THE DICE GAME - calculating probabilities
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Introduction
INTRODUCTION
In the real world events can not be predicted with total certainty. The best we can do is say how likely they are to happen, using the idea of probability.
Probability of an event happening = | Number of ways it can happen | |
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Total number of outcomes |
Probability is the branch of mathematics that studies the possible outcomes of given events together with the outcomes' relative likelihoods and distributions. In common usage, the word "probability" is used to mean the chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed as a percentage between 0 and 100%. The analysis of events governed by probability is called statistics.
Statistics is the study of the collection, organization, analysis, and interpretation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments.
SHORT EXPLANATION: Each *point* is considered as a separate task.
Middle
Ann-2, Bob-3
Ann-3, Bob-3
Ann-4, Bob-3
Ann-5, Bob-3
Ann-6, Bob-3
Ann-1, Bob-4
Ann-2, Bob-4
Ann-3, Bob-4
Ann-4, Bob-4
Ann-5, Bob-4
Ann-6, Bob-4
Ann-1, Bob-5
Ann-2, Bob-5
Ann-3, Bob-5
Ann-4, Bob-5
Ann-5, Bob-5
Ann-6, Bob-5
Ann-1, Bob-6
Ann-2, Bob-6
Ann-3, Bob-6
Ann-4, Bob-6
Ann-5, Bob-6
Ann-6, Bob-6
(The bold combinations are cases in which Ann wins.)
Solution:
A = case where Ann is winning
B = case where Bob is winning
N = case neither Ann and Bob is winning
Since both have dices with 6 sides, the total number of combinations when both rooling is 6 x 6 = 36.
The dice will match 6/36 of the time. Means in 6/36 of the time both will have equal number on the dice.
Each player will have the higher number in 15/36 of the time.
P(Ann)=15/36
P(Bob)=6/36
P(Neither)=15/36
With ties going to Bob, Ann wins 15/36 or total of 41.6666...67% => ~41.7%.
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TASK 2
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Solution:
First:
1-(21/36)^2 is the chance for Ann to win if, when she loses the first time, both of them roll again.
Conclusion
A = case where player is winning. P(A) = 15/36
B = case where bank is winning. P(B) = 21/36
P is pay out, the amount of the bank pays if the player win.
V is betting value, the amount of money the player has to pay each time she/he wants to play.
Now, lets calculate for a fair game
E = P(A) * P - P(B) * V
0 = 15/36 * P - 21/36 * V
so we have the ratio P = 21/15 * V
In case they roll to infinity and the player finally win the game
Winning - 1st roll = 15/36
Winning - 2nd roll = 21/36 * 15/36
Winning - 3rd roll = 21/36 * 21/36 * 15/36
Sum of all (with geometric progression)
S = t_1 / (1-r)
S = (15/36) / (1-21/36)
S = 15/15
This cannot be right because the calculated probability is 1.
Bank wins the game
Winning - 1st roll = 21/36
Winning - 2nd roll = 15/36 * 21/36
Winning - 3rd roll = 15/36 * 15/36 * 21/36
…
Sum of all (with geometric progression)
S = t_1 / (1-r)
S = (21/36) / (1-15/36)
S = 21/21
This cannot be right because the calculated probability is 1.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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