# Human Strategic Decision Making in Parametrized Games

## Abstract

**:**

## 1. Introduction

## 2. Parametrized Games

**Definition**

**1.**

**Definition**

**2.**

**Definition**

**3.**

## 3. Parametric Decision Lists

**Definition**

**4.**

## 4. Parameter Sampling

**Lemma**

**1.**

**Proof.**

**Lemma**

**2.**

**Proof.**

**Lemma**

**3.**

**Proof.**

**Lemma**

**4.**

**Proof.**

**Theorem**

**1.**

**Proof.**

**Theorem**

**2.**

**Proof.**

## 5. Comparison of Approaches in 2 × 2 Games

- If $a\ge e$ and $b\ge d$ then (1,1).
- Else if $c\ge g$ and $d\ge b$ then (1,2).
- Else if $e\ge a$ and $f\ge h$ then (2,1).
- Else if $g\ge c$ and $h\ge f$ then (2,2).
- Else $\left(\right(p,1-p),(q,1-q\left)\right)$ for $p=\frac{h-f}{b-f+h-d}$, $q=\frac{g-c}{a-c+g-e}$.

## 6. Parametrized Game Examples

#### 6.1. Final Jeopardy

- If ${p}_{2}=0$ wager 0.
- Else if ${p}_{1}=0$ wager 0.
- Else if ${p}_{1}=1$ wager 2.
- Else if ${p}_{2}\ge \frac{1}{2}$ wager 2.
- Otherwise wager 1 with probability $x=\frac{(1-{p}_{1})(1-2{p}_{2})}{1-{p}_{1}+{p}_{1}{p}_{2}}$ and wager 2 with probability $1-x.$

- If ${p}_{2}=0$ wager 0.
- Else if ${p}_{1}=0$ wager 3.
- Else if ${p}_{1}=1$ wager 0.
- Else if ${p}_{1}\ge \frac{1}{2}$ and ${p}_{2}\ge \frac{1}{2}$ wager 2.
- Else if ${p}_{2}\ge \frac{1}{2}$ wager 3.
- Otherwise wager 0 with probability $y=\frac{{p}_{1}{p}_{2}}{1+{p}_{1}{p}_{2}-{p}_{1}}$ and wager 3 with probability $1-y.$

#### 6.2. Generalized Kuhn Poker

- Two players: A and B
- Both players ante $1
- Deck containing three cards: 1, 2, and 3
- Each player is dealt one card uniformly at random
- Player A acts first and can either bet $1 or check
- –
- If A bets, player B can call or fold
- ∗
- If A bets and B calls, then whoever has the higher card wins the $4 pot
- ∗
- If A bets and B folds, then A wins the entire $3 pot

- –
- If A checks, B can bet $1 or check.
- ∗
- If A checks and B bets, then A can call or fold.
- ·
- If A checks, B bets, and A calls, then whoever has the higher card wins the $4 pot
- ·
- If A checks, B bets, and A folds, B wins the $3 pot

- ∗
- If A and B check, then whoever has the higher card wins the $2 pot

- 1.
- Player A’s strategy in the first round:
- A always bets if $x\le \lfloor \frac{n-1}{9}\rfloor $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}9$, then A bets with $x=\lceil \frac{n-1}{9}\rceil $ with probability $\frac{n-1}{9}-\lfloor \frac{n-1}{9}\rfloor $
- A always checks if $\lceil \frac{n-1}{9}\rceil <x<\lfloor \frac{2n+4}{3}\rfloor $
- A always bets if $x\ge \lceil \frac{2n+4}{3}\rceil $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3$, then A bets with $x=\lfloor \frac{2n+4}{3}\rfloor $ with probability $\lceil \frac{2n+4}{3}\rceil -\frac{2n+4}{3}$

- 2.
- Player B’s strategy facing a bet:
- B always calls if $y\ge \lceil \frac{n-1}{3}\rceil $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3$, then B calls with $y=\lfloor \frac{n-1}{3}\rfloor $ with probability $\lceil \frac{n-1}{3}\rceil -\frac{n-1}{3}$
- B always folds if $y<\lfloor \frac{n-1}{3}\rfloor $

