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I had this interview question, but I couldn't answer it.

If you have four cards and were to bet based upon suits while flipping them one at a time while trying to minimize variance, if you guess the suits correctly you get +1, otherwise -1. how would you do it?

2 how would you bet on the suits if you want to maximize the return while minimize the variance.

The four cards will be drawn from the same deck of cards.

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  • $\begingroup$ There is no game theory here. $\endgroup$ – Juanito Jun 5 '17 at 5:17
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    $\begingroup$ Are they four cards from one deck of cards or are they four cards from indeterminate many decks? $\endgroup$ – fleablood Jun 5 '17 at 6:03
  • $\begingroup$ these four cards come from the same deck $\endgroup$ – szd116 Jun 5 '17 at 13:57
  • $\begingroup$ You have improved the Question by specifying the four cards are drawn from the same deck, and presumably by this you mean "without replacement". The Question seems to have two parts. It would help to format the Question a little better, presenting the setup of the "game" and then the two parts that request strategies. If you like, I will try my hand at that, but since you are the source of the Question, it would probably work out better in the long run if you did it. $\endgroup$ – hardmath Jun 5 '17 at 21:05
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On the first draw there is an equal chance for all 4 suits to be drawn : 0.25
Thus you should not bet on the first draw.
Knowing the result of the first draw, there is now less chance that the same suit will be drawn : eg if Clubs was drawn on the first draw, then the chance for Clubs to be drawn in the second draw is $\frac{12}{12+13+13+13}$ , whereas the chance for each of the other three suits to be drawn is $\frac{13}{12+13+13+13}$.
Thus you might bet on the other three suits on the second draw, or bet against the suit drawn in the first round.
Logically this continues for the other two draws, where you bet on the suits that have been drawn the least, and against the suits that have been drawn more.

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    $\begingroup$ This answer assumes that the goal is to maximise expected return, but the original OP talks about "minimise variance". $\endgroup$ – mlc Jun 5 '17 at 6:19
  • $\begingroup$ Knowing the result of the first draw, there is now less chance that the same suit will be drawn Not necessarily, the question doesn't specify what pool of cards the original $4$ were drawn from. $\endgroup$ – dxiv Jun 5 '17 at 6:27
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Assume that each suit $i=1,2,3,4$ represents an "asset" which pays $+1$ with probability $p_i$ if you guess right and $-1$ otherwise. The expected return on betting for $i$ is $2p_i-1$ while the expected return on betting against $i$ is $1-2p_i$. In either case, the variance of betting on suit $i$ is $4p_i (1-p_i)$.

Given your current estimate for the $p_i$'s, suppose you spread your bet $x_i$ $(i=1,2,3,4)$ over the four suits, with $x_1+x_2+x_3+x_4=1$. This bet (or "portfolio") has a variance $$\sum_{i=1}^4 x_i^2 [4p_i(1-p_i)]$$ and your job is to find $x_i$'s to minimise this function under the constraint $x_1+x_2+x_3+x_4=1$.

After you flip one card, update probabilities and re-optimize your bet ("portfolio"). For instance, when the game starts with $p_i = 1/4$, the variance-minimising bet is to put $x_i =1/4$ and bet an equal amount on each suit.

Clearly, it is a different matter if you also care about "return". You would then maximise some objective function that takes off the expected value and the variance of your bet.

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  • $\begingroup$ can you explain how is the expected return on betting for i is 2pi - 1 ? $\endgroup$ – szd116 Jun 5 '17 at 14:48
  • $\begingroup$ Win +1 with probability $p$, lose -1 with probability $1-p$. Expected payoff is $p (+1) + (1-p) (-1) = 2p-1$. $\endgroup$ – mlc Jun 5 '17 at 16:35
  • $\begingroup$ for variance, I used E(x^2) - E(x)^2 = E(x^2) - (4p(i)^2 - 4p(i) + 1) how did you get the E(x^2) ? Also, why would you assume the constraint of x1 +x2+x3+x4 =1 ? if p(i) probability is low wouldn't you bet low on x(i) ? thanks. $\endgroup$ – szd116 Jun 5 '17 at 18:45

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