Question

You are dealt a hand of three cards one at a time. What is the probability that you get no spades?

What is the probability you have at least one ace?

Answer 1

Probability of no spades `= 41.35%`
Probability of at least one Ace `=21.74%`

Explanation:

• The "no spade" question

With a standard deck of 52 cards, 13 of them are spades. This means that `(52-13=39)` cards can satisfy the condition of not getting a spade.

We'll use the combination formula to see how many hands in total that is (we'll also use it to see how many hands we can make in total). The general formula for combinations is:

`C_(n,k)=(n!)/((k)!(n-k)!)` with `n="population", k="picks"`

To calculate the probability, we take the number of hands we can get with no spades and divide by the total number of hands we can get.

Number of hands with no spades

From the population of non-spade cards (39), we're picking 3:

`C_(39,3)=(39!)/((3)!(39-3)!)=(39!)/((3!)(36!))=>`

`(cancelcolor(blue)39^13xxcancelcolor(green)38^19xx37xxcancelcolor(red)(36!))/(cancelcolor(blue)3xxcancelcolor(green)2xxcancelcolor(red)(36!))=13xx19xx37=9139`

Total number of 3-card hands possible

`C_(52,3)=(52!)/((3)!(52-3)!)=(52!)/((3!)(49!))=>`

`(52xxcancelcolor(blue)51^17xxcancelcolor(green)50^25xxcancelcolor(red)(49!))/(cancelcolor(blue)3xxcancelcolor(green)2xxcancelcolor(red)(49!))=52xx17xx25=22100`

Probability of no spades

`color(blue)(ul(bar(abs(color(black)(P=9139/22100~=41.35%))))`

• The "at least one Ace" question

There are 2 ways to approach this question:

• we can count up the number of hands with 1, 2, and 3 Aces, or
• we can figure out the number of hands with no Aces and subtract that from the total number of hands possible

I'm going to use the second method.

Number of hands with no Aces

There are 4 Aces in a standard deck, which means there are `(52-4=48)` cards that will satisfy the "no Ace" condition:

`C_(48,3)=(48!)/((3)!(48-3)!)=(48!)/((3!)(45!))=>`

`(cancelcolor(blue)48^8xx47xx46xxcancelcolor(red)(45!))/(cancelcolor(blue)(3xx2)xxcancelcolor(red)(45!))=8xx47xx46=17296`

Number of hands with at least one Ace

We can now subtract the number of hands with no Aces from all those possible to find the number of hands with at least one Ace:

`22100-17296=4804`

Probability of at least one Ace

`color(blue)(ul(bar(abs(color(black)(P=4804/22100~=21.74%))))`