- 3.
- Player B’s strategy facing a check:
- B always bets if $y\le \lfloor \frac{n-1}{6}\rfloor $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}6$, then B bets with $y=\lceil \frac{n-1}{6}\rceil $ with probability $\frac{n-1}{6}-\lfloor \frac{n-1}{6}\rfloor $
- B always checks if $\lceil \frac{n-1}{6}\rceil <y<\lfloor \frac{n+3}{2}\rfloor $
- B always bets if $y\ge \lceil \frac{n+3}{2}\rceil $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}2$, then B bets with $y=\lfloor \frac{n+3}{2}\rfloor $ with probability $\lceil \frac{n+3}{2}\rceil -\frac{n+3}{2}$

- 4.
- Player A’s strategy after A checks and B bets:
- A always calls if $x\ge \lceil \frac{n+5}{3}\rceil $
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3,$ then A calls with $x=\lfloor \frac{n+5}{3}\rfloor $ with probability $\lceil \frac{n+5}{3}\rceil -\frac{n+5}{3}$
- A always folds otherwise

#### 6.3. Weakest Link

## 7. Related Research

## 8. Conclusions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Conflicts of Interest

## Appendix A. Uniform Random 2 × 2 Two-Player Strategic-Form Games

- If $a\ge e$ and $b\ge d$ then (1,1).
- Else if $c\ge g$ and $d\ge b$ then (1,2).
- Else if $e\ge a$ and $f\ge h$ then (2,1).
- Else if $g\ge c$ and $h\ge f$ then (2,2).
- Else $\left(\right(p,1-p),(q,1-q\left)\right)$ for $p=\frac{h-f}{b-f+h-d}$, $q=\frac{g-c}{a-c+g-e}$.

