Probability Modeling for Coin Flips

AI Thread Summary
The discussion focuses on calculating the probability of achieving a negative total score from a series of coin flips, where five different coins yield varying scores for heads and tails. Participants suggest using probability generating functions or tree diagrams to analyze all possible outcomes, emphasizing the independence of each coin flip. The complexity of the calculations is acknowledged, with recommendations for programming solutions to handle the numerous combinations. A Monte Carlo method is also proposed as an alternative for approximating the probabilities. The conversation highlights the need for a solid understanding of probability theory to effectively model the scenario.
phya2020a
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Hi math gurus .
I have probability modeling question.
There are 5 coins A, B, C, D,E,F . Each coin has 2 outcomes on flipping Head (H) or Tail (T). For each coin type there is a different score associated with H or T outcome. The cumulative sum of score will be recorded for the number of flips performed for each coin type.

Trial X :
Coin type ...If H, Score....If T, Score...# of flips
A......110...... -70.....14
B......87......-62......9
C......90......-70......8
D......100......-70.....8
E......94......-62......4

Example :
So if A is flipped and outcome is H then score will be 110 and if outcome is T , the score will be -70. For H, T , H, H, T outcomes the score will be 110-70+110+110-70 = 190

For Trial X
Score A ( 14 flips)=190, Score B( 9 flips)= -25, Score C( 8 flips)= -140, Score D( 8 flips)= -110, Score E ( 4 flips)= 32

Total score for trial X = -53

Ques 1. What will be the probability of 'Total score for trial X ' be negative ?

Ques 2. If 20 iterations of trial X are performed then what will be the probability of sum of 'Total score for trial X ' for 20 iterations be negative ?

Example:
Iteration 1 : Total score for trial X = -53
Iteration 2 : Total score for trial X= 150
Iteration 3 : Total score for trial X = 50
...and so on

Sum ( Iteration 1 through Iteration 20) = 1200 { positive in this case}
 
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Hey phya2020a.

Hint: What is the distribution of the sum?

You can either calculate the probability function by assigning a probability to every single outcome (for n trials you will have 2^n outcomes) or you can use analytic tools like a probability generating function.

I would as a suggestion, find the PDF for every single possibility and then sum all probabilities where the sum is less than 0.
 
Hi chiro,
Thanks for the response .
Sorry I have no background in probability .

Can this be done manually ?

If no , I would appreciate if you could provide the broad approach/ additional details , so that I can request someone with programming background to have it coded & modeled .

All I am looking for is a value 0.01 to 0.99 for the 2 questions

Regards
phya




chiro said:
Hey phya2020a.

Hint: What is the distribution of the sum?

You can either calculate the probability function by assigning a probability to every single outcome (for n trials you will have 2^n outcomes) or you can use analytic tools like a probability generating function.

I would as a suggestion, find the PDF for every single possibility and then sum all probabilities where the sum is less than 0.
 
Basically you draw a tree diagram for all possible combinations and when you have more than one event that has the same outcome, then you merge the probabilities together.

Once you have done this your distribution is done by collecting all different events and their probabilities.

Then you take this distribution and add up all the probabilities for events less than 0 and that is your answer.

If you draw a tree diagram, note that every single flip of the coin is independent and the probability for A = a and B = b (A is one flip, B is another) is P(A=a) 8 P(B=b) so if you have probability of success is p, then probability of failure is 1 - p.

Some examples are:

P(A = success, B = success) = P(A=success) * P(B=success) = p*p = p^2
P(A = success, B = failure) = P(A=success)*P(B=failure) = p*(1-p).

Basically do this for all branches (which is 2^n branches) up to all leaves (i.e. the final branch with no other branches), collect all similar events together probability wise and then form a distribution using this.

The you just add all probabilities up where the event is less than zero and you're done.
 
phya2020a said:
Can this be done manually ?

I don't see any way to do it with a pencil-and-paper calculation. (I think you are asking about a general type of problem and you just gave one specific example.)

If no , I would appreciate if you could provide the broad approach/ additional details , so that I can request someone with programming background to have it coded & modeled .

If your coins are all "fair" and the the results of the tosses are independent then you should say so. Other than those details, I think you have described the problem well enough so that a programmer with a knowledge of probability theory could write a program to determine the answer. As Chiro says, the way to do a deterministic calculation is to generate all possible combinations of total numbers of heads and tails of each coin and note the score that results. To each score, you add the probability of each combination that results in that score.

A programmer who can't implement the deterministic calculation could get an approximate answer using the Monte-Carlo method.
 
Chiro/Stephen Tashi -appreciate all your help.
-p
 
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