In a card game with a standard deck of 52 cards, dealing stops when either two kings appear or at least one king and one ace appear. What is the expected number of cards that will be dealt?
I know that the expected number of cards to see the first king/ace is 44/9. But I don't know how to proceed to the next step.
Any help would be much appreciated !
The $44$ cards that aren’t aces or kings are uniformly distributed among the $9$ bins between and around the $8$ aces and kings. Thus the expected number of them that are dealt is $\frac{44}9k$, where $k$ is the number of kings and aces dealt until $2$ kings or a king and an ace are dealt. Since this is linear in $k$, we can just substitute the expected value of $k$ into this expression.
If the first $j$ ace-or-king cards are aces and the next is a king, then $j+1$ are dealt unless $j=0$, in which case $2$ are dealt. The probability for this is $\frac{\binom{7-j}3}{\binom84}$, since $3$ slots for the remaining kings are chosen from the remaining $7-j$ slots.
Thus the expected value of $k$ is
$$ \binom84^{-1}\left(\binom73\cdot2+\sum_{j=1}^4\binom{7-j}3(j+1)\right)=\frac{23}{10}\;. $$
Adding the expected number of other cards and the expected number of ace-or-king cards yields
$$ \frac{44}9\cdot\frac{23}{10}+\frac{23}{10}=\frac{1219}{90}=13.5\overline4\;. $$