Difference between revisions of "Expected value"

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*Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.
 
*Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.
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==Problems==
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===Introductory===
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*{{problems}}
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([[Expected value|Source]])
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===Intermediate===
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*In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math>
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([[Mock_AIME_2_2006-2007/Problem_13|Source]])
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===Olympiad===
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*{{problems}}
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([[Expected value|Source]])
  
== Example Problems ==
 
 
* [[Mock_AIME_2_2006-2007/Problem_13 | Mock AIME 2 2006-2007 Problem 13]]
 
  
  
 
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[[Category:Definition]]
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[[Category:Combinatorics]]

Revision as of 13:50, 19 January 2008

This is an AoPSWiki Word of the Week for Jan 17-23

Given an event with a variety of different possible outcomes, the expected value is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is not the same as the "most likely outcome."

For example, flipping a fair coin has two possible outcomes, heads (denoted here by $H$) or tails ($T$). If we flip a fair coin repeatedly, we expect that we will get about the same number of heads as tails, or half as many as the total number of flips. Thus, the average outcome is $\frac 12 H + \frac 12 T$. Note that not only is this not the most likely outcome, it is not even a possible outcome for a single flip.


More formally, we can define expected value as follows: if we have an event $Z$ whose outcomes have a discrete probability distribution, the expected value $E(Z) = \sum_z P(z) \cdot z$ where the sum is over all outcomes $z$ and $P(z)$ is the probability of that particular outcome. If the event $Z$ has a continuous probability distribution, then $E(Z) = \int_z P(z)\cdot z\ dz$.

Uses

  • Expected value can be used to, for example, determine the price for playing a probability based carnival game. You first find the expected value per player. Then you can charge a reasonable price so that you gain money, but the price isn't unreasonably high.

Problems

Introductory

  • This page is in need of some relevant examples or practice problems. Help us out by adding some. Thanks.

(Source)

Intermediate

  • In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

(Source)

Olympiad

  • This page is in need of some relevant examples or practice problems. Help us out by adding some. Thanks.

(Source)


This article is a stub. Help us out by expanding it.