## Appendix B. Simplified Two-Player Final Jeopardy

- If player 1 wagers 0 and player 2 wagers 0, then player 1 wins with probability 1. So player 1’s expected payoff is $1-0.5=0.5.$ (Note that we are counting a win as having payoff 0.5, a loss as having payoff -0.5, and tie as having payoff 0, so that the game is zero sum.)
- If player 1 wagers 0 and player 2 wagers 1, then player 1 also wins with probability 1. So player 1’s expected payoff is $1-0.5=0.5.$
- If player 1 wagers 0 and player 2 wagers 2, then player 1 wins with probability $1-{p}_{2}$, and the players tie with probability ${p}_{2}.$ So player 1’s expected payoff is $1-{p}_{2}+0.5{p}_{2}-0.5=0.5-0.5{p}_{2}.$
- If player 1 wagers 0 and player 2 wagers 3, then player 1 wins with probability $1-{p}_{2}$, and player 2 wins with probability ${p}_{2}.$ So player 1’s expected payoff is $1-{p}_{2}-0.5=0.5-{p}_{2}.$
- If player 1 wagers 1 and player 2 wagers 0, then player 1 wins with probability 1. So player 1’s expected payoff is $1-0.5=0.5.$
- If player 1 wagers 1 and player 2 wagers 1, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and the players tie with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is$${p}_{1}+(1-{p}_{1})(1-{p}_{2})+0.5(1-{p}_{1}){p}_{2}-0.5={p}_{1}+1-{p}_{1}-{p}_{2}+{p}_{1}{p}_{2}+0.5{p}_{2}-0.5{p}_{1}{p}_{2}-0.5=0.5{p}_{1}{p}_{2}-0.5{p}_{2}+0.5.$$
- If player 1 wagers 1 and player 2 wagers 2, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is$${p}_{1}+(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+1-{p}_{1}-{p}_{2}-0.5=0.5-{p}_{2}.$$
- If player 1 wagers 1 and player 2 wagers 3, then player 1 wins with probability $1-{p}_{2}$, player 2 wins with probability $(1-{p}_{1}){p}_{2}$, and the players tie with probability ${p}_{1}{p}_{2}.$ So player 1’s expected payoff is$$1-{p}_{2}+0.5{p}_{1}{p}_{2}-0.5=0.5+0.5{p}_{1}{p}_{2}-{p}_{2}.$$
- If player 1 wagers 2 and player 2 wagers 0, then player 1 wins with probability ${p}_{1}$, and the players tie with probability $1-{p}_{1}.$ So player 1’s expected payoff is$${p}_{1}+0.5(1-{p}_{1})-0.5={p}_{1}+0.5-0.5{p}_{1}-0.5=0.5{p}_{1}.$$
- If player 1 wagers 2 and player 2 wagers 1, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is$${p}_{1}+(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+1-{p}_{1}-{p}_{2}+{p}_{1}{p}_{2}-0.5=0.5-{p}_{2}+{p}_{1}{p}_{2}.$$
- If player 1 wagers 2 and player 2 wagers 2, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is$${p}_{1}+(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+1-{p}_{1}-{p}_{2}+{p}_{1}{p}_{2}-0.5=0.5-{p}_{2}+{p}_{1}{p}_{2}.$$
- If player 1 wagers 2 and player 2 wagers 3, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is$${p}_{1}+(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+1-{p}_{1}-{p}_{2}+{p}_{1}{p}_{2}-0.5=0.5-{p}_{2}+{p}_{1}{p}_{2}.$$
- If player 1 wagers 3 and player 2 wagers 0, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 3 and player 2 wagers 1, then player 1 wins with probability ${p}_{1}$, player 2 wins with probability $(1-{p}_{1}){p}_{2},$ and the players tie with probability $(1-{p}_{1})(1-{p}_{2}).$ So player 1’s expected payoff is$${p}_{1}+0.5(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+0.5-0.5{p}_{1}-0.5{p}_{2}+0.5{p}_{1}{p}_{2}-0.5=0.5{p}_{1}{p}_{2}+0.5{p}_{1}-0.5{p}_{2}.$$
- If player 1 wagers 3 and player 2 wagers 2, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is
- If player 1 wagers 3 and player 2 wagers 3, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is
- If player 1 wagers 4 and player 2 wagers 0, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 4 and player 2 wagers 1, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 4 and player 2 wagers 2, then player 1 wins with probability ${p}_{1}$, player 2 wins with probability $(1-{p}_{1}){p}_{2},$ and the players tie with probability $(1-{p}_{1})(1-{p}_{2}).$ So player 1’s expected payoff is$${p}_{1}+0.5(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+0.5-0.5{p}_{1}-0.5{p}_{2}+0.5{p}_{1}{p}_{2}-0.5=0.5{p}_{1}{p}_{2}+0.5{p}_{1}-0.5{p}_{2}.$$
- If player 1 wagers 4 and player 2 wagers 3, then player 1 wins with probability ${p}_{1}+(1-{p}_{1})(1-{p}_{2})$, and player 2 wins with probability $(1-{p}_{1}){p}_{2}.$ So player 1’s expected payoff is
- If player 1 wagers 5 and player 2 wagers 0, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 5 and player 2 wagers 1, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 5 and player 2 wagers 2, then player 1 wins with probability ${p}_{1}$, and player 2 wins with probability $1-{p}_{1}.$ So player 1’s expected payoff is ${p}_{1}-0.5.$
- If player 1 wagers 5 and player 2 wagers 3, then player 1 wins with probability ${p}_{1}$, player 2 wins with probability $(1-{p}_{1}){p}_{2},$ and the players tie with probability $(1-{p}_{1})(1-{p}_{2}).$ So player 1’s expected payoff is$${p}_{1}+0.5(1-{p}_{1})(1-{p}_{2})-0.5={p}_{1}+0.5-0.5{p}_{1}-0.5{p}_{2}+0.5{p}_{1}{p}_{2}-0.5=0.5{p}_{1}{p}_{2}+0.5{p}_{1}-0.5{p}_{2}.$$

- (0,0) is a Nash equilibrium if ${p}_{2}=0.$
- Else (0,3) is a Nash equilibrium if ${p}_{1}=0.$
- Else (2,0) is a Nash equilibrium if ${p}_{1}=1.$
- Else (2,2) is a Nash equilibrium if ${p}_{1}\ge \frac{1}{2},{p}_{2}\ge \frac{1}{2}.$
- Else (2,3) is a Nash equilibrium if ${p}_{1}<\frac{1}{2},{p}_{2}\ge \frac{1}{2}.$
- Else P1 wagers 1 with probability $x=\frac{(1-{p}_{1})(1-2{p}_{2})}{1-{p}_{1}+{p}_{1}{p}_{2}}$ and wagers 2 with probability $1-x$, and P2 wagers 0 with probability $y=\frac{{p}_{1}{p}_{2}}{1+{p}_{1}{p}_{2}-{p}_{1}}$ and 3 with probability $1-y$ is a Nash equilibrium if ${p}_{2}<\frac{1}{2}.$

