What is the expected outcome of a pick 3 lotto if a ticket costs $\$2$?

[aprilryan]

Hey all,

I'm so grateful I found this place! I would like some assistance with a pesky probability word problem. The problem is:

"A pick 3 lotto has a $\$1000$ prize. What is the expected outcome if a ticket costs $\$2$? (The probability of winning is .001)".

So far I got this:

x P(x)

1000 .001
-2 .999

I don't know where to go from here. Do i have to multiply or add?

Any help is appreciated. Thanks.

 

[Ackbach]

Re: Can someone help me with a probability word problem?

You're nearly there! A couple of thoughts:

1. Even if you win, you still paid two dollars. So the net income wouldn't be $\$1000$, would it?
2. Once you have the correct outcomes and probabilities, you would multiply the outcomes times their probabilities, and then add the results to get the expected value.

 

[aprilryan]

Thanks for your assistance.

Is this right?

I multiply 1000 times .001 and .999 times -2

Then after I multiply the two numbers, do I add -.998 and 1?

 

[Siron]

Almost correct! As Ackbach pointed out, your ticket still costs $2 \$ $. Therefore, if you win you still woud not have $1000 \$ $ but $? \$$, since you had to buy the ticket in the first place. Why do you want tot add $-0.998$ and $1$?

To think about: when you have the expected outcome you should always check if it is a realistic answer or not. In this case, the probability to win is very small and hence you expect a fairly low outcome. If your answer would be something like $900 \$ $ then you know there is something wrong in your calculations. However, if the question would have been: the probability to win is $1000 \$ $ is $10 \%$ then your expected outcome will be much higher etc... .

 

[aprilryan]

Thanks Ackbach and Siron. I'll try to figure this one out again. I'll let you know how it goes!

 

[Jameson]

Thanks Ackbach and Siron. I'll try to figure this one out again. I'll let you know how it goes!

Any update?

The general answer would look like:
$$\text{Expected value}=P(\text{win}) \cdot (\text{prize})+P(\text{loss}) \cdot (\text{cost to play}) $$

 

[aprilryan]

Do I multiply .001 times 1000? Sorry can someone break it down step by step? I still don't get it.

 

[aprilryan]

Ok I think I got it. Took a while to process everything.

This would be the setup, correct? .001 * 1000 + 1 * 2