# Probability Assignment: Scratch and Win!

The fast food chain, MacDuff’s is running a competition.  With every order, you obtain a card which has 12 circles covered up, and you can scratch off up to four circles. You win if 3 or more PALM TREES are revealed, but lose if 2 or more CRABS are revealed.  If you win, you can scratch off any one of the three squares to show what you have won. The card above is shown with all of the circles and all of the squares revealed.  Macduff’s want the game to be both fun and relatively easy to win.

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1. Verify that the ratio of PALM TREES to crabs is in fact 2:1 in the picture above.
Palm tree: 8/12  and Crab: 4/12
Ratio is 8:4 = 2:1

The 8 trees and 4 crabs are in the 2:1 ratio on the card, which indicate there is a higher chance of picking a palm tree than a crab palm. The ratio shows that there are twice as many palm trees than crabs on the card.

2. If all the circles are now covered up, on your first choice, what is the probability of revealing
(a)  a PALM TREE=  8/12                             (b)  a CRAB =  4/12
………………………..= 2/3……………………………………..= 1/3
There is a 67% chance you will chose a palm tree on the first choice and about 34% of choosing a crab. Thus, there is a higher probability one will chose a palm tree than a crab on their first try.

3. Given that you obtained a PALM TREE on your first go, what is the probability of
(a) obtaining a PALM TREE on your second go? 7/11
(b) Obtaining a CRAB on your second go? 4/11

Because the palm tree was scratched on the first go, there are only 11 circles left and either the probability of choosing a palm tree or a crab. There is a 7/11 chance of obtaining a palm tree because one palm tree circle was already scratched off on the first go, leaving only 7 palm tree circles left on the card. On the other hand, there is a 4/11 chance of obtaining a crab on the second go because no circles with crabs on it had been scratched off just yet.

4. By drawing a probability tree, or otherwise, list all the winning combinations, and find all their probabilities. Each section of the tree diagram represent each trial a circle has been scratched off, thus the denominator of the fraction decreases for there are less circles to choose from. The probability of crabs and palm trees depend on if the crab of the palm tree was scratched off in the previous try. For each pair, the total should equal 1 because there is an 100% chance that either a crab or a palm tree will be obtained.

5.What is the probability of winning this game?

P(winning the game)
= PPPP + PPPC + PPCP + PCPP + CPPP
= (8/12*7/11*6/10*5/9) + (8/12*7/11*6/10*4/9) + (8/12*7/11*4/10*6/9) + (8/12*4/11*7/10*6/9) + (4/12*8/11*7/10*6/9)
= 1680/11880 + 1344/11880 (4)
= 7056/11880
= 0.59

The overall probability of winning this game is 59%, thus a 41% chance of loosing the game. From observing the percentage of winning, it can be seen that it is not a very high probability of winning. The probability of winning the game was calculated by adding all the possible outcomes in order to win, which were to obtain 4 palm trees or 3 palm trees and 1 crab in different orders. Each outcome was calculated by multiplying the chance of choosing those circles. The outcomes are dependent events and so the denominator will be affected each time because the circles cannot be replaced. In conclusion, there is a close chance of both winning and loosing this game but the chances of winning if the palm trees are in the ratio 2:1 is 59%.

6. Find the following conditional probabilities:
(a)  P(winning | first circle shown is a PALM TREE)

P(Win|First circle shown is PT)
= PPP + PPC +  PCP + CPP
= (7/11*6/10*5/9) + (7/11*6/10*4/9) + (7/11*4/10*6/9) + (4/11*7/10*6/9)
= 210/990 + 168/990 (3)
= 714/990
= 0.72

Given that the first circle shown is a palm tree, there is about 72% probability of winning. The probability of winning is pretty high because there are four various ways in which one can win the game. Thus, the calculation was done by adding all the possible ways in order to win, given that the first circle scratched is a palm tree. As the player obtains a palm tree on their first try, they have a higher chance of winning because the probability in loosing has decreased from the original time (when none of the circles were scratched).

(b)  P(winning | first circle shown is a CRAB)

P(Win|First circle shown is Crab)
= PPP
= 8/11*7/10*6/9
= 336/990
=0.34

The probability of winning is lower when the first circle shown is a crab because there is only one way to win, which is to get palm trees for the rest of the three circles chosen. When the first circle is a crab, the probability to win automatically decreases drastically because there is only one way to win, which is to achieve all palm trees for the next three trials. The probability of winning after first circle shown is the crab was calculated by finding the solution to the one event, that could only lead the player to win the game. Overall, the probability of winning a game when given that the first circle shown is a crab is 34%.

