Expected values probability help

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Homework Help Overview

The problem involves calculating the expected value of a game based on the outcomes of tossing a fair coin three times, with specific winnings and losses associated with the number of tails that occur.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to incorporate the scenario where one tail occurs, questioning how to set up the expected value calculation correctly. There are attempts to clarify the probabilities associated with different outcomes.

Discussion Status

Some participants are exploring the correct setup for the expected value calculation, while others are pointing out errors in the probability assignments. Guidance has been offered regarding the inclusion of zero winnings for the case of one tail occurring, and there is acknowledgment of mistakes in probability calculations.

Contextual Notes

There is an emphasis on ensuring that the probabilities for all possible outcomes add up correctly, as well as a focus on understanding the implications of each outcome in the context of the game.

mtingt
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A fair coin is tossed 3 times, and a player wins $26 if 3 tails occur, wins $13 if 2 tails occur, and loses $26 if no tails occur. If 1 tail occurs, no one wins. what is the expected value?


i don't really understand what does the "if 1 tail occurs , no one wins. how do you set up that?

i only set up up to ( $26*3/8) + ($13* 2/8) -(26* 4/8) how do i set up the if 1 tail occurs?
 
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mtingt said:
i don't really understand what does the "if 1 tail occurs , no one wins. how do you set up that?

Use $0 for the winnings.

i only set up up to ( $26*3/8) + ($13* 2/8) -(26* 4/8) how do i set up the if 1 tail occurs?[/

Your probabilities add up to more than 1, so they are wrong.
 
if i use 0 for the winnings that 1 tail occurs that just means i add 0 to the end of my equation?
($26*3/8) + ($13* 2/8) -(26* 4/8) + 0 (1/8)

isnt that still the same?
so what is wrong with my equation?
 
mtingt said:
if i use 0 for the winnings that 1 tail occurs that just means i add 0 to the end of my equation?

($26*3/8) + ($13* 2/8) -(26* 4/8) + 0 (1/8)
isnt that still the same?
Yes. The zero just shows a grader than you know what you are doing.

so what is wrong with my equation?

You have computed the probabilities incorrectly. For example, the probability of getting 2 heads in 3 tosses isn't 2/8. Of the 8 possible results of tossing 3 coins, there are 3 possible ways that one can get 2 heads and 1 tail.
 
Stephen Tashi said:
You have computed the probabilities incorrectly. For example, the probability of getting 2 heads in 3 tosses isn't 2/8. Of the 8 possible results of tossing 3 coins, there are 3 possible ways that one can get 2 heads and 1 tail.

oh wow, i can't believe i made such a stupid mistake, thank you for helping me out
 

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