# 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 $1 x Value on the card. What is the maximum bet you would place on the game if you played once? If you played 10,000 games, what is the maximum bet you would make on each game? If you don't like the card chosen, you can ask for another card to be pulled from the deck (with the first card replaced), what is the maximum bet you would place if you played once? • Did you try to calculate expected values? Jul 2, 2019 at 10:24 • what I did what calculated$E[return]$for each but that wasn't the maximum bet that should be placed on the game. Jul 2, 2019 at 10:30 • So, to be pedantic: you had an idea but the interviewer said it was wrong? Could you show us your calculations? Jul 2, 2019 at 10:31 • Trent, how is the size of the bet related to the return? If I bet 1USD, would I total between 0 and 11USD, if I bet 1000USD, is the total then between 0 and 11000USD? Also, is it possible to change bet size between repetitions among the 10000 games? Jul 2, 2019 at 10:36 • Trent, you pay 1USD to play and get 1 to 12USD? Isn't the total gain between 0 and 11USD then? Or is it so that you pay the bet-size, let's say 6USD to get 1 to 12 USD and then the net gain is -5 to 6USD. I am confused about whether the question is about how much you are willing to stake or what you are willing to pay for a fixed set of potential gains? Is it a combination of both? Jul 2, 2019 at 12:53 ## 4 Answers It all depends on your level of risk aversion and degree of intertemporal substitution. Let's assume you are risk neutral: 1. Game is played once: you are willing to pay $$6.5 = \sum^{N=12}_{i=1} \frac{1}{12} i$$ 2. Game is played 10,000 times. Still willing to play 6.5$ for each game.
3. Card replaced: well you replace everytime your first draw is lower than 6.5. So you replace with 0.5 probability. And you are willing to pay 8usd to play this 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 $$\gamma=2$$.

1. 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 $$W$$.
2. 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.
3. 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;  • if this is a trading interview then almost surely they are expecting to hear that the game played 10000 times is less risky and therefore the amount paid per game is very close to 6.5 for any reasonable utility function. – dm63 Jul 2, 2019 at 20:58 • That is simply not true. That’s a good way to fail at trading. Jul 2, 2019 at 21:35 • @phdstudent Now you have me curious. Why is that not true? Jul 2, 2019 at 21:51 • if you play a single time the max potential loss is much smaller than the max potential loss than playing the same game 10k times. However if the notional on the single game was 10k larger than the notionals on each of the individual games then you have a larger risk on the single. This is interview question so I would not expect these values to be worked out "on-the-fly", a broad discussion and perhaps estimate of the ideas would be satisfactory. – Attack68 Jul 3, 2019 at 6:55 • The people who think playing multiple times is less risky should also read about "the gamblers' fallacy" Jul 3, 2019 at 8:13 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? • As asked you are absolutely right! I wonder if the question was supposed to mean that you are betting against other people to earn the right to play (Which would actually make the game really interesting). So if there are 3 players, only the highest bidder gets the card but everyone loses their bet (which changes the question a LOT--the number of betters is critical). The question makes no sense as asked. Jul 2, 2019 at 23:59 • @BillK think context is relevant. If you are applying for a financial trading job and the question is representative of financial market trading a reasonable assumption would be that the market is competitive and if the question is interpreted as open, rather than a poorly phrased closed one, you might well score points with the interviewer for discussion. The objective is not necessarily to get the question right but impress the interviewer, also they might be similar! – Attack68 Jul 3, 2019 at 6:43 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. • Yes. But with 1% chance you lost 64,000. A risk neutral investor wouldn’t be willing to take that risk. A risk averse investor even less. Think about it. You have 64,000usd. Would you play the game? I would not. The math is in my answer above. Jul 3, 2019 at 22:17 • Ok, make it 63,000. That is 6 standard deviations below the mean, so 1 in 10^8 chance of losing 63,000 versus (1-10^-8) chance to win 2,000usd. I would definitely play that game. – dm63 Jul 3, 2019 at 23:43 • I have clarified the above. The question is what is the maximum you would be willing to pay. Of course any rational person would be willing to pay 1 usd to play this game. 2usd to play this game... up to a maximum of 65,000usd. So of course you would be willing to pay 63,000 or 64,000. That's not the interesting point. Jul 4, 2019 at 10:46 • Noted , Thanks@phdstudent – dm63 Jul 4, 2019 at 13:55 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.

• Regarding your last point, I think the expected value should be around 8.4861 (I assume the probability mass is (2x-1)/144) Jul 3, 2019 at 2:03
• TY - fat fingered the evaluation Jul 3, 2019 at 23:45