# Math Help - hard probability question

1. ## hard probability question

suppose you have a standard 52 card deck of playing cards and 1 joker. you spread these cards face down over a table. now one by one, you flip over cards and you can only flip over each card once. What is the probability that you will flip over all 4 aces before you flip over the 1 joker?

2. probability:

numerator: ways to get 4 aces before 1 joker in a deck of 53 cards
denominator: ways to deal 53 cards

A successful dealing will look like this:

aAbAcAdAeJf where the sum of a through f is 48.

So the question is really asking you how many ways you can have 6 integers sum to 48, then divide by the number of ways to deal 53 cards.

3. Let A denote the event that in first (j-1) draws 4 aces have been drawn and let B denote the event that the j'th card is the joker.

$
P(A\cap B)=P(A)P(B|A)=\sum_{j=5}^{53}\frac{\binom{4}{4}.\b inom{48}{j-5}}{\binom{53}{j-1}}.\frac{1}{54-j}
$

$
=\frac{1}{5}
$

4. Hello, oblixps!

I'd like to see your teacher's solution to this problem.
I don't think it can be solved during a timed exam.

Suppose you have a standard 52-card deck of playing cards and 1 joker.
You spread these cards face down over a table and flip them over, one by one.
What is the probability that you will flip over all 4 aces before you flip over the joker?
There are $53!$ possible permutations of the cards.
Then I have a long long list of cases . . .

If the Joker $(J)$ is in one of the first four positions {#1, 2, 3, 4},
. . it is impossible to find the four Aces first.

$J$ at #5
The four Aces are among the first four cards: . $4!$ ways.
The other 48 cards have $48!$ possible permutations.
Hence, there are: . $(4!)(48!)$ ways for $J$ at #5.

$J$ at #6.
The four Aces are among the first five cards
. . along with one of the other 48 cards: . $_{48}C_1$ choices.
These 5 cards can be ordered in $5!$ ways.
The other 47 cards can have $47!$ permutations.
Hence, there are: . $(_{48}C_1)(5!)(47!)$ ways for $J$ at #6.

$J$ at #7.
The four Aces are among the first six cards
. . along with two of the other 48 cards: . $_{48}C_2$ choices.
These 6 cards can be ordered in $6!$ ways.
The other 46 cards can have $46!$ permutations.
Hence, there are: . $(_{48}C_2)(6!)(46!)$ ways for $J$ at #7.

. . . $\vdots$

$J$ at #51.
The four Aces among the first 50 cards
. . along with 46 of the other 48 cards: . $_{48}C_{46}$ choices.
These 50 cards can be ordered in $50!$ ways.
The other 2 cards can have $2!$ permitations.
Hence, there are: . $(_{48}C_{46})(50!)(2!)$ ways for $J$ at #51.

$J$ at #52.
The four Aces are among the first 51 cards.
. . along with 47 of the other 48 cards: . $_{48}C_{47}$ choices.
These 51 cards can be ordered in $51!$ ways.
The last card can have only $1!$ order.
Hence, there are: . $(_{48}C_{47})(51!)(1!)$ ways for $J$ at #52.

$J$ at #53.
The rest of the deck occupies the first 52 positions.
. . And it has $52!$ permutations.
Hence, there are: . $52!$ ways for $J$ at #53.

Task: add these numbers and divide by $53!$ . . .

I'll wait in the car . . .
.

5. Originally Posted by oblixps

suppose you have a standard 52 card deck of playing cards and 1 joker. you spread these cards face down over a table. now one by one, you flip over cards and you can only flip over each card once. What is the probability that you will flip over all 4 aces before you flip over the 1 joker?
This is not as hard as you think.

Ignore all the other cards (they don't matter) and consider only the aces and joker. All 5! sequences of the 5 cards are equally likely. Now, how many of the sequences end in a joker? 4!. So the probability is

$\frac{4!}{5!} = \frac{1}{5}$.

Or, even more simply, one of the 5 cards must be the last turned over. They are all equally likely. So the probability the last card is the joker is 1/5.

6. Thank you 'awkward'.

I knew there had to be some simple explanation but couldn't give much thought to it in the morning as there was an early morning class.