00:01
Okay, so in this question, we're looking at a card game.
00:04
This is a little different because typically we're looking at regular, you know, normal cards, but this one's not normal.
00:09
This one's actually, we're looking at a deck that has 13 ranks, but there's no, like, face cards or anything like that.
00:17
So just numbers 1 to 11 and 8 different suits.
00:22
So we have a total of 88 cards.
00:26
That's important to keep in mind.
00:27
So how many different ways are there to get any five cards? hand? well, let's look at each of the cards individually.
00:39
First of all, we don't have any repeating cards.
00:41
None of them look the same.
00:43
You may have, let's say, draw two cards and they're both fives, but then they would be a different suit.
00:48
So you're not going to have any, like, duplicates.
00:51
So that makes each card unique.
00:53
So for the first card i hand you, there's 88 choices.
00:58
But then if i hand you one of those cards, i only have 87 left to draw a second card from, 86 left to draw a third card from 85 left to draw a fourth card from and 84 left to draw the final card from so let's go ahead and multiply these together 88 times 87 times 86 times 85 times 84 is a very large number 4 billion, 701 ,000, sorry, 701 million, 90 ,240.
01:53
That's how many options there are for drawing five cards from an 88 card deck.
02:02
So that's the number of ways.
02:06
Now, how many different ways are there to get exactly one pair? so we want exactly one pair.
02:13
The first thing that we want to do is we want to think about choosing a card of a certain rank, right? because that's what makes something a pair.
02:21
It has nothing to do with the suit.
02:22
It has to do with the number that's on there.
02:25
And we have 13 ranks.
02:27
So let's just pick a card with a rank.
02:30
13, choose one.
02:32
Let's say like we pick a 7.
02:36
Of course, we're randomly drawing these.
02:39
Second, we're going to choose another card of the same rank in order to make a our pair.
02:45
Well, assuming, so this is where it gets a little bit tricky.
02:49
Let's just look at an example of what it would look like for a five card hand.
02:55
So if i have like four, four, three, two, one.
03:00
This is just an example, right? to have exactly one pair, that means that only one number is matching and the rest have to be all different.
03:10
So look at this.
03:11
How many ranks do we have, aka how many different numbers? how many different numbers? cards do we have we have a one a two a three and a four and since we're repeating one of the cards in this case we have another four we're only going to have four different ranks so i pick my first card let's say i pick it it's a four then the second card i need something to match it and i can choose from any one of these three numbers and it's going to match or sorry these four numbers four three two and one so i have four four three two and one so i have four four ranks that would match with one of the cards that i have, and i only want one to match.
03:53
So four, choose one.
03:57
Now, the other ones cannot match.
04:00
We do not want them to match with those cards.
04:03
So let's look at how many cards we have left over.
04:06
Total, we have 88 cards, right? so 88 minus, we've already picked two cards.
04:12
We've already drawn two cards.
04:13
We have 86 cards left over to choose from.
04:18
And we can choose any one of them.
04:20
So we're just going to do 86, choose one.
04:30
So now we have three cards, and we need two more.
04:32
And again, they are not supposed to match at all.
04:37
So we're going to do 85, choose one.
04:44
And then the last card is going to be 84, choose one.
04:49
Which i'm running out of space just a little bit for it.
04:51
Oops.
04:54
Hang on one second.
04:55
Let me just put an arrow.
04:56
This is supposed to be 84 choose one.
05:00
I just ran out of space.
05:05
Now when it comes to evaluating the choices, whenever you do like a number choose one, it's just itself.
05:10
So it's really just 13 times four times 86 times 85 times 84 ways of getting exactly one pair.
05:30
So not two pairs, exactly just one pair.
05:34
So let's calculate that.
05:36
13 times 4 times 86 times 85 times 84 is 31 ,930 ,080.
06:04
So that's our answer for part b.
06:06
How many different ways are there to get exactly one pair right there? now, what is the probability of being dealt exactly one pair? well, to get probability, all that we do is we take.
06:18
The number of ways of getting exactly one pair divided by total number of ways of selecting a card so what we have to do is 31 ,930 ,080 divided by 4 billion 701 million 90 ,240 .0 .0.
06:50
Okay, so it says round to the nearest seven decimal places.
06:52
So let's do that.
06:54
Six, seven, nine, two, and one.
07:11
So it's about like a little bit more than half a percent chance of getting exactly one pair.
07:25
All right.
07:26
Now the next part of the question is up here.
07:29
How many different ways are there to get exactly two pairs? okay.
07:35
So a number of ways of getting exactly two pairs.
07:37
You know what? let's just compute that down here because it's kind of similar.
07:41
Similarly modeled.
07:45
But let's think of an example.
07:47
So maybe i could get a seven, a seven, a five, a five, and a one.
07:53
This is just one example.
07:55
Of course, it could be interchanged with any of the numbers between one and 11.
07:59
But, but this is going to model the same way we did before.
08:02
Okay, there's 13 ranks.
08:04
Think about how many ranks am i going to, sorry, one second.
08:13
So 13 choosing a rank.
08:17
Sorry, i meant 13 times two, or 13 choose two.
08:20
So we want two cards to be pairs.
08:23
So that's why we're going to do 13 choose two.
08:29
Now we can look at how many ranks do we have.
08:34
Since we're doing, since we're doing five cards, you have one rank here that's going to be a pair, another rank here that's going to be a pair, and then just the other one that's by itself.
08:50
We're going to have actually three ranks that we're going to be choosing from.
08:54
So we have three ranks to choose from, and we're going to choose one of them to be the first pair, and we're going to choose the other one to be the second pair, and then, of course, our last card.
09:13
So we've got one, two, three, four, and then our last card, we have, let's see, how many cards have we already picked four? so we have 84 cards left to choose from.
09:25
It can be literally anything else.
09:32
But actually, we're going to need to go back and fix the last one too.
09:35
But it actually can't be just anything because we don't want it to match with the other numbers.
09:43
So how many, we have eight different suits, right? so we have six more sevens.
09:52
This is just theoretically.
09:54
And we have six more fives.
09:56
We could possibly choose from.
09:58
So we should actually do 86 minus 12 because we couldn't have it match with that.
10:04
So we only have 74 cards left to choose from because we cannot choose any of the leftover sevens and fives that haven't been picked.
10:12
There's 12 of those.
10:13
And obviously we can't pick any of the four cards we've already drawn.
10:17
So that takes 16 cards out of the equation.
10:19
So we only have 74 cards left to choose from and we can just pick one.
10:25
So before we calculate that, let's just go back to this last one and fix it real quick because for the same reasons um everything was fine until we got here we did 86 choose one 85 choose one and 84 choose one but in reality we don't want another pair right so um we had already picked two cards so we're down to 86 cards left to choose from and that's why originally we had put 86 there but we don't want it to pay with the other two cards...