(c)   P(winning | first two are one of each)

P = 8/12 C = 4/12
P(Win|First two circles are one of each)
= PP
= 7/10*6/9
= 42/90
= 0.47

When the first two circles scratched were one of each (palm tree and crab), there is only one possible way to win which is to obtain two palm trees for the following trials. Since the first two circles were one of each, the only solution left to win is by obtaining two palm trees and so by calculating the probability of getting two palm trees on the following trials helped reach the probability to win if the first two are a palm trees and a crab. So when given that the first two circles are one crab and one palm tree, there is a 47% probability that you will win the game.

(d)  P(winning | first two are PALM TREES)

First try palm tree -> 8/12 Second try palm tree -> 7/11
P(Win|First two are palm trees)
= PC + CP + PP
= (6/10*4/9) + (4/10*6/9) + (6/10*5/9)
= 24/90 + 24/90 + 30/90
= 78/90
= 0.87

The probability of winning, given that the first two circles are palm trees is 87%, which is relatively high due to the many different ways one can win. Overall, as there are more possibilities to win, there is a higher probability to win. The probability to win if the first two are palm trees can be calculated by adding all the possible outcomes to win the game.

7. Now find the probability of winning if the card were slightly different:  if it had 9 PALM TREES and 3 CRABS.

P (winning)
= PPPP + PPPC + PPCP + PCPP + CPPP
= (9/12*8/11*7/10*6/9) + (9/12*8/11*7/10*3/9) + (9/12*8/11*3/10*7/9) + (9/12*3/11*8/10*7/9) + (3/12*9/11*8/10*7/9)
= 3024/11880 + 1512/11880 (4)
= 9072/11880
= 0.76

When there are 9 palm trees and 3 crabs on the card, the probability of winning is 76%. Compared to a card with only 8 palm trees and 4 crabs, the chance of winning on this card is about 17% higher because there are less crabs which results to a higher chance of obtaining a palm tree for each trial. As the amount of palm trees increase and the amount of crabs decrease, the probability of winning a game increases. When the card has the palm tree in ratio of 2:1 the chance to win is lower than a card with the  palm tree in the ratio of 3:1. Thus when the ratio of palm trees increases, the probability to win the game increases as well.

8. What is the probability of winning if there are 6 PALM TREES and 6 CRABS?

P (winning)
= PPPP + PPPC + PPCP + PCPP + CPPP
= (6/12*5/11*4/10*3/9) + (6/12*5/11*4/10*6/9) + (6/12*5/11*6/10*4/9)  + (6/12*6/11*5/10*4/9)  + (6/12*6/11*5/10*4/9)
= 360/11880 + 720/11880 (4)
= 3240/11880
= 0.27

When the ratio between the palm trees and crabs are even, the probability of winning is 27%. The probability is very low because for each trial, there is less than a fifty fifty chance of picking a palm tree. The more the ratio between the two are close together or if there are more crabs, it results with a very low probability of winning the game due to the very low fraction of chance on picking a palm tree three times out of four scratches. In short, when the ratio of palm trees 0f the card is 1:1, the probability of winning is very low compared to the chances of winning of a card with palm trees that are in the ratio 2:1.

9. What if there were 11 PALM TREES and 1 CRAB?

P (winning)
= PPPP + PPPC + PPCP + PCPP + CPPP
= (11/12*10/11*9/10*8/9) + (11/12*10/11*9/10*1/9)+ (11/12*10/11*1/10*9/9) + (11/12*1/11*10/10*9/9) + (1/12*11/11*10/10*9/9)
= 7920/11880 + 990/11880 (4)
= 11880/11880
= 1.0

When there are 11 palm trees and 1 crab, the probability of winning the game is 100%. This is because it only takes three palm trees in order to win the game and since there is only one crab, there are no possible ways to obtain any circle other than a palm tree if the only crab is obtained. Thus, when there is only one crab or the ratio of palm trees on the card is 11:1 , the probability of winning the game is 100%.

10. Draw accurately by hand, or by calculator/computer,  a graph which shows the probability of winning with n PALM TREES out of 12. By looking at this graph, the probability of winning the game increased as the number of palm trees on the card is larger. When there are 11 palm trees, there is a 100% probability to win the game and so the closer the number of palm trees on the card is to 11 has a higher chance of winning. The ratio between the palm tree and the crab also effect the probability of winning the game. When the ratio of palm trees is high, the more likely to win the game because there are more options to win.

