Expectanion value probabilities

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Homework Help Overview

The discussion revolves around calculating the expected value of profit from a game involving flipping a coin 10 times, where the probability of getting heads (crown) is 0.57, and the profit is determined by the number of heads obtained.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's attempt to calculate the expected value using a summation formula, questioning the validity of the probability used for obtaining heads. Others suggest the need for a proper understanding of distributions and the correct method to calculate the probability of getting a specific number of heads in multiple flips.

Discussion Status

The discussion is ongoing, with participants providing insights into the correct statistical approach to the problem. There is a recognition of the need for understanding binomial distributions and the importance of checking the plausibility of results. Some participants express uncertainty about their knowledge of distributions, indicating a potential gap in understanding.

Contextual Notes

Some participants mention that they have not yet learned about distributions in school, which complicates their ability to solve the problem effectively. There is also a note on the importance of permutations and combinations in calculating probabilities for multiple outcomes.

ParisSpart
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Flipping a coin 10 times. The currency brings crown with probability 0.57. If you bring a total of k heads they add 2 ^ k euros.

What is the average profit in this game?

i must find the expectanion value and i do this:

∑from k=1 to 10 --->((0.57)^k)*(2^k) =22.0445164

i think that my solve is not correct any ideas?
 
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The probability of getting k heads in 10 flips is not 0.57^k.
 
0.43?
 
No, and guessing will not help.
This is a basic problem in statistics, you should know which distribution you have to use.

As a general remark: you should always check if your values, results, ... are plausible. Please do this before you ask questions.
As an example, the probability to get k times heads in 10 flips is certainly different than the probability to get k times heads in 1 flip. Therefore, the value has to depend on the number of flips in some way, otherwise it cannot be right.
 
we don't have learned distrinutions in school yet...
 
in your sum, you need the probability of exactly k heads. So you have to take into account that 10-k are not heads. Also, you don't care which of the 10 are heads. how many different possibilities are there for which are?
 
ParisSpart said:
we don't have learned distrinutions in school yet...
That makes this problem very difficult! I imagine you have learned enough to know that the probability of all 10 being crowns will be (.57)10 and the probability that all ten are not crowns will be (1- .57)10= (.43)10. But the probability that the first flip is a crown and the rest are not is (.57)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)= (.57)(.43)9 while the probability of "first not a crown, second a crown, then the rest not crowns" is (.43)(.57)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)= (.57)(.43)9 so that the probability of "one crown, nine non-crowns" is 10(.57)(.43)9.

And things like "four crowns, 6 non-crowns" get much more complicated as the possiblities increase. Are you sure you have not learned things like "permutations and combinations", "binomial coefficents", and "binomial distributions"?
 
how i can solve it with distributions way?
 

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