## Appendix C. Generalized Kuhn Poker

#### Appendix C.1. Three Card Kuhn Poker

- Two players: A and B
- Both players ante $1
- Deck containing three cards: 1, 2, and 3
- Each player is dealt one card uniformly at random
- Player A acts first and can either bet $1 or check
- –
- If A bets, player B can call or fold
- ∗
- If A bets and B calls, then whoever has the higher card wins the $4 pot
- ∗
- If A bets and B folds, then A wins the entire $3 pot

- –
- If A checks, B can bet $1 or check.
- ∗
- If A checks and B bets, then A can call or fold.
- ·
- If A checks, B bets, and A calls, then whoever has the higher card wins the $4 pot
- ·
- If A checks, B bets, and A folds, then B wins the $3 pot

- ∗
- If A checks and B checks, then whoever has the higher card wins the $2 pot

- A bets with a 1 in the first round with probability $\frac{\alpha}{3}$
- A always checks with a 2 in the first round
- A bets with a 3 in the first round with probability $\alpha $
- If A bets in the first round, then:
- –
- B always folds with a 1
- –
- B calls with a 2 with probability $\frac{1}{3}$
- –
- B always calls with a 3

- If A checks in the first round, then:
- –
- B bets with a 1 with probability $\frac{1}{3}$
- –
- B always checks with a 2
- –
- B always bets with a 3

- If A checks and B bets, then:
- –
- A always folds with a 1
- –
- A calls with a 2 with probability $\frac{\alpha}{3}+\frac{1}{3}$
- –
- A always calls with a 3

- There are infinitely many equilibria
- There are no pure strategy equilibria
- Equilibrium strategies contain some elements of deceptive behavior. For example, player A sometimes checks with a 3 as a trap or slowplay, and both players sometimes bet with a 1 as a bluff.

#### Appendix C.2. Solution to Generalized Kuhn Poker

- Player A’s strategy in the first round:
- –
- A always bets if $x\le \lfloor \frac{n-1}{9}\rfloor $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}9$, then A bets with $x=\lceil \frac{n-1}{9}\rceil $ with probability $\frac{n-1}{9}-\lfloor \frac{n-1}{9}\rfloor $
- –
- A always checks if $\lceil \frac{n-1}{9}\rceil <x<\lfloor \frac{2n+4}{3}\rfloor $
- –
- A always bets if $x\ge \lceil \frac{2n+4}{3}\rceil $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3$, then A bets with $x=\lfloor \frac{2n+4}{3}\rfloor $ with probability $\lceil \frac{2n+4}{3}\rceil -\frac{2n+4}{3}$

- Player B’s strategy facing a bet:
- –
- B always calls if $y\ge \lceil \frac{n-1}{3}\rceil $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3$, then B calls with $y=\lfloor \frac{n-1}{3}\rfloor $ with probability $\lceil \frac{n-1}{3}\rceil -\frac{n-1}{3}$
- –
- B always folds if $y<\lfloor \frac{n-1}{3}\rfloor $

- Player B’s strategy facing a check:
- –
- B always bets if $y\le \lfloor \frac{n-1}{6}\rfloor $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}6$, then B bets with $y=\lceil \frac{n-1}{6}\rceil $ with probability $\frac{n-1}{6}-\lfloor \frac{n-1}{6}\rfloor $
- –
- B always checks if $\lceil \frac{n-1}{6}\rceil <y<\lfloor \frac{n+3}{2}\rfloor $
- –
- B always bets if $y\ge \lceil \frac{n+3}{2}\rceil $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}2$, then B bets with $y=\lfloor \frac{n+3}{2}\rfloor $ with probability $\lceil \frac{n+3}{2}\rceil -\frac{n+3}{2}$

- Player A’s strategy after A checks and B bets:
- –
- A always calls if $x\ge \lceil \frac{n+5}{3}\rceil $
- –
- If $n\ne 1\phantom{\rule{0.277778em}{0ex}}mod\phantom{\rule{0.277778em}{0ex}}3,$ then A calls with $x=\lfloor \frac{n+5}{3}\rfloor $ with probability $\lceil \frac{n+5}{3}\rceil -\frac{n+5}{3}$
- –
- A always folds otherwise

