1.
Dora and Her Game
of Go Fish
1

2. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
Okay, in order for Boots to win in one draw. He must
already have 2 pairs and one single card. So all he has
to do is just pick up and win. To solve this we must
find the probability of getting 2 pairs and a single or
getting a 4 of a kind and a single card.
2

3. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
Okay, in order for Boots to win in one draw. He must
already have 2 pairs and one single card. So all he has
to do is just pick up and win. To solve this we must
find the probability of getting 2 pairs and a single or
getting a 4 of a kind and a single card.
3

4. So this is a combinatoric question. We need to use the
choose formula for this. Lets work with the 2 Pairs. There
are 13 face values in one deck of cards and we choose 2.
And for each of the face values chosen, we need 2 suits.
So it is 4 choose 2. It is squared because we chose 2 face
values.
4

5. Now for that single card. There are 11 face values left. So
out of the 11 we chose 1. That for that one card we choose 1
suit out of the 4.
5

6. Now to find the values of each choose event. You can just plug
it into the calculator or do it the long way. For the short way
skip to slide 8.
6

7. To solve for a choose situation the formula is above. n represents the total
number of objects and r is the number of things we chose. Choose formula
include 2 indistinguishable objects (Yes or No). So when we choose 2 we
are saying yes to 2 and no to 11. Notice all of the variables have a factorial.
That is because each one represents how many ways the variable can be
arranged within themselves.
7

8. So when we plug the choose events into the calculators or do
the long way by actually calculating using the Choose Formula
we get the numbers shown above. After multiplying the values
we end up with 123552. There are 123552 ways to make a 5
card Poker hand of 2 Pairs.
8

9. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
Okay now that we have found out how many ways that
we can get a 5 card Poker hand of 2 Pairs, how many
ways can we can get a 4 of a Kind.
9

10. Similar to the 2 Pairs calculations, the 4 of a Kind involves
finding multiple suits of 1 face value. In this case out of the 13
face values we only need one hence the 13 Choose 1. Plus we
want all of the suits of that card so we are choosing 4 out of 4
suits. The single card is just 12 Choose 1 because we want only
1 card out of the 12 remaining cards and only 1 suit for that card
as well.
10

11. Just like the other hand, we just plug the Choose Events into the
calculator if possible or use the Choose Formula to solve each
one. Then we get the numbers above and the answer is 624.
11

12. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
We've found the number of ways we can get a 2 Pair or
a 4 of a Kind. So the probability would be the number of
ways to get the hands over the number hands that can
possibly be made from a deck of cards (sample space).
The total number of possible hands is 52 Choose 5
cards.
12

13. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
Now that we have a sample space, we need to find the
probability of getting a 2 Pair Hand or a 4 of a Kind
Hand from one deck of cards.
13

14. Number of Favourable Outcomes
Probability =
Number of All of the Possible Outcomes
Remember that to find the probability of an outcome of an event we have
the number of favourable outcomes divided by the sample space or the
number of all of the possible outcomes that could happen in the event.
14

15. #'s of Ways to Make 2 Pairs
P(2 Pairs) =
Total # of Hands That Can Be Made from 1 Deck
= 123552
2598960
#'s of Ways to Make a 4 of a Kind
P(4 of a Kind) =
Total # of Hands That Can Be Made from 1 Deck
= 624
2598960
15

16. 123552 624
P(2 Pairs or 4 of a Kind) = +
2598960 2598960
= 124176
2598960
= 0.0478
Now we want the probability of getting a 2 Pair or a 4 of a Kind. So we
add the probabilities together.
16

17. What is he going to do? What is the probability of Boots winning by
drawing only one card if: i) No one has the card he needs ii) One
person has the card he needs iii) Two people have the card he
needs?
Wow we've done so much but it is time to begin
answering the questions. We now know that there is a
4.78% of getting 2 pairs of cards. For the first question, if
no one has the card that Boots needs to win that means
that there are 3 cards out of the remaining 22 that would
let him win.
17

18. If no one has any of the cards that Boots needs to win
that means that there are 3 chances out 22 cards to win
(hence the 3 over 22). So we just multiply that to the
probability of getting 2 pairs. Then we get that there is a
0.6% chance of Boots winning (0.0065 x 100 = 0.6%).
18

19. For the second question, some one has one of the cards
that Boots needs to win. This means that there are 2 left.
It is pretty much the same question but the numerator or
number of favourable outcomes is now 2 not 3. If we go
through the same process we get 0.4% chance of Boots
winning.
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20. The final question says that 2 people has the card that he
needs. This means that there is only one more card left.
Again same process. We multiply the probability of
getting 2 pairs by the probably of getting that card. We
now get an answer of 0.2%. As you can see Boots
doesn't have much luck of winning :P. Poor Boots...
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