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I deal a standard deck of 52 cards to you face up, one card at a time. Before any deal of any card you can shout out "NOW". If you shout out "NOW" and the next card I deal is a queen, then the game finishes and I give you $100. If the next card isn't a queen, then the game finishes and I give you nothing. What is a fair price for you to pay me in order to play this game with me?

What I think of is to use the optional sampling theorem and show it is a martingale. But I am a bit lost on the setup.

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...$\frac{1}{13}$??? – Kolmo May 3 '12 at 9:29
up vote 2 down vote accepted

For this game this recursion holds: $$f(a,n)=\max(\frac{a}{n}*f(a-1,n-1)+\frac{n-a}{n}f(a,n-1), \frac{a}{n})$$ where $a$ is the number of queen left in the deck, and n the cards left (the first part is the expected value of drawing a card, and the second part the expecting value of say "now!")

Say for hypothesys that the value of the game in each state is $g(a,n)=a/n$, you can put in the previous equation and see that: $$f(a,n)=\max(\frac{a}{n}*\frac{a-1}{n-1}+\frac{n-a}{n}\frac{a}{n-1}, \frac{a}{n})=\max(\frac{(n-1)a}{n(n-1)}, \frac{a}{n})=\frac{a}{n}$$

We can use induction over the game three, fom the leaf node (when only 1 cards remain) the expectation for the game is the number of queen (1 or 0), so the formula holds, and if in the children node the formula holds, holds in the parent node too for the previous lemma.

So this game has expectation $\frac{queen}{cards}$ for every strategy that say "now!" at least when only one card was left.

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This is offered as a hint on how to set up your problem.

The fair price is probably going to be $\$100\cdot\frac4{52}=\$7.69$, independent of the player's strategy, which corresponds to the expected win of an a priori strategy to say "now" right before the $n$th card, where $n$ is chosen at random beforehand.

Waiting until $n-1$ cards has been revealed gains some information, but potentially loses some chances to win, so there is a tradeoff. On average, the base price above could very well be the fair price.

The wait time (number $N$ of cards to draw) to see the $k$th queen can be derived from the hypergeometric probability distribution.

wait time for k queens: hypergeometric

The expected values are $8,~17,~26,~35,~44$ for $0\le k\le 4$. Can we construct an optimal strategy as a bayesian decision algorithm? Suppose we choose to wait for the first $n$ cards to be revealed. At this point, the number of queens shown has probability distribution $$ \mathbb{P}\{K=k\}=\frac{{4\choose k}{48\choose n-k}}{{52\choose n}} $$ and, assuming $K=k$ queens have been revealed, the chances of the next card being a queen are $\frac{4-k}{52-n}$.

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