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?
Best Answer
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.