Fun with underspecified games

Thanks a lot to Andrew Gelman for bringing this gem to my attention.

In this particular battle between physicists and economists, I’m taking the economists’ side. https://t.co/fFHGB4owld

— Andrew Gelman (@StatModeling) December 19, 2020

I've been thinking about a particular game of chance. Here is how the game is described in the article to which Gelman links:

Starting with $100, your bankroll increases 50% every time you flip heads. But if the coin lands on tails, you lose 40% of your total. Since you’re just as likely to flip heads as tails, it would appear that you should, on average, come out ahead if you played enough times because your potential payoff each time is greater than your potential loss. In economics jargon, the expected utility is positive, so one might assume that taking the bet is a no-brainer.

Yet in real life, people routinely decline the bet. Paradoxes like these are often used to highlight irrationality or human bias in decision making. But to Peters, it’s simply because people understand it’s a bad deal.

I do not agree. My problem with this “paradox” is that the game (as described) is not entirely clear. What is missing from this description?

• First, it starts the player with a \$100 bankroll. From where did this \$100 appear? Does the player buy in for \$100? Or do they buy in for only \$20?
• Second, the desciption reads “every time” the player flips heads, but how many times must the player flip? Is the player free to cash out at any time, or does the game never end? If the latter, then of course it is a bad deal— the player never has a chance to cash out winnings!
• Third, what happens when the game is over? Can the player buy in to play again?

The answers to these questions matter to whether the game is really a “bad deal” or a rather money-making machine for the player.

Let us investigate some possible rules and evaluate the “deal”

The Simple (One-Shot) Game

Suppose that the game consists of just one flip of the coin with a \$100 buy-in. After that flip, the player will walk away with either \$150 or \$60 for either a \$50 profit or a \$40 loss.

More generally, we might suppose a buy-in generally of \$$x$, so that the player walks away with either \$${x}/{2}$ profit or \$${2x}/{5}$ loss. These outcomes are equally likely, so the expected profit (in dollars) from the one flip is given by $$ \mathrm{E}\!\left[\pi_1\right]=\frac{1}{2}\frac{x}{2}+\frac{1}{2}\frac{-2x}{5}=\frac{x}{20} $$ In other words, for every \$20 put up, the player can expect \$1 in profit. However, the profit is uncertain. For every \$20 at risk, actual profits will be not \$1 but either \$10 or -\$8. For better or for worse, actual profits will miss expectations by exactly \$9 for every \$20 put up. Put another way, the expected squared difference between the actual and expected profit is given by $$ \mathrm{E}\!\left[\left(\pi_1-\mathrm{E}\!\left[\pi_1\right]\right)^2\right]=\frac{1}{2}\left(\frac{x}{2}-\frac{x}{20}\right)^2+\frac{1}{2}\left(\frac{-2x}{5}-\frac{x}{20}\right)^2=\frac{81x^2}{400} $$ The square root of this comes to $\${9x}/{20}$— the amount by which we expect the player to miss expected profits.

The Repeated Game

Suppose that the game consists of just one flip of the coin with a \$100 buy-in. After that flip, the player will cash out up with either \$150 or \$60 for either a \$50 profit or a \$40 loss. If the player then buys in again, them the player will again cash out another \$50 profit or \$40 loss.

Without putting too fine a point on it, the expected profit after $N$ flips would be $N$ times the expected profit from a single flip. With a \$100 buy-ins, $$ \mathrm{E}\!\left[\pi_N\right]=N\mathrm{E}\!\left[\pi_1\right]=N\frac{100}{20}=5N $$ with expected squared difference $$ \mathrm{E}\!\left[\left(\pi_N-\mathrm{E}\!\left[\pi_N\right]\right)^2\right]=N\mathrm{E}\!\left[\left(\pi_1-\mathrm{E}\!\left[\pi_1\right]\right)^2\right]=2025N $$ for an expected miss of profits of $\$45\sqrt{N}$.

With more flips, the expected profits grow faster than the expected miss. Over time, the chance that the players loses a significant amount simply vanishes. However, the player needs a modest bankroll in order to withstand any bad luck early in the game.

Figure 1: Evolution of Profits in the Repeated Game

With only a few flips, the chance of losses is high. By 100 flips, it is equally likely that the player has profited by \$1000 as it is that they have any loss. By 1000 flips, there is less than 1 in 4500 chance that the player will still be down. So why does Peters view the game as a bad deal for the players?

Letting It Ride

Suppose that instead of buying in for the next coin flip that the player may only cash out or “let it ride.” If the player puts up \$100 and wins, the next buy-in is more expensive: \$150 to win \$75 or lose \$60. If after the first flip, the player loses they may risk the remaining \$60 to win \$30 or lose \$24.

Note that in this latter case, a subsequent win will not make the player whole. Before, a loss of \$40 could be followed by a win of \$50 for a \$10 profit. Here, a \$40 loss could be followed by a win of \$30 for an overall loss of \$10.

So what do profits in this game look like if the player stays in for a large number of flips? In the simple repeated game, each flip added or subtracted from the player’s total by fixed amounts (\$50 or -\$40). Here, each flip multiplies the player’s money on the table by fixed amounts (1.5 or 0.6.) Recalling that $\log{ab}=\log{a}+\log{b}$, we see that multiplication by fixed amounts is equivalent to addition of different fixed amounts. So we may analyze the player’s money in the same way we analyzed the profits before. Starting with \$100, the expected amount of money on the table after one flip is $$ \mathrm{E}\!\left[\log{W_1}\right]=\log{W_0}+\frac{1}{2}\log{\frac{3}{2}}+\frac{1}{2}\log{\frac{3}{5}}=\log{W_0}+\frac{1}{2}\log{\frac{9}{10}}\lt\log{W_0} $$ In other words, the player’s expected log wealth shrinks with each flip. This is the crux of Peters’ paradox: expected profits are positive and grow in the long run yet expected wealth shrinks. I maintain that there is no paradox at all beacuse the two games are distinct. It is not even clear that “letting it roll” is even a bad deal.

