1. In poker, a royal flush is getting 10,j,q,k and A of the same suit. Show that there is 1 chance in 649 740 of achieving this.
2. Three digits are randomly chosen from 5,9,8, and 1 to make a three digit number. what is the probabily that this number is
a) Greater than 500? b) even? c) divisible by 3?
Thank you in advance. I want to learn the more mathematical way instead of like tree diagram (visual, as it takes too long).
Random note: Just about the whole chapter of probability, i did not understand...
Probability trees are stupid. They work for some people, but they've never made sense to me. I'll offer a more abstract and general way of thinking about probability.
In general, the kind of probability you'll encounter is an
equiprobable space, i.e. we're assuming that every outcome in the
sample space (all possible outcomes) has the
same probability/chance of happening. In this kind of probability, the chance of an
event (which is a set of
particular outcomes) happening is given by
(the number of outcomes that event can happen) / (total number of possible outcomes). The numerator is the
size of the event (the number of outcomes classified as that event), and the denominator is the
size of the sample space (total number of outcomes). Remember, an event and the sample space are both
sets of outcomes.
An example of this is
what is the probability of picking an ace from a deck of cards? Our event is
picking any ace. Our deck of cards is the sample space. The size of our event (picking an ace) is 4, because there are 4 aces in the deck, so picking any one of the aces is an outcome within our event. The size of our sample space is 52, because there are 52 possible outcomes.
So therefore, the probability of picking an ace from a deck of cards is
(size of event)/(size of sample space) = 4/52 = 1/13.
Now getting to your questions,
1. Before I go through this problem in detail, please note that this is not an easy problem to learn probability with. In fact, this is the kind of problem 3 unit maths students study (permutations and combinations), so this would be a really tough problem for general maths. This problem uses the a technique that isn't obvious if it's the first time you've seen it. I will use and explain this technique in question 1 without referring to it, and name it in question 2.
It's tempting to say that our sample space is 52, since we have a deck of cards, but that would be
incorrect. The question asks for 5 cards, so our sample space should be
all possible ways of drawing 5 cards. We only need the size of the sample space. To find the size...
How many ways can we draw the first card? We have a deck of 52 cards, so there are 52 ways of drawing the first card.
How many ways can we draw the second card? Now that we only have 51 cards left in our deck, there are 51 cards of drawing the second card.
Repeating this reasoning, there are 50 ways of picking the third card, 49 ways for the fourth, and 48 ways for drawing the fifth.
Now how many ways can we draw 5 cards all together? The answer is 52×51×50×49×48. Why are we multiplying those numbers? I'll give you an explanation by considering an easier example: how many ways can be draw 2 cards? Suppose my first card is the ace of spades. Then there are 51 ways for me to choose the second card to get a
unique pair of cards. Now suppose my first card is the ace of hearts. Then there are
another 51 ways of drawing a
unique pair of cards, so that means so far, we have 51+51=102 unique pairs of cards. Now if we do this for all 52 possibilities of drawing the first card, then we have 52×51 ways of drawing a pair of cards. So for 5 cards, we have 52×51×50×49×48 number of ways of drawing unique hands. (In general, and while this isn't exact, you can use the rule of thumb that when you see "and" in a probability question, you multiply, and when you see "or" in a probability question, you add.)
Now that we have the size of our sample space, we need the size of our event space. In a single suit, there is only
one way of getting 10, J, Q, K and A in a random draw of 5 cards
in that order. But in our case,
the order does not matter, so we need to count the number of ways to draw those 5 cards in particular. But don't freak out, because we can solve this minor problem exactly like how we found the size of the sample space.
We have 5
unique cards. (the suit doesn't matter; we'll deal with it afterwards.) The number of ways we can draw the first card is 5. The number of ways we can draw the second card is 4. The number of ways we can pick the third card is 3, for the fourth card we have 2 ways, and the last card has just one way because we only have one card left. So we have 5×4×3×2×1=120 ways of rearranging 5 cards (or, if you like, we have 120 ways of drawing 5 cards from a deck of five; probability problems can be interpreted in many ways).
Remember there are 4 suits, so the size of our event is 4×120 = 480.
Now we have our size of event, and size of our sample space. So therefore, the probability of a royal flush is 480/(52×51×50×49×48)=1/649740. In other words, we have 1 in 649,740 chance of getting a royal flush.
2. Our sample space is all possible ways of making 3-digit numbers with those 4 digits. By the same technique that we have used in the previous question (called the
multiplication principle), the number of ways to draw the 4 digits (or, if you prefer, the number of ways of drawing 3 cards from a deck of 4) is 4×3×2=24. Note that the multiplication principle takes into account
ordering, which is relevant in this question. All we need to do for the following subproblems is to find the size of the events.
(a) As long as the 3-digit number doesn't start with 1, we're good. That means we need to find all possible 3-digit numbers with 5, 1, 9 and 8 such that the number doesn't start with 1. We're trying to fill _ _ _. How many numbers go into the first spot? 3, because we don't include 1. Second? 3, because we now have 3 numbers left, and remember that 1 can go into here. Last? 2. 3×3×2 = 18.
So the answer is 18/24=3/4.
(b) The number is even as long as the last digit is 8. So we need to find the number of ways we can make numbers of the form _ _ 8 with the other digits. The problem is the same as the number of ways we can draw 2 cards from a deck of 3. So 3×2=6 number of ways. Another way to think about the problem is this: How many numbers can we put into the first digit? 3. How many for the second digit? We've used a digit already, so 2. Thus 3×2=6 ways.
So our answer is 6/24=1/4 again.
(c) A number is divisible by 3 if its digits add to 3. Thus we can have numbers with digits {5, 9, 1} or {9, 8, 1}. The number of ways of making a 3-digit number with 3 digits is 3×2×1=6 (again, it's the same as the number of ways of drawing 3 cards with a deck of 3 cards). Now we have 2 sets of 3 digits, so we multiple our previous answer by 2 to get the event size: 2×6=12.
So our answer is 12/24=1/2.
Hope this helps!
