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Author Topic: Probability question for D or No D  (Read 5000 times)

Gromit

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Probability question for D or No D
« on: December 21, 2005, 12:52:33 AM »
Ok, had a discussion tonight about the first night of the show. Karen had 3 low amounts left, and one big one ($75, $500, $50,000, $500,000). She has an offer from the Banker of $138K.

So, the question is, just what exactly are the odds of her improving her situation if she continues forward one more time? On the surface, it appears to be 75%, if she can remove anything other than the 500K, her offer will go up.

*But*, in reality she's only choosing from 3 cases, not four. One is hers already and set aside, and that one is out of the equation so to speak. So that leads us to this, she is in one of these situations:

Case 1: She holds the $500K case. Only low cases are left on stage to choose from, she has a 100% chance of improving her situation (3/3).
Case 2: She holds the $75 case. She has a 2/3 chance of improving.
Case 3: She holds the $500 case. She has a 2/3 chance of improving.
Case 4: She holds the $50K case. She has a 2/3 chance of improving.

She has a 75% chance of being in a situation where she has a 66% chance of improving, and a 25% chance of being in a 100% situation.

So, what are the exact odds for her improving? Somewhere between 66% and 75%?

Matt Ottinger

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Probability question for D or No D
« Reply #1 on: December 21, 2005, 12:57:18 AM »
Oh, if only we had a professional math teacher with an understanding of game show theories and strategies on this board. Even one who doesn't post much.
This has been another installment of Matt Ottinger's Masters of the Obvious.
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Gromit

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Probability question for D or No D
« Reply #2 on: December 21, 2005, 12:59:39 AM »
Yeah, that would be useful.  I have a feeling I may be overthinking it.

Speedy G

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Probability question for D or No D
« Reply #3 on: December 21, 2005, 01:26:34 AM »
You don't need a math professor to see how the math works out.  (Will someone with a math minor work?  Even though it is basic probability...)

Quote
She has a 75% chance of being in a situation where she has a 66% chance of improving, and a 25% chance of being in a 100% situation.
Thus, her overall odds of improving are:
(3/4)*(2/3) + (1/4)*(1)
(2/4)+(1/4)
= 3/4 = 75%

Same odds, longer way of expressing the situation.
Solar-powered flashlight, hour 4 of the Today show, the Purple Parrots.  *rips open envelope, blows into it*

Unrealtor

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Probability question for D or No D
« Reply #4 on: December 21, 2005, 01:47:38 AM »
I'm not a teacher, but I'm one course shy of getting a math degree. The short answer is that it doesn't matter whether you worry about what the contestant has.

Let k = the number of cases more than the mean value, and n = the number of cases left unrevealed (including the contestant's case).

Without considering what the contestant has:

You have n-k cases smaller than the mean distributed over n cases in total, so you have a (n-k)/n chance of eliminating a case with a lower value than a mean. (This is also the probability that any of the remaining cases is lower than the mean.)

(n-k)/n = (n/n)-(k/n) = 1-(k/n) (this becomes important later)

When you do consider what the contestant has:

You have an (n-k)/n chance of having a case that's below the mean. If you do, there are n-k-1 cases left below the mean to choose from out of n-1, so your probability of eliminating one is (n-k-1)/(n-1).

You have a k/n chance of having a case that's above the mean. If you do, there are n-k small cases to choose from out of n-1.

So, putting all of this together:

n-k   n-k-1   k   n-k
--- x ----- + - x ---
  n     n-1    n   n-1

n²-2nk-n+k²+k   kn-k²
------------- + ----
    n²-n         n²-n

n²-n+k-nk
---------
  n²-n

n²-n   k-nk
---- + ----
n²-n   n²-n

     k(n-1)
1 - ------
     n(n-1)

1 - k/n

(In case it shows up differently on your system, ² is supposed to be a superscript 2, or "squared.")

Both of the probabilities work out to 1-(k/n), regardless of whether we consider the box or not.
« Last Edit: December 21, 2005, 01:48:59 AM by Unrealtor »
"It's for £50,000. If you want to, you may remove your trousers."

Gromit

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Probability question for D or No D
« Reply #5 on: December 22, 2005, 02:52:10 AM »
Thanks guys. Guess I have to apologise to the wife.