#### Appendix C.3. Proof of Correctness

- B is facing a bet from A.Note that for all $n\ge 3,$ we have$$\u2308\frac{n-1}{9}\u2309\le \u230a\frac{n-1}{3}\u230b\le \u2308\frac{n-1}{3}\u2309\le \u230a\frac{2n+4}{3}\u230b$$
- –
- B is dealt $y\ge \lceil \frac{n-1}{3}\rceil $A is bluffing with probability$$\frac{\left(\frac{n-1}{9}\right)}{\left(\frac{n-1}{9}\right)+\left(\frac{n-1}{3}\right)}=\frac{1}{4}.$$B wins the pot whenever A is bluffing, and either wins or loses when A is value betting. Therefore, his expected payoff of calling is at least$$\frac{1}{4}\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}}\$3-\frac{3}{4}\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}}\$1=\$0.$$Since his expected payoff of folding would be $0, calling is a best response.
- –
- B is dealt $y=\lfloor \frac{n-1}{3}\rfloor $The analysis of the previous case shows that B will obtain an expected payoff of $0 by both calling and folding, and is therefore indifferent between the two actions.
- –
- B is dealt $y<\lfloor \frac{n-1}{3}\rfloor $Now B loses whenever A is value betting, and either wins or loses when A is bluffing. So his expected payoff of calling is at most $0, and folding is a best response.

- A checked in the first round and is facing a bet from B. Note that for all $n\ge 3,$ we have$$\u2308\frac{n-1}{6}\u2309\le \u230a\frac{n-1}{3}\u230b\le \u2308\frac{n-1}{3}\u2309\le \u230a\frac{n+3}{2}\u230b$$
- –
- A is dealt $x\ge \lceil \frac{n-1}{3}\rceil $B is bluffing with probability$$\frac{\left(\frac{n-1}{6}\right)}{\left(\frac{n-1}{6}\right)+\left(\frac{n-1}{2}\right)}=\frac{1}{4}.$$A wins the pot whenever B is bluffing, and either wins or loses when B is value betting. Therefore, his expected payoff of calling is at least$$\frac{1}{4}\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}}\$3-\frac{3}{4}\phantom{\rule{0.166667em}{0ex}}\xb7\phantom{\rule{0.166667em}{0ex}}\$1=\$0.$$Since his expected payoff of folding would be $0, calling is a best response.
- –
- A is dealt $x=\lfloor \frac{n-1}{3}\rfloor $The analysis of Case Appendix C.3 shows that A will obtain an expected payoff of $0 by both calling and folding, and is therefore indifferent between the two actions.
- –
- A is dealt $x<\lfloor \frac{n-1}{3}\rfloor $Now A loses whenever B is value betting, and either wins or loses when B is bluffing. So his expected payoff of calling is at most $0, and folding is a best response.

## Appendix D. Weakest Link

- P1 votes for you, P2 votes for P1:If you vote for P1, you will go head-to-head against P2 and obtain expected payoff ${p}_{2}W$.If you vote for P2, it will be a three-way tie and you will obtain expected payoff $\frac{W({p}_{1}+{p}_{2})}{3}$, which was calculated above.
- P2 votes for you, P1 votes for P2:If you vote for P2, you will go head-to-head against P1 and obtain expected payoff ${p}_{1}W$.If you vote for P1, it will be a tie and you obtain $\frac{W({p}_{1}+{p}_{2})}{3}$.

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**Figure 1.**Exploitability vs. number of training games for two-player zero-sum games with uniform-random payoffs in [−1, 1], with results averaged over 10,000 test games for each number of training games.

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Ganzfried, S. Human Strategic Decision Making in Parametrized Games. *Mathematics* **2022**, *10*, 1147.
https://doi.org/10.3390/math10071147

**AMA Style**

Ganzfried S. Human Strategic Decision Making in Parametrized Games. *Mathematics*. 2022; 10(7):1147.
https://doi.org/10.3390/math10071147

**Chicago/Turabian Style**

Ganzfried, Sam. 2022. "Human Strategic Decision Making in Parametrized Games" *Mathematics* 10, no. 7: 1147.
https://doi.org/10.3390/math10071147