You play a game where all number cards (2-10) have a value equal to their number, ace has a value of 1, face cards have a value of 10 and joker can have the same value as any card of your choice. To play the game, you draw three cards one after the other, and you win if your hand total (sum of values) exceeds a stated goal number. If the goal number is 29, what is your probability of winning if you play a single time?
I suggested the following solution:
For 10, we can draw 16 possible card: (10, J, Q, K) for four suits.
Also, we may draw one of the two jokers.
In order to get more than 29, we need to get the following:
$$P(10, 10, 10)=\frac{16}{54}*\frac{15}{53}*\frac{14}{52}=\frac{140}{6201}
\\P(Joker, 10, 10)=\frac{2}{54}*\frac{16}{53}*\frac{15}{52}=\frac{20}{6201}
\\P(10, Joker, 10)=\frac{16}{54}*\frac{2}{53}*\frac{15}{52}=\frac{20}{6201}
\\P(10, 10, Joker)=\frac{16}{54}*\frac{15}{53}*\frac{2}{52}=\frac{20}{6201}
\\P(Joker, Joker, 10)=\frac{2}{54}*\frac{1}{53}*\frac{16}{52}=\frac{4}{18603}
\\P(10, Joker, Joker)=\frac{16}{54}*\frac{2}{53}*\frac{1}{52}=\frac{4}{18603}
\\P(Joker, 10, Joker)=\frac{2}{54}*\frac{16}{53}*\frac{1}{52}=\frac{4}{18603}$$
Resulting a total of $\frac{28}{2067}\approx0.032898$
It turns out it's the wrong answer. Is there any simpler solution? What am I missing here?
Best Answer
Obtaining a value strictly greater 29 can only occur by obtaining a value of 30. This can only happen in a hand consisting of all jokers, face cards, and 10 cards. There are exactly 18 such cards. Therefore, the probability is $$ \binom{18}{3} \,/\, \binom{54}{3} \approx 0.032898, $$ which is the same as your solution.