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Say you have a standard deck of 52 cards with only 1 Ace of Spaces. How many times would you have to "draw and replace" a card to have a 90% chance for the card you drew to have been an Ace of Spades 5 individual times.

I'll try to rephrase it:

Event: You draw a card. If it was the Ace of Spaces, make a mark on your wall. Replace the card and shuffle the deck.

How many times would you need to repeat this event before you have at least a 90% chance of having 5 marks on your wall. (doesn't need to be in a row. Assume that you stop drawing cards once you make the 5th mark on your wall.)

I'm getting into probability and I have no idea how I would approach this problem, would really appreciate if someone could explain to me how this is done!

RobPratt
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Daniel Maybach
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  • Step 1: answer the following: what are the chances of drawing the Ace of Spades exactly twice, given that you drew $n$ times, where $n$ = 3,4, or 5. Note that here, the $n^{\text{th}}$ card drawn does not need to be the Ace of Spades. Compute the three separate answers for $n$ = 3,4, or 5. – user2661923 Oct 09 '20 at 22:53
  • @user2661923 Event A is AoS, P = 1/52. Event B is Not, P = 51/52. The chance of drawing an Ace of Spaces twice in a row is (1/52)(1/52.) The chance of drawing an Ace of Spades at least twice in three tries is... [(1/52)(1/52)]+1/52? I guess that also assumes that we have to draw the Ace of Spades twice in a row though. I'm not sure how to approach this ^^" I think that the chance of drawing an Ace of Spades at least once in 3 tries is 1-[[1-(1/52)]^3]? – Daniel Maybach Oct 09 '20 at 23:02
  • Good work. Adopting the terminology that $p = 1/52$, and letting $q = (1 - 52)$, you have that the chance of A-s, A-s, is $p^2 \times q$. When $n = 3$, what about (for example) A-s, , As or , As, As. For $n=3$, don't you have to sum all three possibilities? Are you familiar with Pascal's triangle? What are the # of ways that you can select $n$ things, taken $k$ at a time. Can you use this comment to complete Step 1, re $n=3$, $n=4$, and $n=5$? – user2661923 Oct 09 '20 at 23:07
  • Hmm, I'm not familiar with Pascal's Triangle but I have come across "Binomial distribution" while searching on StackOverflow before asking my question. I see that when n = 3, there are 3 possible ways to draw exactly 2 Aces, and to calculate the probability, we'd have to sum the probabilities of each way. [(1/52)(1/52)(51/52)] is one way... Wait, but then all 3 combinations would be equally likely to one another? I've messed up somewhere. I believe to select n things taken k at a time, we'd use nCk = n!/(k!(n-k)!) ? – Daniel Maybach Oct 09 '20 at 23:20
  • Yes, and that is enough work. I'm going to provide an answer. – user2661923 Oct 09 '20 at 23:21
  • I appreciate you taking the time to do so, thank you! – Daniel Maybach Oct 09 '20 at 23:23
  • Another editing that I think will facilitate computing the minimum value for $n$. – user2661923 Oct 10 '20 at 02:14

1 Answers1

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Capitalizing on the OP's work, re the comments:

Let $p = (1/52)$ and let $q = (1 - p).$

Given fixed $n \in \mathbb{Z^+},$
Let $F_n(k) \equiv $ chance of drawing the A-spades exactly $k$ times out of $n$
where $k \in \{0,1,2, \cdots, n\}.$

Let $G_n(k) \equiv $ chance of drawing the A-spades at least $k$ times out of $n$

You want the minimum positive integer $n$ such that $G_n(5) \geq 0.90.$

$$F_n(k) = \binom{n}{k}p^k q^{(n-k)}.$$
$$G_n(k) = \sum_{i=k}^n F_n(i)$$

That is, $$G_n(5) = F_n(5) + F_n(6) + \cdots + F_n(n).$$

At this point, for $n$ fixed, you have two choices:

you can compute $G_n(5)$ via the above formula, or you can deduce that

$$G_n(5) = 1 - \left[F_n(0) + F_n(1) + \cdots + F_n(4)\right].$$

This begs the question, is there a cleaner formula that can be used to compute the smallest value of $n$.

