Card Betting: Expected Value, Risk Aversion, and Redraw Choices

# Interview Question - Card betting


# Interview Question - Card betting












I had to answer questions for a job interview today and I got these questions. I had no idea how to answer them.

There is a deck of 12 cards numbered 1 to 12. One card is pulled from the deck at random. The return is $1 x Value on the card.

What is the maximum bet you would place on the game if you played once?

If you played 10,000 games, what is the maximum bet you would make on each game?

If you don't like the card chosen, you can ask for another card to be pulled from the deck (with the first card replaced), what is the maximum bet you would place if you played once?

## Answer by phdstudent (score 12)

https://quant.stackexchange.com/a/46410

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

Let's assume you are risk neutral:



- Game is played 10,000 times. Still willing to play 6.5$ for each game.



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 $\gamma=2$.

- If the game is played once, your expected utility is -0.2586. This translates to a willingness to pay 3.86usd to play the game. (edit: the way to get this value is to compute the certain equivalent: $-0.25 = \frac{W^{1-\gamma}}{1-\gamma})$, replace $\gamma$ with 2 and solve for $W$.

- If the game is played 10,000 times, and with no time discounting, still you are willing to pay 6.5usd every time to play the game, because the variance of the payoff decreases.

- If you can choose to put a card back, on this scenario you would for sure be willing to pay more than 3.86$ but less than 8usd. But I would need to draw the decision tree to be sure.

Of course as $\gamma$ increases your willingness to pay decreases.

The trickiest case is when your coefficient of intertemporal of substitution also matters. On this case the answer to 1, is similar but the answer to 2 is completely different. You would need to use an Epstei-Zin utility function and evaluate the outcome.

After thinking a bit I have edited point 2 above. The truth is if the game is repeated and there is no time discounting you even if you are risk averse the result comes closer to risk neutrality.

The intuition is simple if you think about mean-variance preferences your utility is:

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

As you increase the draws, variance decreases as @dm correctly pointed out and the second term starts to vanish. Still the max amount you would be willing to pay to play the game 10,000 times is 65,000usd, or 6.5usd per game.

Here's the code to crunch the numbers you can run it in matlab with a CRRA utility function which is somewhat more realistic.

```
% One draw and repeat experiment 100.000 times
N  = 1;
random_draws = randi([1 12],N,100000);

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

% Willingness to pay

% Risk_neutral
W2P_neutral = Expected_value;

% Risk-Averse
gamma = 2;
Utility = ((random_draws).^(1-gamma))./(1-gamma);
Expected_utility = nanmean(Utility);

W2P_averse = ((1-gamma)*Expected_utility).^(1/(1-gamma));

% 10,000 draws and repeat experiment 100.000 times
N = 10000
random_draws = randi([1 12],N,100000);

total_money = sum(random_draws,1);

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

% Willingness to pay per draw

% Risk_neutral
W2P_neutral = Expected_value/N;

% Risk-Averse
gamma = 2;
Utility = ((total_money).^(1-gamma))./(1-gamma);
Expected_utility = nanmean(Utility);

W2P_averse = (((1-gamma)*Expected_utility).^(1/(1-gamma)))/N;
```

## Answer by Jesse (score 3)

https://quant.stackexchange.com/a/46416

If the game is played in exactly the way you stated it, why would you ever bet more than 1 dollar? Assuming you bet 1\$, then you get 1\$ x value on card. And if you bet 12\$, you get 1\$ x Value on card. What's the point of betting more than 1 dollar?

## Answer by dm63 (score 2)

https://quant.stackexchange.com/a/46426

To elaborate on my comment: with respect to questions 1 and 2, the distribution of the payoff for one game is discrete uniform with a mean of 6.5 and a SD of about 3.5 according to my calculations. Now, if you are guaranteed to play this 10,000 times, then you are enititled to consider the distribution of the sum of the outcomes, which by the Central Limit Theorem is approximately Normal with mean 65,000 and SD 3.5*sqrt(10,000) = 350. Hence, even the most risk averse investor should be willing to pay 64,000 since the probability of making money in that case exceeds 99%. I think this is what the interviewer is expecting.

## Answer by Partly Cloudy (score -1)

https://quant.stackexchange.com/a/46418

It depends how long the game takes.

One assumes the game is about as much fun as it sounds, so the only reason to play would be to make a profit. If you said \$6.50 for the first and second questions, half-hearted congratulations, you should get the job on the basis you're decent at maths ... but don't expect to get paid — you've just told your interviewer you're willing to do boring crap like this game for no money.

Assuming you expect a salary, divide that salary by the number of games that could be played in your working year and subtract that from the \$6.50. Say you want $50k/yr working 40h/wk for 50 weeks/yr, if games take 1 minute, that's about \$0.41667 (desired profit per minute or equivalently per game); you should be willing to pay up to \$6.08 to play once, and no more than \$6.08333 on average for a set of games.

For the ten-thousand games question, the interviewer is probably wanting to hear an argument that the risk is small, as others have said; that's straightforward: the standard deviation $\sigma \approx \$350$, dm63 says, and at just $6\sigma \approx \$2000$ the odds are one in a billion against losing that much.

For the last question, a sensible person would redraw if the first card is below-average ($\le6$). So expected value is $9 {1\over2}$ half the time (when they don't redraw) and $6 {1\over2}$ half the time (when they do redraw), for an average of exactly \$8, less the required average profit per game.

In my opinion you should probably be hired (with salary) if you explain the answer to the last question clearly or have a good explanation of SD or if you bring the value of your time into it unprompted. So a good interview question.

But don't take the job. Whatever company this is neglected to ask the interesting question. You don't want to work for those sorts.

The interesting game is where you always draw two cards and the payout is the higher value in dollars.

Draw a matrix of each outcome (where the row and column indices are the two cards):

\begin{bmatrix} 1&2&3&...&12\\ 2&2&3&...&12\\ 3&3&3&...&12\\ &&...&&\\ 12&12&12&...&12\\ \end{bmatrix}

Compute the average:

$${{\Sigma_{i=1}^{N} i \cdot (2i-1)}\over{N^2}} = {{2\cdot\Sigma_{i=1}^{N} i^2 - \Sigma_{i=1}^{N} i} \over{N^2}} = {{2N(N+1)(2N+1)}\over{6N^2}} - {{N(N+1)}\over{2N^2}} = {{N(N+1)(4N-1)}\over{6N^2}}$$

Thus don't pay more than $\$ 8.486111$ to play, less your desired profit.