Deal or No Deal - Odds of a better deal and most likely deal?

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In "Deal or No Deal," the average value of the remaining cases is $12,003.002, with five cases left. To determine the odds of receiving a better offer in the next round, one must calculate the number of cases with values below the median, which is $50. The odds of a better deal are essentially 50-50, as the average reflects expected outcomes over many games. Players should focus on their personal goals rather than just the expected earnings, as individual needs can influence the decision to accept or reject offers. Ultimately, the value assigned to potential winnings should align with personal financial goals, such as purchasing a house.
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Deal or No Deal -- Odds of a better deal and most likely deal?

Homework Statement


Suppose I am playing "Deal or No Deal." Assume that the banker's offers always equal the average value of the remaining cases. I have 5 cases remaining. The values of the cases are as follows:

$.01
$5
$10
$50,000
$100,000

thus resulting in an average of $12,003.002. If played another round, I'd have to choose 4 cases. What are the odds of getting a better offer on the next round? What offer would I be most likely to receive?

2. The attempt at a solution

First, I would figure out how many cases had values that were below the median (which is $50, in this case). Call that number x. Then the odds of a better deal on the next round equals x / n! * 4, where n is the number of remaining cases.

As for my other question, I don't know the answer.
 
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moonman239 said:

Homework Statement


Suppose I am playing "Deal or No Deal." Assume that the banker's offers always equal the average value of the remaining cases. I have 5 cases remaining. The values of the cases are as follows:

$.01
$5
$10
$50,000
$100,000

thus resulting in an average of $12,003.002. If played another round, I'd have to choose 4 cases. What are the odds of getting a better offer on the next round? What offer would I be most likely to receive?

2. The attempt at a solution

First, I would figure out how many cases had values that were below the median (which is $50, in this case). Call that number x. Then the odds of a better deal on the next round equals x / n! * 4, where n is the number of remaining cases.

As for my other question, I don't know the answer.

Well, the average is also the expectation.
That means that if you play a zillion games, you will on average earn $12,003.002.
In other words, to optimize your earnings it doesn't matter what you do.

What are the odds to receive a better offer?
Well, that would be 50-50, because that is what an expectation means.

If you want to play an interesting game, you shouldn't worry about your expected earnings.
You should worry about what you want to earn.

So if you would need for instance $20,000 to buy a house, you can play for that.
You wouldn't care whether you would $5 or $10 or zip, because those are all equally worthless.
And also, you wouldn't accept the offer, because it simply isn't enough to buy your house.
So you'd bet on the less good chance to earn what you really want.

If, after the first round one of the worthless prices is taken away, you'll get an offer of about $38,000 and you should take that right away, because it pays for your house!

You shouldn't go for the $100,000, because you don't need it to pay for your house.

Concluding, I'd say that you would assign different values to the prices than the actual monetary value, because of what you can or cannot do with it.
 
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