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Expected value

I don't know anything about "expected value". I am moderately good in combinatorics, but I never heard this term before today. Like I have got 6 coins numbered a, b, c, d, e & f; and I pick up a coin randomly, what is the expected value (I mean a to f) of the coin?. Can anyone help me?

Note by Joseph Gomes
3 years, 9 months ago

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Comment deleted Apr 11, 2013

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@Hero P. Thanks a lot Now I can solve Expected Raffle Value Problem Rushikesh Jogdand · 3 years, 9 months ago

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@Hero P. Thanks a lot. I had a misconception that expected value referred to the outcome having highest probability. Sambit Senapati · 3 years, 9 months ago

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@Hero P. X=46; Y=91. I would choose the 5-envelope version, because there is more chance of getting the envelope with more money when the number of envelopes is smaller. My explanation is probably wrong, you are requested to solve the last part of your problem. Joseph Gomes · 3 years, 9 months ago

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Comment deleted Apr 11, 2013

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@Hero P. I would preface your post with a statement on risk-neutrality. Given 2 games with the same expected outcome, a risk-adverse person will prefer a lower variance, and a risk-loving person will prefer a higher variance. It is not immediately obvious that a person must always want to choose a game with lower variance. Even in the financial sector, not everyone picks the option with lower variance due to lack of information (or can't be bothered to figure out variance), agency problem, focus on kurtosis, etc.

Your statement "larger chance of seeing a net gain or net loss" is not true, and doesn't follow from simply knowing about the variance. \(P(X \neq 0 ) \) is not influence by the variance. It is easily to construct games with a higher variance but a low probability of change in outcome (e.g. the game that with probability \( \frac{1}{N} \), pays out \(M\), with probability \( \frac {1}{N} \) pays out \(- M\) and with probability \( 1 - \frac {2}{N} \) pays out \(0\). has a high variance as M increases, but a low chance of seeing a net gain or net loss, especially as we increase \(N\).)

Your statement on 'ruin probability' is also not necessarily true. Variance is independent of direction of movement. As a counter example, consider the 10-envelope game where 7 of them give you 1, 2 of them give you 10 and 1 of them gives you 73. Then, this has a higher variance then your 5-envelope game (with \(X=46\)) (453.6 against 324). However, the probability of ruin in bounded below by the probability of ruin in the 5-envelope game.

It is a peeve of mine that people have overly simplified all risk to just a few numbers (expected value and variance), and they do not understand the actual / valid implications, or even the real implications. Another example of a misleading statistic is the "Value at Risk" measure, which failed to account for compounding events, leading to the current financial crisis. Calvin Lin Staff · 3 years, 9 months ago

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@Hero P. Thank you, with the help of your explanation I have solved the Expected Raffle Value problem of this week. Joseph Gomes · 3 years, 9 months ago

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I solved the Expected Raffles Value problem using this method. Soham Dibyachintan · 3 years, 9 months ago

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