Risk Aversion and the Value of Repeated Coin-Flip Bets

# Flip a coin $5$ times in a row. If you get $5$ Heads, you get $\$28$. How much would you pay to play?


# Flip a coin $5$ times in a row. If you get $5$ Heads, you get $\$28$. How much would you pay to play?












In preparation for quant interview season, I tried out interview questions and came across the following one:

> Flip a coin $5$ times in a row. If you get $5$ Heads, you get $28$ dollars. How much would you pay to play? How about if you get to play this $5000$ times?

My simple answer to this would be I would pay $\\\$0.875$, and if I play it $5000$ times, I would pay $\\\$4375$. However, my answer seems a little off in the sense that the game is highly skewed. Meaning, most of the time, I am actually going to lose my $\\\$0.875$ for each play, but only when I play the game many times, would I get close to the expected value. Meaning, my value for $5000$ games of $\\\$4375$ seems to be closer to the right answer but perhaps the single-play answer needs to be adjusted even further down because the risk-reward is so low? Does anyone have any idea if this thought process is correct and if so, how should the adjustment be made?

## Answer by user name (score 3)

https://quant.stackexchange.com/a/80530

I think you could also find the variance of the payoff of a single game, and argue that since 5000 is a large number, the average payoff will be approximately Gaussian with the appropriate mean and variance. Then you could set a level of risk-averseness $k$, and say you will pay 5000 $\times$ (the payoff of $\mu - k\sigma$).

I think this is exploiting the property of CLT that no matter how skewed your initial distribution is, when it is repeated independently for a large number of iterations, the average will be pretty symmetric

## Answer by phdstudent (score 3)

https://quant.stackexchange.com/a/81854

It all depends on your level of risk aversion and degree of intertemporal substitution.

Let's assume you are risk neutral:



- Game is played 5,000 times. Still willing to play 0.8750 for each game or $5,000 \times 0.8750$ in total.

Now if you are risk-averse you need to assume a coefficient of risk-aversion and utility function. Let's say $U = \frac{W^{1-\gamma}}{1-\gamma}$ and $\gamma=2$. Also note, that getting 5 times heads, is the same as flipping a 32 faces dice and getting a particular roll.

With this particular utility function, you would never be willing to pay to play this game. You truly hate the zero payoff scenarios.

But we can try with different preferences, say mean variance:

$U = E_t[W] - \frac{\gamma}{2} Var(W)$

If $\gamma = 2$ you are willing to pay -8.81 to play the game one time (which literally means that you dislike risk so much that you need to get paid to play the game) and 0.73 to play the game 5,000 times. As you play the game more times it converges to the risk-neutrality case. As you increase the draws, variance decreases. Eventually, for sufficiently large number of games it converges to risk neutrality.

Here's the code to crunch the numbers you can run it in matlab with mean-variance utility.

```
% One draw and repeat experiment 100,000,000 times
Payment = 28;
N  = 1;
random_draws = randi([1 32],N,1000000);
Payoff = NaN(N,1000000);
Payoff(random_draws == 1) = 28;
Payoff(not(random_draws == 1)) = 0;

Expected_value = nanmean(Payoff);
Std = std(Payoff);

% Willingness to pay

% Risk_neutral
W2P_neutral = Expected_value;

% Risk-Averse
gamma = 2;
Utility = nanmean(Expected_value) - gamma * Std;

W2P_averse = Utility;

% 5,000 draws and repeat experiment 100.000 times
N = 5000;
random_draws = randi([1 32],N,1000000);
Payoff = NaN(N,1000000);
Payoff(random_draws == 1) = 28;
Payoff(not(random_draws == 1)) = 0;

Payoff = sum(Payoff,1);

Expected_value = nanmean(Payoff);
Std = std(Payoff);

% Risk_neutral
W2P_neutral = sum(Expected_value)/N;

% Risk-Averse
gamma = 2;
Utility = nanmean(Expected_value) - gamma * nanmean(Std);

W2P_averse = Utility
```

Note: Here's a similar case, where CRRA delivers positive utility: Interview Question - Card betting