1
$\begingroup$

Given a deck of 40 cards, which consists of cards numbered from 1-7 [inclusive], I am trying to calculate the probability that:

  • the 1st card drawn is numbered 1
  • the 2nd card is numbered 2,

... (continues for 3, 4, 5, 6,

  • the 7th card is numbered 7

For the case where there are 6 cards in the deck for 1, 2, 3, 4, 5 and 5 cards in the deck for 6, 7 - I believe I figure out the probability like so:

$${{6\over40} * {6\over39} * {6\over38} * {6\over37} * {6\over36} * {5\over35} * {5\over34} = {194400\over93,963,542,400} = {9\over4,350,164}}$$

is this correct?

Then, lets say I want to try and get the greatest probability of drawing the cards in order by choosing how many of each number gets added to the deck.

Do I just try different numbers and see if the probability goes up/down, or is there an easier method?

$\endgroup$
  • $\begingroup$ But $40$ is not a multiple of $7$. Exactly how is that pack of cards? $\endgroup$ – ajotatxe Aug 25 '15 at 21:09
  • $\begingroup$ @ajotatxe "For the case where there are 6 cards in the deck for 1, 2, 3, 4, 5 and 5 cards in the deck for 6, 7" - but note my second question is about choosing how many of each card is in the deck. $\endgroup$ – DoubleDouble Aug 25 '15 at 21:10
  • $\begingroup$ Silly question, sorry. Then, esentially you want to maximize the product $x_1\cdot\ldots\cdot x_7$ given that all the variables are integers and that $\sum_{k=1}^7x_k=40$. Correct? $\endgroup$ – ajotatxe Aug 25 '15 at 21:13
  • $\begingroup$ "from 1-7 [inclusive]" or exclusive??? $\endgroup$ – barak manos Aug 25 '15 at 21:32
  • $\begingroup$ @barakmanos inclusive includes 1 and 7, from my understanding? I mean to include 1 and 7. $\endgroup$ – DoubleDouble Aug 25 '15 at 21:34
1
$\begingroup$

Suppose we change the number of occurrences of the various card types, leaving the total number of cards at $40$. In the computation of the probability of your event, The denominator remains at $(40)(39)\cdots(34)$, so we want to maximize the numerator.

Suppose one of the card types occurs $a$ times, and another occurs $b$ times, where $b\ge a+2$. We will show that $(a+1)(b-1)\gt ab$. Equivalently, we show that $(a+1)(b-1)-ab\gt 0$, that is, that $b-a-1\gt 0$. This is clear, since $b-a-2\ge 0$.

So if $b\ge a+2$ we can always increase the product without changing the sum by making sure that any two card types differ in occurrence by at most $1$. We therefore want to find $7$ numbers with sum $40$ such that all numbers are $k$ or $k+1$ for some $k$.

If $s$ of them are $k+1$ and $7-s$ are $k$, we want $s(k+1)+(7-s)k=40$. That gives $s+7k=40$. Solutions are $k=5$, $s=5$; $k=4$, $s=12$, and so on. Taking $k$ as big as possible maximizes the numerator.

We conclude that $2$ of the numbers from $1$ to $7$ must each occur on $5$ cards, and the remaining $5$ numbers must each occur $6$ on $6$ cards. If that condition is met, any choice of $2$ numbers to appear $5$ times yields the same maximal probability.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.