Nah.

davidhammett

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Probability question for D or No D
« Reply #6 on: December 22, 2005, 10:49:07 AM »
[quote name=\'Matt Ottinger\' date=\'Dec 21 2005, 12:57 AM\']Oh, if only we had a professional math teacher with an understanding of game show theories and strategies on this board. Even one who doesn't post much.
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I thought I heard my phone ringing, but decided it was probably that pesky banker...  :-)

The basic question has already been answered; if there are four cases left (three on stage, one for the contestant), then all probabilities should be based out of four suitcases.  Thus, if there are 3 out of 4 that will improve the contestant's situation, the odds are with him or her to continue on in the game.

But that's where this game gets tricky.  First, there is the concept of the contestant's mathematical expectation in the game.  The expectation is computed by multiplying all possible probabilities and corresponding payouts and then summing them.  For example, if someone has four cases left with amounts of $75, $500, $50,000, $500,000, then the contestant's expectation is (1/4)(75)+(1/4)(500)+(1/4)(50000)+(1/4)(500000) = 137,643.75.  Since the bank offer basically equaled that expectation (rounded to the nearest thousand), an argument can be made for the contestant taking it.

Then again, that's not totally correct, because the truer expectation should be based not on what's in the player's suitcase, but rather how much higher the deal could potentially go.  That's difficult to predict, of course, since the bank does not always offer something close to the exact average of the remaining amounts... but let's assume it does.  Then, if the contestant chooses one more suitcase, there's a 1/4 chance of uncovering each of the amounts; the resulting deals would end up approximately as follows:

If $75 is chosen, the deal will be about 550500/3 = $183,000
If $500 is chosen, the deal will be about 550075/3 = $183,000
If $50000 is chosen, the deal will be about 500575/3 = $167,000
If $500000 is chosen, the deal will be about 50575/3 = $17,000

Thus, the expectation if the contestant chooses one more case is:
(1/2)(183000)+(1/4)(167000)+(1/4)(17000) = $137,500.  In fact, if you work this scenario out in general, the expectation will always be the same as it is if you just consider the individual cases as I did originally above.  However, this assumes that the bank will offer something close to the average.  Still, I (and some others of us) have noticed that the bank offers do tend to approach the true mean as more and more suitcases are removed, thus confirming that the producers want the contestant to stay in the game for as long as possible (since the deals do tend to get better from a percentage sense).

But perhaps the most important consideration of all has nothing to do with complicated mathematics; it has to do with the raw emotion of it all... namely, if you're standing there with essentially $138,000 to call your own, how hard does that become to turn down?  In that case, other factors work their way into the expectation; it may not just be a case of how much money you stand to lose, but what sort of an effect that might have on you.  Of course, the same can be said about what sort of an effect winning more money would have as well, but the point is that suddenly any sort of expectation calculation takes on a whole new perspective; it's not just about the numbers any more.

clemon79

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Probability question for D or No D
« Reply #7 on: December 22, 2005, 11:47:41 AM »
[quote name=\'davidhammett\' date=\'Dec 22 2005, 07:49 AM\']In that case, other factors work their way into the expectation; it may not just be a case of how much money you stand to lose, but what sort of an effect that might have on you.  Of course, the same can be said about what sort of an effect winning more money would have as well, but the point is that suddenly any sort of expectation calculation takes on a whole new perspective; it's not just about the numbers any more.
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And this is why the game as played "for funsies" sucks. At that point, it's all about maximizing score, and as David just explained in his usual excellent fashion, the math to figure out where things maximize isn't all that hard, and the way it's being presented in the US, it comes down to a coin flip, essentially, at the later stages, since they offer you the exact statistical middle-of-the-road.

Sit down with a buddy and play a game. Each of you get a quarter. Start flipping. The one who flips a tail first loses. Isn't that a gas?

The game is all about the emotion of WINNING REAL MONEY. Yeah, odds ARE that you won't knock out a case that has The Big One in it (and let's thank Michael O'Keefe for giving that a nifty catchphrase), but are you willing to turn down a significant amount of real money to find that out? If it's played for points or cyberdollars or whatever, who the hell cares? You shrug, and get 'em next time. But when it's money? That's interesting.
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