11. Your expected winnings,  known as E(X),  is found by the formula: E(x) = P(win) x Prize money – Cost of Game.
(a) Find the expected winnings if there are 8 Palm trees and 4 crabs, where the cost of a card is \$2 and the prize is \$4.  Is this profitable for the card makers?  Explain.

8 Palm Trees and 4 Crabs
E(x) = 0.59 x \$4 – \$2
= 2.36 – 2
= +0.36

This card is not profitable for the card makers because they loose \$0.36  per game, while the player gains \$0.36 each game. When the outcome of the expected winnings is a positive number, it shows that the player gains \$0.36 for each game they play and the card maker looses that much money per game.

(b) What are the expected winnings if, for the same scenario, the prize is \$3?  Explain whether this is profitable.

\$3 Prize
E(x) = 0.59 x \$4 – \$3
= 2.36 x \$3
= – 0.64

This card is profitable for the card makers because they gain \$0.64 per game, while the player looses \$0.64 each game. This is because the difference between the price of the card and the prize is very small, making it harder for the player to gain back all the money the player has lost.

(c)  Find out whether it would be profitable to make a card cost \$2 and then pay a prize of \$5, if there were 6 palm trees and 6 crabs.

6 Palm Trees and 6 Crabs. Costs \$2 and a \$5 prize.
E(x) = 0.27 x \$5 – \$2
= 1.35 – 2
= -0.65

It would be profitable for the card makers to make a card that costs \$2 and then pay a prize of \$5, if there were 6 palm trees and 6 crabs because they would gain \$0.65 per game, while the player looses \$0.65 each game. Since the probability to win is very low, the cost of the card can be very low and the prize very high without the card makers having to loose money.

12. Use information from your graph to help you design your own scratch card, with a different total number of circles.  Decide on a price to charge for each card and the prize money so that the card is both profitable and enticing/popular.

Total number of circles = 18
Banana = 11  (get 3 bananas to win)
Monkey = 7 ( total of 4 circles to open)
Cost of game = \$3
Prize money = \$5

P (win)
= BBBB + BBBM + BBMB + BMBB + MBBB
= (11/18*10/17* 9/16* 8/15) + (11/18*10/17* 9/16*7/15)  + (11/18*10/17*7/16*9/15)  + (11/18*7/17* 10/16*9/15) + (7/18*11/17* 10/16*9/15)
= 7920/73440 + 6930/73440 (4)
= 35640/73440
= 0.49

E(x) = P(win) x Prize money – Cost of Game.
=0.49 x 5 – 3.
= 2.45-3.
= -0.55

13.  Reflect on the work you have done and answer the following questions:

• Do you think your scratch card is both profitable and enticing?  If so, how have you managed to do this?  What do you have to think about when designing such a card?
I think my scratch card is profitable and slightly enticing. I was able to manage this by making sure the total of expected winnings is a negative number, in order for me to profit from the scratch card. Since there is a good difference between the cost of the game and the prize money, it may attract many players to want to play the game. When designing these cards, I think one should think about how we, as card makers, can profit as well as make many people want to play the game. I did not think about the fairness but thought of ways in which only the card makers can profit from the game because we want to earn money. So the ratio between the banana and monkey was made as large, yet  not too different or else the card makers will not profit much. As you design the cards, the game has to be pretty fair and not impossible to win or else the players will not want to play the game.
• How large a role does the design of such games have in society?
The design of such games have a large role in society because card makers make sure that they make cards that are profitable and allow players to loose much money when they become addicted to the game. If the card is actually not profitable for the player and they do not realize this, they could end up loosing much money. These types of games are sold everywhere in society and it is important for the card makers to produce games that would be profitable to themselves and not the players. With these kind of games, it usually brings many people to loose their money yet they keep playing because they really want the prize money.
• Is there a better or more efficient way to design a profitable yet enticing game?
There is a better and more efficient way to design a profitable yet enticing game by either changing the amount of bananas so that the players have a higher chance of winning, which would make the game more enticing. Also, if the prize of the money was higher, the game would be enticing. It is hard to make the cards both profitable and enticing. For example, by changing the amount of money for the prize to \$6, more people are likely to play the game because they can make double the amount they pay. Although the money we profit as card makers will be very low:
E(x) = P(win) x Prize money – Cost of Game.
=0.49 x 6 – 3.
= 2.94-3.
= -0.6

## One thought on “Probability Assignment: Scratch and Win!”

1. eadurkin says:

You explain the mathematics very clearly, Yurika. Your mathematical working is accurate and detailed. You make some good points in the reflection section as well.
More detailed feedback will be posted on Moodle and in Powerschool.