## Penalty Shootout Wager

Having been inspired by the recent World Cup, A approaches B with a proposition:

A: Hey B, let’s play a game of sudden-death penalties. If I score and you miss, you pay me $10. If you score and I miss, I’ll pay you$10. If we both score or both miss, we play another round. How about it?
B: That’s not a fair game! Your scoring rate against me is twice my scoring rate against yours! How about this: 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. That sounds fairer!

Should A accept B’s counter proposal?

The first thing we should work out is the probability of A winning the game. If we let $p$ be the probability that A wins the game, then the probability that B wins the game is $1 - p$. A’s expected winnings from the game, which we will denote with $E$, would then be determined as below:

\begin{aligned} E &= \mathbb{P}\{\text{A wins} \} \times \text{Amount A wins} \\ &\hspace{2em} + \mathbb{P} \{\text{A doesn't win} \} \times \text{Amount A wins} \\ &= p \times \text{Amount A wins} \\ &\hspace{2em} - (1-p) \times \text{Amount A loses}.\end{aligned}

A should accept the counter offer if his expected winnings is positive, and reject it if his expected winnings is negative.

As with this AMC problem, we could play one round of the game and examine the possible states we find ourselves in. Let the probabilities of A scoring and B scoring be $p_a$ and $p_b$ respectively. After one round, we find ourselves in one of 3 possible states:

Case 1: A scores and B misses.
In this case, A wins. The probability of this case happening is $p_a(1-p_b)$.

Case 2: A misses and B scores.
In this case, A loses. The probability of this case happening is $(1-p_a)p_b$.

Case 3: A & B both score or both miss.
In this case, we find ourselves back at square one. Hence, A has probability $p$ of winning the game. The probability of this case happening is $p_ap_b + (1-p_a)(1-p_b)$.

Summarising the 3 cases, we have an equation for $p$:

\begin{aligned} &p &&= p_a(1-p_b) + p [p_ap_b + (1-p_a)(1-p_b)], \\ &p &&= p_a(1-p_b) + p(1 - p_a - p_b + 2p_ap_b), \\ &p(p_a + p_b - 2p_ap_b) &&= p_a(1-p_b), \\ &p &&= \frac{p_a(1-p_b)}{p_a + p_b - 2p_ap_b}. \end{aligned}

In the problem we are not told what the values of $p_a$ and $p_b$ are, just that $p_a = 2p_b$. Substituting this into the expression for $p$ above:

\begin{aligned} p &= \frac{2p_b(1-p_b)}{2p_b + p_b - 2(2p_bp_b)} \\ &= \frac{2p_b(1-p_b)}{3p_b - 4p_b^2} \\ &= \frac{2 - 2p_b}{3-4p_b}.\end{aligned}

Putting it all together in our initial expression for A’s expected winnings, we get

\begin{aligned} E &= p \times \text{Amount A wins} \\ &\hspace{2em} - (1-p) \times \text{Amount A loses} \\ &= \left(\frac{2 - 2p_b}{3-4p_b}\right) (10) - \left(1 - \frac{2 - 2p_b}{3-4p_b} \right) (20) \\ &= \left(\frac{2 - 2p_b}{3-4p_b}\right) (10) - \left( \frac{1-2p_b}{3-4p_b} \right) (20) \\ &= \frac{(20 - 20p_b) - (20 - 40p_b)}{3-4p_b} \\ &= \frac{20p_b}{3 - 4p_b}, \end{aligned}

which is definitely non-negative since the denominator is always positive (why?). Hence, under the counter proposal A’s expected winnings are non-negative, so he should accept the counter proposal.