Author Topic: Probability of the Price is Right  (Read 44453 times)

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Offline Cyclone

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Re: Probability of the Price is Right
« Reply #165 on: January 03, 2012, 08:45:53 PM »
No offense, but this chart looks so confusing it might be useless.  What's the point of calculating odds of beating two players for an optimal wheel strategy?  After all, if you're last, you either beat it or go home.

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Offline JokerFan

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Re: Probability of the Price is Right
« Reply #166 on: January 03, 2012, 09:24:08 PM »
I might have missed something, but I don't think it's wheel strategy that's being calculated.  I thought it means that if you spin a 10, then you have a 1.99% chance of beating your previous two spinners if you spin again.

Offline PriceFanArmadillo

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Re: Probability of the Price is Right
« Reply #167 on: January 03, 2012, 09:55:56 PM »
Army, hasn't the math already been taken to this and borne out different answers?

He's ignoring the fact that it's possible to spin again, stay under $1.00, and still lose.  And several other things, I'm sure.

Now that I'm thinking about it, I'm going to try to put together a spreadsheet with the actual math...give me a while.
Armadillo is exactly right - ClockGameJohn
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Offline PriceFanArmadillo

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Re: Probability of the Price is Right
« Reply #168 on: January 03, 2012, 10:20:41 PM »
Here's what I came up with...



I'm not entirely sure I'm handling ties correctly, and I completely ignored the possibility of a three-way tie, and I'm ignoring second-player strategy, and probably a few other things, but this is what I've got.
« Last Edit: January 03, 2012, 10:23:00 PM by PriceFanArmadillo »
Armadillo is exactly right - ClockGameJohn
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Offline Cyclone

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Re: Probability of the Price is Right
« Reply #169 on: January 04, 2012, 12:21:49 AM »
So you have a 0.13% chance of winning with 5 cents?  You can only tie with that, I believe...so there is still a formula flaw.

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Offline JokerFan

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Re: Probability of the Price is Right
« Reply #170 on: January 04, 2012, 12:26:02 AM »
There are 2 ways (although the first might not be entirely correct given the formula)
1) You and someone else tie at 5 cents.  The two of you have a spin-off and you get the higher number.
2) The first two spinners go over $1.  You, the third and final spinner, spin 5 cents.

Offline Cyclone

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Re: Probability of the Price is Right
« Reply #171 on: January 04, 2012, 12:47:43 AM »
There are 2 ways (although the first might not be entirely correct given the formula)
1) You and someone else tie at 5 cents.  The two of you have a spin-off and you get the higher number.
2) The first two spinners go over $1.  You, the third and final spinner, spin 5 cents.
I later realized the second of those.  I'm not sure tying and going to a spinoff factors into it, however.

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Online SteveGavazzi

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Re: Probability of the Price is Right
« Reply #172 on: January 04, 2012, 02:22:26 AM »
There's also the possibility of an incredibly stupid third spinner who beats you, spins again to try for a dollar, and goes over.
"Every game is somebody's favorite." -- Wise words from Roger Dobkowitz.

Offline redhawkhockeyfan

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Re: Probability of the Price is Right
« Reply #173 on: January 04, 2012, 03:03:17 AM »
There's also the possibility of an incredibly stupid third spinner who beats you, spins again to try for a dollar, and goes over.

Please tell me this has never happened before.  I'm not as familiar with the show's history as most here, but this would be insane.  It almost seems like Bob would not have allowed it to happen.  Would be painfully awkward tv.

I assume this was just proposed as a hypothetical scenario, but it still piqued my interest late at night!

Offline themountainclimber

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Re: Probability of the Price is Right
« Reply #174 on: January 04, 2012, 09:26:52 AM »
Please tell me this has never happened before.  I'm not as familiar with the show's history as most here, but this would be insane.  It almost seems like Bob would not have allowed it to happen.  Would be painfully awkward tv.
Unfortunately, it has. And he went over too.
I'll see if I can find the video.

Offline PriceFanArmadillo

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Re: Probability of the Price is Right
« Reply #175 on: January 04, 2012, 10:01:41 AM »
So you have a 0.13% chance of winning with 5 cents?  You can only tie with that, I believe...so there is still a formula flaw.

