MHB Expected Value of Gambling Game: Solve It Now!

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The discussion focuses on calculating the expected value of a gambling game involving rolling a die and drawing a card from a deck. Participants note the probabilities: a 1 in 6 chance of rolling a 3 and a 1 in 4 chance of drawing a heart. The expected value can be computed using the formula E[X]=100 * P[X=3 and a heart], assuming no buy-in for the game. Clarification is sought on whether the expected profit or just the expected value is needed, but the main calculation hinges on determining the combined probability of both successful outcomes. Understanding these probabilities is essential to solving the problem correctly.
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Here is the question: You are playing a gambling game (silly, yea I know). The first part of the game is to throw a die. If it comes up a 3, you move on. Otherwise, you lose. The second part of the game entails pulling a card out of a standard deck. If it is a heart, you win $100. Otherwise, you lose. What is the expected value of the game?

Here is what I know: you have a 1 out of 6 chance of rolling a 3 and a 13 out of 52 (i hope) chance of pulling a heart.

I don't know how to get the answer that is needed. Can anyone help me?
 
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Re: help, I am confused

Are you supposed to find the expected profit? If so, what is the buy-in to the game?

edit: I have edited the topic title to reflect the nature of the problem.
 
MarkFL,
I don't know. that is the question word for word that I have on my paper from my Professor. That is what is confusing, it just says "What is the expected value of the game?"
 
Are you given an answer that you are expected to be able to compute?
 
$$E[X]=\sum_{x} x \cdot P[X=x]$$
That seems complicated but it's not so bad. Assuming there is no buy-in to play the game and that the only payout occurs when you get a 3 and a heart, the above equation simplifies to:

$$E[X]=100 \cdot P[X= \text{3 and a heart}]$$

Now you just need to figure out that probability of getting a 3 and a heart and you'll be done. :)
 

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