Expected values probability help

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SUMMARY

The discussion focuses on calculating the expected value of a game involving a fair coin tossed three times, where winnings are determined by the number of tails. The player wins $26 for three tails, $13 for two tails, and loses $26 for no tails, while one tail results in no winnings. The correct expected value calculation incorporates the probability of each outcome, specifically addressing the mistake in the probability of getting two heads and one tail, which is not 2/8 but rather 3/8. The final expected value formula is correctly represented as ($26*3/8) + ($13*3/8) - ($26*1/8) + 0*(1/8).

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  • Understanding of probability theory, specifically binomial distributions.
  • Familiarity with expected value calculations in gambling scenarios.
  • Basic knowledge of combinatorial outcomes from multiple coin tosses.
  • Ability to interpret and manipulate algebraic expressions.
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  • Study the binomial probability formula to calculate outcomes from multiple trials.
  • Learn how to derive expected values in various gambling games.
  • Explore combinatorial mathematics to understand the arrangement of outcomes in probability.
  • Practice problems involving expected value calculations with different scenarios and rules.
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Students studying probability, educators teaching statistics, and anyone interested in understanding expected value in games of chance.

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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