Six Card Sum-Quant Guide

Question

Jamie is told there are 3 aces and 3 jacks in a pile. Each turn, a card is drawn without replacement; Jamie earns $1 if he guesses the drawn card correctly. Jamie plays 6 turns under the optimal strategy. How much money should Jamie expect to earn. I am having difficulty in approaching this problem. I can find out for the first round to be 0.5. Then onwards I am stuck.

Answer 1

The concept of symmetry is extremely useful in this problem. For example, the expected earning in the case where $1$ ace and $2$ jacks are left is the same as the expected earning in the case where $2$ aces and $1$ jack are left. i.e. $E[1A, 2J] = E[2A, 1J]$

We can simplify any subproblem by ignoring As and Js. Let $x$ be the number of cards of type $1$ (either As or Js) and let $y$ be the number of cards of the type $2$ (Js if x is As or vice versa).

Hence, we want $E[nxmy]$, where there are n cards of type x and m cards of type y. In your problem, you require $E[3x3y]$. It is easy to define any subproblem recursively. $$E [3x3y] = \frac12 + E[2x3y]$$ $$=> E[2x3y] = \frac35 \cdot \left( 1 + E[2x2y] \right) + \frac25 \cdot E[1x3y]$$ $$=> E[2x2y] = \frac12 \cdot (1 + E[1x2y]) \quad and \quad E[1x3y] = \frac3 4 \cdot(1+E[1x2y]) + \frac14 \cdot E[0x3y] $$ $$=> E[1x2y] = \frac23 \cdot(1 + E[1x1y]) + \frac13 \cdot E[0x2y] \quad$$ Now, the smaller subproblems are pretty obvious: $E[1x1y] = \frac 32, E[0x2y] = 2, E[0x3y] = 3$

Keep substituting back to get $E[3x3y] = \frac{41}{10} = 4.1$

Answer 2

Define $f(a,b)$ to be the expected amount of additional money gained when the deck currently contains $a$ aces and $b$ jacks.

$f(0,0)=0,~f(1,0)=1,~f(0,1)=1$ should be obvious. $f(a,b)=f(b,a)$ should also be obvious, as should the strategy of always guessing that the next card is the one which there is more of remaining (doesn't matter if ace or jack if equal amount, so go ahead and always guess ace in those cases)

We have that when $a\geq 1$ that $f(a,0)=a$, since at that point we know the exact contents remaining in the deck and so can say with certainty for each subsequent draw that the card will be an ace. Similarly $f(0,b)=b$.

Lastly, we can note that when $a\geq b\geq 1$ we have that $f(a,b) = \underbrace{\frac{a}{a+b}\left(1+f(a-1,b)\right)}_{\text{guessed right}} + \underbrace{\frac{b}{a+b}\left(0+f(a,b-1)\right)}_{\text{guessed wrong}}$. The other case of when $b>a\geq 1$ is covered by $f(a,b)=f(b,a)$

We have all information we need now to work backwards to find $f(3,3)$

$f(0,0) = 0$

$f(1,0)=1,~~f(0,1)=1$

$f(2,0)=2,~~f(1,1)=1.5,~~f(0,2)=2$

$f(3,0)=3,~~f(2,1)=\frac{7}{3},~~f(1,2)=\frac{7}{3},~~f(0,3)=3$

$f(3,1)=3.25,~~f(2,2)=\frac{17}{6},~~f(1,3)=3.25$

$f(3,2)=3.6,~~f(2,3)=3.6$

$f(3,3)=4.1$

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In more detail for example, $f(3,1) = \frac{3}{3+1}(1+f(2,1))+\frac{1}{3+1}(1+f(3,0)) = \frac{3}{4}\cdot (1+\frac{7}{3})+\frac{1}{4}\cdot (0+3) = 3.25$