Statistics and "Deal Or No Deal"
Mar 28, 2006 at 1:27 AM Thread Starter Post #1 of 14

bahamaman

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I was/am watching this inane TV show this evening and a question arose. In the game show, contestants select 1 briefcase out of 26, with each briefcase containing various amounts of money from one cent to one million dollars.

Assume that a contestant selects briefcase number 7 (but, of course, he/she isn't allowed to see how much is in the briefcase). Assume further that, in the course of the game, the contestant has opened up 13 briefcases, none of which contains the million dollar prize.

At this point in the game, would the contestant's odds of winning $1,000,000 be 1 in 26 or would it be 1 in 13? I am tempted to say 1 in 13. But, in another sense, it seems that her odds of winning shouldn't change as new information is revealed.

I nearly flunked stat, so I will apologize in advance if the answer is obvious.
 
Mar 28, 2006 at 1:39 AM Post #2 of 14

Squeek

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I've never taken stats, but logic dictates that her odds improve as more cases are opened.

What started out as 1 in 27 becomes 1 in 13 if there are only 13 cases left.

While her original odds never changed, meaning that when she first stood up her odds were 1 in 27, the odds continue to improve as you go down the line if the $1mm case is not opened.

Perhaps I'm way off base, but that makes sense to me!
 
Mar 28, 2006 at 1:39 AM Post #3 of 14

Oski

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Let's take it further. Say that 24 of the briefcases have been opened, and 2 remain. The person now has quite obviously a 50% chance. So going back to your example above after 13 of the 26 briefcases have been opened, there should be a 1 in 13 chance with the remaining briefcases.
 
Mar 28, 2006 at 1:44 AM Post #4 of 14

bahamaman

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Is there no argument (not even a philosophical one) that the contestant's odds remain constant throughout the process and that "knowledge" cannot have an effect on those odds?
 
Mar 28, 2006 at 2:10 AM Post #7 of 14

IstariAsuka

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Quote:

Originally Posted by bahamaman
Ouch, my head hurts! I know now why I do so poorly at math. Care to provide a simple explanation?


Basically, when you initially choose the door, there is a 2/3rds chance that you choose a goat.

Given this situation, as soon as the announcer eliminates one of the other 2 doors as the goat, you're better off switching, because it is twice as likely that you initially chose one of the two initial goats than it is you chose the car, and thus win by switching. On the other hand, there is only a 1/3 chance that you initially chose the car, and so win by not switching.
 
Mar 28, 2006 at 2:26 AM Post #8 of 14

K2Grey

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The best way to illustrate the Monty's Haul problem (IMHO) is to take it to an extreme.

Let us say there are 1000 doors. You pick a door and then 998 doors are revealed. Is the chance of the remaining door being the one 1/1000? No, the chance of the remaining door being correct is 999/1000 - you can treat it as a choice between 1 door, or another door which "represents" 999 doors.
 
Mar 28, 2006 at 2:27 AM Post #9 of 14

Unclewai

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Quote:

Originally Posted by TMHBAT
http://math.ucsd.edu/~crypto/Monty/monty.html


It's different for the Monty Hall problem. In Deal or No Deal, the contestant chooses any of remaining suitcases that can contain the million dolalr prize. For the Monty Hall question, Monty Hall always shows the dud.

1/26 is your prior probability. As you open more cases, you get more information, hence your conditional probability changes.
 
Mar 28, 2006 at 3:24 AM Post #10 of 14

plus_c

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Quote:

Originally Posted by bahamaman
Is there no argument (not even a philosophical one) that the contestant's odds remain constant throughout the process and that "knowledge" cannot have an effect on those odds?


The only argument I could see would be what I would call the "tinfoil hat" argument - that the show producers are moving the prize around.

The Monty analogy is not quite the correct one here. Rather, you should be thinking about pulling marbles out of a bag without putting them back when you pull them. You start with 26 marbles, or briefcases in this case. One of them represents $1 million. At the beginning, your probability is 1 in 26. As you pull marbles, none of which are the winners, the pool of marbles that is left decreases. So on your second pull, the chance of hitting the $1 million marble is 1 in 25, and so on. As you choose briefcases and find out that they don't contain the prize, you are eliminating that briefcase from the pool of possible briefcases. So as you make your 13th pick, your chance is 1 in 13.
 
Mar 28, 2006 at 3:34 AM Post #11 of 14

dan1son

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As they said. In the case of Deal or no Deal the announcer, Howie Mandel, doesn't pick the case knowing that it does not hold the large amount of money, so it would be like the marbles. When the contestant picks them the odds will most definitely change as you go along.

However another thing with deal or no deal is that they give you a deal based on what is left... so the amount of money you can win varies depending on what cases have been chosen and how much total amount of money remains. I'm sure with little work anyone with sufficient math knowledge could figure out their algorithm for choosing the amount to offer.
 
Mar 28, 2006 at 4:09 AM Post #13 of 14

PhilS

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Quote:

Originally Posted by dan1son
However another thing with deal or no deal is that they give you a deal based on what is left... so the amount of money you can win varies depending on what cases have been chosen and how much total amount of money remains. I'm sure with little work anyone with sufficient math knowledge could figure out their algorithm for choosing the amount to offer.


What's kind of interesting, or amusing, if you watch the show, is that they seem to believe (perhaps with good reason) that every contestant is risk averse at every amount of money. Thus, they always seem to offer as the "deal" an amount less than the expected value of the cases left. Thus, if there were only two cases left, and one is a $500,000 case and the other is a $.01 case, it would seem the expected value is $250,000, give or take a penny, yet the banker typically offers less than that. If you were one of those Texas Hold'em poker players who bases every move on the pot odds, you would basically never take the "deal." Of course, people do, because they are in fact risk averse, especially when large amounts of money are involved. So they would rather have $200,000 for certain if the other choice is a 50% chance of $500,000. Anyway, it's sort of interesting to study human nature and see how people figure out what to do. Although some just seem to go with whatever the crowd tells them to do.
 
Mar 28, 2006 at 4:12 AM Post #14 of 14

PhilS

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Quote:

Originally Posted by Unclewai
I didn't really do the calculation, but I thought the banker's offer is the expected value of the brief case?


I thought that too at first, but if you watch and calculate, you can see pretty quickly that it's not. And it's because the banker (the producers) are smarter than that. They realize they don't need to offer the expected value.
 

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