Roll the Die: Win \$10 or Lose \$2?

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In summary, the conversation discusses a game where paying \$2 gives the player a chance to win \$10 by rolling a one on a balanced die. The two possibilities for the game are winning \$10 or losing \$2, with a probability of 1/6 for rolling a one. The expected value of the game is discussed and the question is posed whether one would pay \$2 to play.
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lilly2
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someone help me understand these questions please:
1) Would you pay \$ 2 to play this game? if you throw a one on a balanced die, you receive \$10, otherwise you lose you \$2.
a) what are the two possibilities for this game?
b) what are the corresponding probabilities associated with these possibilites? and would you pay 2 dollars to play this game?
 
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lilly said:
someone help me understand these questions please:
1) Would you pay \$ 2 to play this game? if you throw a one on a balanced die, you receive \$10, otherwise you lose you \$2.
a) what are the two possibilities for this game?
b) what are the corresponding probabilities associated with these possibilites? and would you pay 2 dollars to play this game?

Welcome to MHB lilly! :)

The two possibilities are that either you win \$10 or you lose \$2.

A balanced die is a die with 6 sides that each have the same probability of coming up.
So the probability on a one is 1/6.

How much do you think you would earn if you play this game?
This is called "expected value".
To find it you need to multiply the probabilities with the outcomes and add them up.
What do you think it is?
 

Related to Roll the Die: Win \$10 or Lose \$2?

1. How does the game "Roll the Die: Win $10 or Lose $2" work?

The game involves rolling a standard six-sided die. If the player rolls a 1 or 2, they lose $2. If they roll a 3, 4, 5, or 6, they win $10. The player can choose to play multiple times in a row or stop after each roll.

2. What are the odds of winning in "Roll the Die: Win $10 or Lose $2"?

The odds of winning in this game are 4 out of 6, or 66.67%. This means that for every six rolls, the player is expected to win four times and lose twice.

3. Is there a strategy that can increase the chances of winning in "Roll the Die: Win $10 or Lose $2"?

No, this game is based purely on chance and there is no strategy that can increase the chances of winning. Each roll of the die is independent and the outcome cannot be predicted.

4. Can the game "Roll the Die: Win $10 or Lose $2" be modified to increase the chances of winning?

Yes, the game can be modified to increase the chances of winning by changing the payout or altering the rules. For example, if the player wins $10 for rolling a 1 or 2 and loses $2 for rolling a 3, the odds of winning would increase to 50%.

5. How is the game "Roll the Die: Win $10 or Lose $2" related to probability and expected value?

This game is related to probability because it involves calculating the likelihood of certain outcomes occurring. The expected value in this game is calculated by multiplying the probabilities of winning and losing by the corresponding payouts and then subtracting the losing outcomes from the winning outcomes. In this case, the expected value is ($10 x 4/6) - ($2 x 2/6) = $2.67, meaning that on average, the player can expect to win $2.67 per roll.

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