I do not know of any way to use these formulas other than via a computer.

Note that attempting to find a shortcut through logarithms should fail, because $G_n(k)$ is equal to the sum of some factors, not their product.

Edit
Reaction to "Assume that you stop drawing cards once you make the 5th mark on your wall."

This is a perfectly natural constraint. Unfortunately, the constraint that you will stop drawing cards adds an unnecessary complication.

The cleanest approach, as represented by my answer, is to assume that for a fixed $n$, you are going to draw $n$ times, and then ask whether you have drawn the Ace of Spades at least 5 times.

Edit-1
A case can be made that I have misinterpreted the question that you asked. The alternative interpretation would be the same as the original, except that you are requiring that the last card drawn be the Ace of Spades.

I would compute the answer to the alternative question as

$$F_{(n-1)}(4) \times p.$$

That is $F_{(n-1)}(4)$ represents the probability of drawing the Ace of spades exactly 4 times out of the first (n-1) drawings. The $p$ factor represents also drawing the Ace of spades on the $n^{\text{th}}$ drawing.

Then, you would be looking for the smallest value of $n$ where the above formula computes to $0.90$

However, this is impossible, because of the final factor in the formula, $p$.

This conclusion may seem counter-intuitive. For example, if you decide that you are going to draw 10,000 times, the chances of drawing the Ace of Spades at least 5 times out of 10,000 is well over 0.90. However, the chance of drawing the Ace of Spades exactly 4 times in the first 9,999 times and then drawing the Ace of Spades on the 10,000-th (i.e. the last) drawing is less that (1/52).

In other words, if you are asking how many times do you have to draw so that the combined chances of drawing exactly 4 times out of the first (n-1) draws, and the Ace of Spades on the n-th draw is $\geq$ 90%, the answer is that this is impossible.

Edit-2
Simplification of the determination of $n$ with respect to the primary interpretation of the problem in my answer.

It occurs to me that the algebra can be manipulated to facilitate computing the minimum value of $n$, so that your chances of drawing the Ace of Spades at least 5 times is > 90%.

Suppose we maintain my interpretation of the problem but simultaneously adopt the idea that you will stop on the 5th drawing of the Ace of spades.

Then, the chance of your stopping on the $n^{\text{th}}$ drawing is

$$F_{(n-1)}(4)\times p.$$

This means that the chances of stopping on or before the $n^{\text{th}}$ drawing is

$$p \times \sum_{k=5}^n F_{(k-1)}(4).$$

Since you want this probability to be $\geq 0.90$, you have to determine the smallest value of $n$, such that

$$\sum_{k=5}^n F_{(k-1)}(4) \geq \frac{0.90}{p}.$$

Based on my understanding, determining the minimum value of $n$, based on the above formula should match perfectly with determining the minimum value of $n$ re

$$G_n(5) \geq 0.90.$$

user2661923
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  • Wow, thanks so much for the insights and walking me through the basic steps in comments! I'm certain I cannot yet grasp it, but it allows me a reference to be able to explore the topics of the individual steps further in an attempt to round out my understanding of probability.

    If I may ask, what exactly is that form of (n k) in the definition for function F? I understand the Sigma summation and the series following to some extent though. wait, it could just be nCk or nPk?

     – Daniel Maybach Oct 09 '20 at 23:41
  • I believe the more complicated question is the one I want to be explored, even if I lack the tools right now to understand it. The idea is the "goal" is to reach 5 marks on the wall, and my question is what is the n number of trials would it take before the probability is 0.9 for reaching the goal. – Daniel Maybach Oct 10 '20 at 00:00
  • However, I welcome your initial interpretation for my purposes of learning as well: That it is not required that the last card drawn be the Ace of Spades, (ignoring the constraint that you stop when you have reached 5 Marks on the wall.) – Daniel Maybach Oct 10 '20 at 00:03
  • This is still one of my favourite exchanges I've ever had online - thanks so much for the time and effort you put into indulging a bored engineering student's curiosity! – Daniel Maybach Jan 17 '23 at 15:22