Expectanion value probabilities

AI Thread Summary
The discussion centers on calculating the expected profit from flipping a coin 10 times, where the probability of getting heads (crown) is 0.57 and the payout is based on the number of heads. The initial calculation presented is incorrect because it does not account for the probability distribution of obtaining k heads in multiple flips. Participants emphasize the need to use binomial distribution to accurately determine the probabilities of different outcomes. They suggest considering permutations and combinations to solve the problem correctly. Understanding these statistical concepts is crucial for finding the expected value in this scenario.
ParisSpart
Messages
129
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
Back
Top