# Thread: Problem from my test! Help!

1. ## Problem from my test! Help!

Hey have a look at these

You have 2 standard 52 card decks of cards that have been shuffled randomly. You turn over the first card from the first deck, and the first card from the second deck,
(i) What is the probability they match?
(ii)You keep turning over the top card on each deck, until you have gone through the whole deck. How many matches do you expect to find?
(iii) You have a thousand letters and envelopes addressed to people, the envelopes get mixed up and the letters get randomly put into envelopes. How many do you expect to end up in the right envelope?

I thought there was a 1/52 chance for the first part but the other parts were confusing. Can anyone help? I'd really appreciate it

2. Hello, konix!

I hope this wasn't a homework problem.
The theory required is quite advanced.
. . In fact, I won't into the details here . . .

You have 2 standard 52 card decks of cards that have been shuffled randomly.
You turn over the first card from the first deck, and the first card from the second deck.
(i) What is the probability they match?

(ii)You keep turning over the top card on each deck, until you have gone through
the whole deck. How many matches do you expect to find?

(iii) You have a thousand letters and envelopes addressed to people,
the envelopes get mixed up and the letters get randomly put into envelopes.
How many do you expect to end up in the right envelope?
(i) The first card can be any card: .$\displaystyle \frac{52}{52} \:=\:1$
. . .The other card must match: .$\displaystyle \frac{1}{52}$
The probability that the two top cards match is: .$\displaystyle (1)\left(\frac{1}{52}\right) \:=\:\frac{1}{52}$ . You were right!

{ii) and (iii) This is a classic (very old) problem.
It has been called "The Mis-addressed Envelopes" or "The Careless Hatcheck Girl".

A permutation of objects in which no object is in its proper place
. . is called a "complete permuatation" or, more commonly, a disarrangment.

You might try to discern a pattern, but it's very elusive . . .

. . . $\displaystyle \begin{array}{ccc}\text{objects} & \text{disarrangements} \\ 1 & 0 \\ 2 & 1 \\ 3 & 2 \\ 4 & 9 \\ 5 & 44 \\ 6 & 265 \\ \vdots & \vdots \end{array}$

With a deck of 6 cards, there are 6! = 720 possible orders for the first deck.
There are 265 orders for the second deck to have no matches.
Hence, there are: .$\displaystyle 720 - 265 \:=\:455$ orders with some match.

Therefore: .$\displaystyle P(\text{at least one match}) \:=\:\frac{455}{720} \:=\:0.63194 \:\approx\:\frac{5}{8}$

With 52 cards, we need the number of possible disarrangements, $\displaystyle d(52)$
. . and divide by 52!, the number of possible orderings of the cards.
This is the probability of no matches.
. . Subtract from one (1), to get the probability of some matches.

Skipping all the underlying theory, as the number of objects gets larger,
. . the probability of getting no matches approaches $\displaystyle :\frac{1}{e} \:=\:0.367879441 ...$

Maybe you can apply all this to your problem . . .

Edit: I was reminded that the proper term is derangement
. . . . That's true . . . *sigh*

3. thanks a million, should be able to do it now
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}