2
$\begingroup$

it's been a (long) while since I've taken a math / probability class, so my memory is rather hazy.

Suppose I have a deck of 100 cards, 50 of which are aces, and 50 of which are kings.

How would I model (with some sort of mathematical expression) the probability of drawing (at least) ONE Ace card after N draws, if:

  • 1 Card is drawn each time.
  • If the card is not an Ace, it is put back (replaced) in the deck (total cards remains 100)
  • The deck then has 3 King cards removed and replaced by 3 Ace cards (n + 3 ace cards, n - 3 king cards).
  • The deck is then shuffled/randomized.

I thought it might be expressed by some sort of hyper-geometric distribution, but I'm not sure how to go about the "increase the flat chance of success by 3%" (the replacement of the 3 Kings with 3 Aces).

$\endgroup$

1 Answer 1

1
$\begingroup$

The probability that the $N$th draw is an Ace given that no previous draw is an ace is just the number of aces in the deck over $100$: $$P(N\text{th draw an Ace}|1,\cdots ,N-1\text{ not aces})=\frac{50+3\cdot(N-1)}{100}$$

The probability that none of the first $N-1$ draws were aces is the opposite probability

$$P(1,\cdots ,N-1\text{ not aces})=\prod_{i=1}^{N-1}\left(1-\frac{50+3\cdot(i-1)}{100}\right)$$

$$=\frac{1}{100^{N-1}}\prod_{i=1}^{N-1}\left(53-3\cdot i\right)$$

Now we combine these expressions:

$$P(N\text{th draw is first Ace})=P(N\text{th draw an Ace}|1,\cdots ,N-1\text{ not aces})\cdot P(1,\cdots ,N-1\text{ not aces})$$

$\endgroup$
2
  • $\begingroup$ Thankyou! (good thing that Pi symbol could be copy/pasted into google search too...) $\endgroup$
    – Arcanum
    Jan 3, 2018 at 2:28
  • $\begingroup$ Ah, sorry. $$\prod$$ is just like $$\sum$$ but for products instead of sums. Eg $$N! = \prod_{i=1}^Ni$$ $\endgroup$
    – rikhavshah
    Jan 4, 2018 at 17:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .