better strategy. I will explain through some of the reasoning I had with this problem:
First attempt, Strategy 1:
Let us say we double if we reach $+1$ for the first time. If we reach the score $+1$ our probability of winning has now risen to $\frac{3}{4}$. The probability that we reach $+1$ ever is $\frac{2}{3}$ (Simple Markov Chains) Let $S_1$ denote Strategy 1. And the event that we reach $s=1$ be denoted $J$
Hence $\mathbb{E}[S_1] = \mathbb{P}(J^c)\mathbb{E}[S_1|J^c] + \mathbb{P}(J)\mathbb{E}[S_1|J] =\frac{1}{3} \cdot -\$1 + \frac{2}{3} \cdot ( \$2 \cdot \frac{3}{4} - \$2 \cdot \frac{1}{4}) = \$\frac{1}{3} $
Amazing! We now have reached a positive expected value of this game. Surely if the wager simply started at $2$ instead of $1$ i.e we are playing for twice us much then our expected value has indeed doubled! Hence if we modify $S_1$ slightly by doubling before the first toss, so the game is twice as valuable then our expected payoff of now $\frac{2}{3}$ You can check that the maths works out for this but it is intuitive. Expectation is linear.
Second attempt, Strategy 2:
Why only double the wager if we reach $s=1$ for the first time. Since we have this advantage we ought to double every single time we reach $s=1$. I.e if the score goes $0 \to 1 \to 0 \to 1$ why not double again? We showed it was profitable before. Let us define $S_2$ by doubling every single time we reach $s=1$. We will calculate its expected payoff by conditioning on the first 2 tosses, using recurrence. Instead of recording the tosses we show simply write the sequence of the score. i.e Heads Tails shall be represented as $(0,1,0)$
We have four possible states for the first two tosses: $ \{ (0,1,2), (0,1,0) , (0,-1,-2) , (0,-1,0) \} $
Under the first we simply doubled once when we reached $s=+1$ and then won $\$2$
Under the second we doubled the value of the game and then are back at the beginning.
Under the second we simply lost $\$1$. We never doubled the game and lost.
Under the fourth nothing of significance has happened. The game is at $s=0$ and we have done nothing.
Hence: $\mathbb{E}[S_2] = \frac{1}{4}\cdot \$2 + \frac{1}{4} \cdot 2\mathbb{E}[S_2] + \frac{1}{4} \cdot (-\$1) + \frac{1}{4} \cdot \mathbb{E}[S_2] $
Solving this yields $\mathbb{E}[S_2] = \$1$
Great so now we have made even more money! And like before if we choose to make the game twice as valuable, think of this as simply the starting wager is $\$2$. This happens if we choose to simply double before the first toss. Then we have a strategy, $S_2^{'}$, with expected payoff of $2$.
I have ran both of these strategies in python and the empirical values of EV match our "theoretical" answers.!
The problem
Now you might think I am just getting greedy but lets say now we double the wager ALWAYS, unless $s=-1.$ This means we double initially, we double if $s=1$ and we now also double on $s=0$. My reasoning is this: If you are player $A$ and I told you that you could pick any initial value for the wager, you should always want a larger value. Ultimately as long as Player $A$ is not "loosing" they should always want to have the game very expensive because of there innate advantage. We soundly justified doubling the game initially so why not always double on $s=0$ too ? Lets call $S_3$ this strategy. I cannot seem the calculate the expected value of $S_3$ Python simulations show nothing at all helpful, it oscillates from about -20000 to 20000 ?
Could anyone show a method for calculation $\mathbb{E}[S_3]$ ?
Thanks a lot!
Oskar
With your third strategy the expected value is undefined because it does not converge. Let us assume there is some expected value $E$ when the stake on the first toss is $2$ because you doubled already. If we get two heads we win $4$ because we double once more. If we get two tails we lose $2$ because we don't double any more. If we get tails and then heads we are back to a score of $0$, but we double again, so the expected value is $2E$. If we get heads and then tails we are back to a score of $0$ but we have doubled twice, so the expected value is $4E$ The contribution of each round of two flips increases without bound, as the chance we return to $0$ is $\frac 12$ but the expected value from then is $\frac 32E$ (plus the bit that comes from terminating the game).