Penalty Shootout Wager 2

B, having made a fool of himself by proposing a penalty shootout format that disadvantaged him (see this post— B’s expected winnings were ≤ 0), goes back and thinks about how he can get his money back. After some thought, he proposes the following game to A:

B: A, let’s play another game. Penalties as well, but this time the game is over the moment someone scores, and the person who scored wins. As before, your scoring rate against me is twice my scoring rate against yours, so let’s stick with the original payouts. If you score and I miss, I’ll pay you $10, but if I score and you miss, you have to pay me double-$20. What do you say?

Has B redeemed himself by offering a game where his expected winnings is non-negative?

We will use the same general approach as in Part 1, but let me take this chance to introduce the state diagram. The state diagram maps out all possible states the game can be in, and shows the probability of moving from one state to another with each passing time step. Letting the probabilities of A scoring and B scoring be p_a and p_b respectively, the state diagram for the game is as follows, where the orange circle indicates the starting state:


(Note: We have drawn a state diagram on this blog before, have a look!) Hence, to find B’s expected winnings of the game, we need to find the probability that B wins given that the current state is B’s turn, i.e. \mathbb{P}\{ \text{B wins } | \text{ B's turn}\} .

Looking at the state diagram, there is 2 ways to reach state “B wins” from the starting state. Either B scores (in which case the game ends), or B misses and A misses (in which case we are back to the starting state). Hence,

\begin{aligned} \mathbb{P} \{ \text{B wins } | \text{ B's turn} \} &= p_b + (1-p_b)(1-p_a) \mathbb{P} \{ \text{B wins } | \text{ B's turn}\}, \\  \left[ 1- (1 - p_a - p_b + p_ap_b) \right]\mathbb{P} \{ \text{B wins} | \text{ B's turn} \}&= p_b, \\  \mathbb{P} \{ \text{B wins } | \text{ B's turn} \} &= \frac{p_b}{p_a + p_b - p_ap_b}. \end{aligned}

Using the fact that p_a = 2p_b, we can simplify the above:

\begin{aligned} \mathbb{P} \{ \text{B wins } | \text{ B's turn} \} &= \frac{p_b}{2p_b + p_b - 2p_b^2} \\  &= \frac{1}{3 - p_b}. \end{aligned}

We can finish now by calculating B’s expected winnings using the definition of expected value:

\begin{aligned} \text{B's expected winnings } &= \mathbb{P} \{ \text{B wins } | \text{ B's turn} \} \times \text{ Amount B wins} \\  &\hspace{2em} - \mathbb{P} \{ \text{B loses } | \text{ B's turn} \} \times \text{ Amount B loses} \\  &= \left( \frac{1}{3-p_b} \right)(20) - \left( 1- \frac{1}{3-p_b} \right) (10) \\  &= \frac{20}{3-p_b} - \frac{10(2-p_b)}{3 - p_b} \\  &= \frac{10p_b}{3 - p_b}, \end{aligned}

which is non-negative as p_b \leq 1. Hence, B has redeemed himself– the game gives B non-negative expected winnings.

Note: In some sense B redeems himself only partially. Even though he now shows that he is capable of producing a game which gives him the edge, that edge is not enough to overcome the edge that he gave up in Part 1. If A & B were to play one game of Part 1 and one game of Part 2,

\begin{aligned} \text{B's expected winnings} &= - \frac{20p_b}{3-4p_b} + \frac{10p_b}{3-p_b} \\  &= \frac{-20p_b(3-p_b) + 10p_b (3-4p_b)}{(3-4p_b)(3-p_b)} \\  &= \frac{-20p_b^2 - 30 p_b}{(3-4p_b)(3-p_b)},\end{aligned}

which can never be positive.

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