4
$\begingroup$

In an interview I received the follow question: We have 3 cards face down, and we give each card in a deck of 52 a numeric score ( A = 1, 2=2, .... , J=11, Q=12, K = 13). The interviewer asked me to find the expected sum of these 3 cards, I approximated that each card has an expected value of (1+2+...+13)/13 = 7, and so the sum is approximately going to be 21 (or adjusted down slightly to 20.5). Then the game begins as follows.

You begin with $1000. I will quote a price to play this game, and you are able to either buy x units, or sell me x units, then we turn over the cards and you realize your profit or loss. For example:

Lets say you are given a price of 25, clearly this is greater than 21, so I would choose to sell (Short) 10 units at that price. The cards turn out to have a sum of 23, so I make a profit of (25-23)*10 = 20. So my total money now is 1020, and so on.

We played several rounds of this, and I was just wondering what is an appropriate strategy for this game? I tended to increase the units I would buy when quoted a very low price, and increase the number I would sell when quoted a very high price (relative to 21), but now that I think about it, is there a more sophisticated way of approaching this problem?

$\endgroup$
  • 1
    $\begingroup$ Why adjust downwards to $20.5$? The expected sum of the $3$ cards is exactly $21$. $\endgroup$ – André Nicolas Apr 9 '15 at 15:41
  • $\begingroup$ @AndréNicolas I was assuming that after the first card is played, then cards 2 and 3 are dependent, and the number of possible cards is reduced, so I thought this would mean that we'd have a total sum of less than 21? $\endgroup$ – dimebucker Apr 9 '15 at 15:50
  • 2
    $\begingroup$ One way of seeing it is to let $X_i$ be the number on the $i$-th card drawn. Then we want $E(X_1+X_2+X_3)$, which by the linearity of expectation (which holds even when we do not have independence), is equal to $E(X_1)+E(X_2)+E(X_3)$. Each $E(X_i)$ is $7$. $\endgroup$ – André Nicolas Apr 9 '15 at 15:56
  • $\begingroup$ @AndréNicolas What's the intuition here? For example if the first card is a 10, then we are restricting slightly cards 2 and 3 to lower cards. So is this just a symmetric argument that if the first card is a low card, then the others are somewhat constricted to higher cards, and it cancels out? $\endgroup$ – dimebucker Apr 9 '15 at 15:57
  • 1
    $\begingroup$ You have given an intuitive argument that there is some "cancellation," if the first card is high the mean of the next two is lowered a bit, and if the first card is low, the mean of the next two is raised a bit. That these lowerings and raisings cancel exactly is I think not obvious, though we could compute a bit with conditional expectations and pin it down. The indicator random variable argument I used above bypasses all that. $\endgroup$ – André Nicolas Apr 9 '15 at 16:01
1
$\begingroup$

Mathematically, the problem is a bit ill-defined.

  1. We're not certain of your goal: Do you care about your strategy's expected value of winnings only? Or do you want to end up above $2000 at the end of the interview but are very averse to busting? Etcetera.

  2. We're also not certain what sorts of prices you're going to be given in the future.

That said, it makes sense to bet more when the price is very high or very low, as you're more certain to win in those cases, so your risk is lower, and furthermore your expected return is higher.

How to go about this in a coherent manner? I think there are several ways. One I'll mention is to have some sense of "acceptable risk." For example, you could decide that on any given round, you don't want more than an $X\%$ chance of losing more than $Y\%$ of your holdings, i.e. use a value at risk metric for deciding how much to wager on any given round.

From a pure expectation point of view, you just buy/sell as much as you can each round.

$\endgroup$
  • $\begingroup$ I agree that the question is not well defined, I think they wanted to see if i could understand what a market maker does. so for the VaR we would require some distribution for the price right? Also, is there no way to attempt to hedge the risk here, I was trying to think of a way to price this game. Like a binomial tree approach (introducing some risk free asset), but in this case we would have multiple branches and one time step, it just seems like the game is pointless as expectation pricing is not used in finance $\endgroup$ – dimebucker Apr 9 '15 at 15:54
  • $\begingroup$ The question of "how should you act when you're at least somewhat risk-averse and you have a stream of investment opportunities of differing value/risk" is an interesting one. You can't just price the game unless you take risk into account otherwise the problem is boring (just buy/sell as much as possible at all times). $\endgroup$ – aes Apr 9 '15 at 16:01
  • 1
    $\begingroup$ There is a distribution for the VaR computation (I was suggesting using it for each individual game): Given a price, you know exactly the distribution of possibilities for the cards, so you know what your distribution of winnings/losses is. $\endgroup$ – aes Apr 9 '15 at 16:04
  • $\begingroup$ couldn't incorporating the risk appetite be done as follows: Say we receive a price of 25, then the P(X1 + X2 +X3 >25) could be calculated, and we can set some limits of how many units we'd buy or sell depending on this probability $\endgroup$ – dimebucker Apr 9 '15 at 16:05
  • 1
    $\begingroup$ For $X\%$ VaR, say with a price of $25$, what you want is not necessarily to calculate $P(X1+X2+X3 > 25)$ but rather to find $M$ such that $P(X1+X2+X3 > M) = X\%$. Then if you sell $S$ units, then your $X\%$ VaR is $S(M-25)$. $\endgroup$ – aes Apr 9 '15 at 16:12
0
$\begingroup$

Another way to look at it. It is a uniform distribution which has a mean of $$ (max+min)/2 $$

The range of outcomes are (3,39) so the mean is:$$ (39+3)/2 = 21 $$

You could even calculate the standard deviation which is given by:

$$ (max-min)/2\sqrt{3} $$

$$ (39-3)/2\sqrt{3}\approx10.3923 $$

Your strategy would be to increase units when price is outside one standard deviation so anything less than 10.6 or greater than 31.4.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.