Consider: in the simple repeated game a player could keep playing with money on the table as long as that money does not fall below \$100. In the latter game, the player can keep playing forever. Assuming increasingly tiny fractions of dollars are accepted, the player will never run out of cash on the table. The same cannot necessarily be said of the house. In the long run, the player should still break the house. Unfortunately, if the player fails to do so quickly the chance of doing so in any reasonable amount of time becomes very small. To illustrate this, let us set a more reasonable goal for the player: to double their money. If at any point the player has \$200 on the table, they walk away; otherwise they let it ride.

Perhaps surprisingly, after just 20 flips, 54% of players will have met their goal; after 200 flips, 65%; after 2000, 65.6%. Clearly, those that do not meet their goal early tend to struggle. However, the game is still a “good deal” in that nearly 2/3 of players will double their money. In fact, if the player gives up after 2000 flips, the average amount of money among all players is over \$150. And no, the winnings are not too poorly distributed. Nobody winds up with as much as \$300, because they would need to have \$200 on the table to get there— meaning they already met their goal and quit.

For this game to be genuinely bad for the player, it requires one important hidden rule.

• The player cannot cash out.

It is not hard to see how that would make it bad for the player if they can never cash out. However, the game is not only not-bad but a money making machine for the player unless there is one more hidden rule:

• The player cannot return and play again.

Why is this rule important? If the player can cash out and then return, they may easily break the house. One way of seeing this is to simply set the player goal to \$150 within one flip. If the player wins, they cash out \$50 in profits and return with a new \$100. If the player loses, they cash out their remaining \$60 and return with an new \$100. This is exactly the same as the simple repeated game. If, on the other hand, the player will only cash out short of their goal after 200 flips, the game slightly changes. To simplify, let us assume that 65% of the time they walk away with exactly \$200. (The actual average should be higher.) Let us assume that if they give up short of their goal they surrender the money on the table. Win or lose, this simplification leads to an underestimate of the player’s profits. $$ \mathrm{E}\!\left[\pi_1\right]=0.65\times 100+0.35\times\left(-100\right)=\$30 $$ with expected squared difference $$ \mathrm{E}\!\left[\left(\pi_N-\mathrm{E}\!\left[\pi_N\right]\right)^2\right]=0.65\times\left(100-30\right)^2+0.35\times\left(-100-30\right)^2=9100 $$

Figure 2: Evolution of Profits in the Repeated 200-Flip Game

Clearly, for the game to be genuinely bad for the player, there must be hidden rules that disallow the player from taking advantage in this way. However, just for fun let us assume the player may put however much money they wish at risk with each flip.

Optimal player control

Suppose the player starts with just \$100. If they can both play repeatedly and control the amount of money at stake in each flip, they will definitely break the house without any risk at all. If the player puts a fraction $f$ of their money $W$ on the table, then the player’s expected (log) money after the next flip is $$ \mathrm{E}\!\left[\log{W'}\right]=\frac{1}{2}\log{\!\left[\left(1-f\right)\times W+\frac{3}{2}fW\right]}+\frac{1}{2}\log{\!\left[\left(1-f\right)\times W+\frac{3}{5}fW\right]} $$ This may be rewritten $$ \mathrm{E}\!\left[\log{W'}\right]=\log{W}+\frac{1}{2}\left[\log{\!\left(1+\frac{1}{2}f\right)}+\log{\!\left(1-\frac{2}{5}f\right)}\right] $$ These expected log money is at an extremum when $$ 0=\frac{\mathrm{d}}{\mathrm{d}f}\mathrm{E}\!\left[\log{W'}\right]=\frac{{1}/{2}}{1+{f^*}/{2}}-\frac{{2}/{5}}{1-{2f^*}/{5}}=\frac{1}{2+f^*}-\frac{1}{{5}/{2}-f^*} $$ or $f^*={1}/{4}$. That is, if the player always stakes 1/4 of their money on every flip, the player will maximize the speed at which their money will grow.

Figure 3: Money Growth is Maximized By Betting 1/4 of Bankroll

By putting only one-quarter of the player money at stake, the potential winnings are much lower (1/8 of money, rather than 4/8) but so are the losses (1/10 of money, rather than 4/10). Importantly, the expected log money grows with each flip. $$ \mathrm{E}\!\left[\log{W'}\right]=\log{W}+\frac{1}{2}\left[\log{\frac{9}{8}}+\log{\frac{9}{10}}\right]=\log{W}+\log{\sqrt{\frac{81}{80}}}>W $$ The expected squared difference from this amount is $$ \mathrm{E}\!\left[\left(\log{W'}-\mathrm{E}\!\left[\log{W'}\right]\right)^2\right]=\log^2{\!\sqrt{\frac{5}{4}}} $$ Note the scale in Figure 4 below. After 1500 flips, the player is a millionaire more than half the time compared to losing 1.5 percent of the time; with another 1000 flips, the player has half the time accumulated half a billion compared to suffering losses only once in about 400. A player so unfortunate to be down to their last cent is only 2500 flips away from a 50/50 short at turning it into $55,000.

Figure 4: Betting 1/4 of Bankroll On Each Flip Makes Millions

Conclusion

It is not the gamble itself which is a bad deal. It is the existence of unspecified rules surrounding the deal that makes the game bad for the player. Given half a chance to play the game repeatedly; to walk away; to return; to control the size of the bets— the player’s ability to bankrupt the house is nearly unlimited. None of these restictions are explicitly placed, but are critical to understanding Peters’ offer.