Question

You have scattered a deck of 52 cards and one joker face down on the table so then you don’t know which card is where. Next you start turning the cards over one by one. Assuming that you cannot flip the same card twice, what are the percentage odds that you will turn over all four aces before you turn over the joker?

Answer 1

We solve this problem by considering only aces and the joker. This is because there are a total of 53 cards in which the 4 aces and 1 joker need to be fixed in a certain order that 4 aces come before the joker. This order of 4 aces before joker can be placed at any place in the set of 53 cards.
So, we select 5 places from the set of 53 places and consider these 5 cards of 4 aces and joker to be placed in a certain order and we find the probability that 4 aces come before the joker by using the permutations to get the required answer.
We use the condition that the number of ways of selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
We also use the formula that the number of ways of placing $'n'$ objects in $'n'$ places is $n!$
For finding the probability, we use the formula of probability that is
$\Rightarrow P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$
Then we find the percentage of odds by using the formula
$\Rightarrow \text{Percentage odds}=P\times 100$

Complete step-by-step answer:
We are given a set of 52 cards and a joker.
So, we can say that there are a total of 53 cards.
Let us assume that there are a total of 53 places where the 53 cards need to be placed.
We are asked to find the percentage of odds such that the aces come before the joker.
We know that there are a total of 4 aces in a deck of 52 cards.
Here, we can see that we need to place the 4 aces and a joker in certain order such that the joker comes at last because 4 aces should come before the joker.
Let us select 5 places from the 53 places where we need to place the 4 aces and a joker in a certain order.
We know that the number of ways of selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$
By using the above formula we get the number of ways of selecting 5 places from 53 places as ${}^{53}{{C}_{5}}$
Now, let us place the order of these 5 cards in the 5 places in a certain order such that the last place is occupied by the joker. So, we need to place 4 aces in the first four places.
We know that the number of ways of placing $'n'$ objects in $'n'$ places is $n!$
By using this formula we get the number of ways of placing 4 aces in 4 places as $4!$
Now, let us place the remaining 48 cards in the remaining 48 places after selecting 5 places from the 53 places.
So, we can say from the above formula that the number of ways of placing 48 cards in 4 places as $48!$
Let us assume that the number of possible outcomes of required order as $'k'$
We know that the number of ways of selecting 5 places from 53 places, placing the 5 cards in the selected places and placing the remaining cards in remaining places all are permutations.
So, we get the number of possible outcomes as
$\Rightarrow k={}^{53}{{C}_{5}}\times 4!\times 48!$
We know that the formula of combinations that is
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using this formula to above equation we get

& \Rightarrow k=\dfrac{53!}{5!\left( 53-5 \right)!}\times 4!\times 48! \\\ & \Rightarrow k=\dfrac{53!\times 4!}{5\times 4!} \\\ & \Rightarrow k=\dfrac{53!}{5} \\\ \end{aligned}$$ Now, let us assume that the total number of possible outcomes as $$'N'$$ We know that the total number of outcomes is placing 53 cards in 53 places. So, by using the formula that is the number of ways of placing $$'n'$$ objects in $$'n'$$ places is $$n!$$ we get $$\Rightarrow N=53!$$ Let us assume that the probability that 4 aces come before joker as $$'P'$$ We know that the formula of probability that is $$\Rightarrow P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$$ By using the above formula we get the required probability as $$\begin{aligned} & \Rightarrow P=\dfrac{k}{N} \\\ & \Rightarrow P=\dfrac{\left( \dfrac{53!}{5} \right)}{53!} \\\ & \Rightarrow P=\dfrac{1}{5} \\\ \end{aligned}$$ Now, we know that the percentage odds can be calculated from the probability as $$\Rightarrow \text{Percentage odds}=P\times 100%$$ By using the above formula we get the percentage odds as $$\begin{aligned} & \Rightarrow \text{Percentage odds}=\dfrac{1}{5}\times 100\% \\\ & \Rightarrow \text{Percentage odds}=20\% \\\ \end{aligned}$$ Therefore we can say that the percentage odds is 20% **Note:** We have a shortcut for this problem. Here, we can see that the 4 aces and a joker need to be placed in a certain order which is not affected by the placement of the remaining cards order because it is an independent event. So, we consider there are 5 places where 4 aces and a joker need to be placed in such a way that the joker comes at last. Here, we can see that the number of ways that joker comes at last is 1 way and there are a total of 5 ways to place the joker in the 5 places. So, by using the formula of probability we get that probability of getting 4 aces before joker as $$\Rightarrow P=\dfrac{1}{5}$$ Now, we know that the percentage odds can be calculated from the probability as $$\Rightarrow \text{Percentage odds}=P\times 100%$$ By using the above formula we get the percentage odds as $$\begin{aligned} & \Rightarrow \text{Percentage odds}=\dfrac{1}{5}\times 100\% \\\ & \Rightarrow \text{Percentage odds}=20\% \\\ \end{aligned}$$ Therefore we can say that the percentage odds is 20%