Cyclone

I valued a tie as half a win, since you win half of all spinoffs (note my earlier assumption ignoring three-way ties).

That said...I still made some mistakes calculating.  In the scenario that you have a nickel as the second spinner and stay, ignoring any formula and reasoning it out, you should win .25%, not .13% (.125%, really, but the spreadsheet rounds)...the only way for you to win is to have the third spinner also get a nickel, spin again (since a spinoff is foolish here), and get the dollar.  The probability of this is .05 * .05 = .0025.  There is no possibility of a tie from a rational opponent.

I'll try again tonight now that I have an idea of a better approach.  Or maybe not, since other people have already done the math better than I and anyone can find it with a simple Google search.
« Last Edit: January 04, 2012, 10:05:25 AM by PriceFanArmadillo »
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Offline Cyclone

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Re: Probability of the Price is Right
« Reply #176 on: January 04, 2012, 05:06:30 PM »
Unfortunately, it has. And he went over too.
I'll see if I can find the video.
?What was the first spin - 10 cents? :D

Would love to see the vid.

Cyclone

Offline bduddy

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Re: Probability of the Price is Right
« Reply #177 on: January 05, 2012, 02:16:12 AM »
I have done far too much work and convinced myself at least that, for the second spinner, the breakpoint is between 50 and 55 cents; 50 and below he should spin again, 55 and above he should stand (although it is very close at 55). This is, of course, assuming he beats (not ties) the first spinner. I'd love to show my work, but it's on a sheet and a half of graph paper...

Now for the first spinner, and the "tie" scenario... lots more work...

Offline Flerbert419

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Re: Probability of the Price is Right
« Reply #178 on: January 05, 2012, 02:46:34 AM »
In this thread, some of Pay The Rent's analysis was examined, but I'm moving over here since I think it has more in common with this thread.

With this post, I showed that it was impossible to only have one correct solution with the least expensive item in the mailbox.  The question I then posed was: What is the minimum number of possible solutions (out of 180) you would need to allow in order to have at least one solution with the least expensive item in the mailbox?  Based on the previous post, this must be greater than or equal to two.

Let's examine a Pay The Rent setup with the following prices for the prizes: $21, $16, $11, $6, $4, and $3.  Although these numbers were chosen for a particular reason, it will hopefully become clear as the analysis is discussed.  We are going to start with the attic, and work our way down to the mailbox.

Clearly, the $21 prize has to go in the attic, as all possible winning solutions have the most expensive prize in the attic.  For the second floor, note that when the $16 prize is combined with any other, the total is more than the combination of any other two prizes.  For this reason, the $16 prize must belong on the second floor so that, when combined with one other prize, it is more expensive than the first floor.  Since the $21 prize is in the attic, the prize that it must be combined with must be less than $5.

There are two possible choices: the $4 prize and the $3 prize.  This is important because putting the $4 prize in this slot will allow the $3 to drop down to the mailbox.  If the $16 prize was increased to, say, $17.50, it would force the $3 prize to the second floor and not be free for the mailbox.  Therefore, two different choices for this slot is the least amount possible.

Placing the $4 prize on the second floor leaves us with $11, $6, and $3 for the first floor and mailbox.  So $21//$16 + $4//$11 + $6//$3 is a winning combination, as well as $21//$16 + $4//$11 + $3//$6.  Luckily, $21//$16 + $4//$6 + $3//$11 is not a solution, and so we have the minimum possible number of solutions here based on my previous work.

However, we also need to consider that the $3 prize could be placed on the second floor, leaving us with $11, $6, and $4 for the first floor and mailbox. So $21//$16 + $3//$11 + $6//$4 is a winning combination, as well as $21//$16 + $3//$11 + $4//$6.  Luckily, $21//$16 + $3//$6 + $4//$11 is not a solution, and so we have the minimum possible number of solutions here as well based on my previous work.

In conclusion, placing the $3 prize on the second floor yields two possible solutions, and placing the $4 prize on the second floor yields two different possible solutions (including the one with the cheapest item in the mailbox).  Therefore, we would need to allow four solutions (out of 180) in order to have at least one solution with the least expensive item in the mailbox.

Feel free to tear apart the above as you see fit.
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Offline bduddy

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Re: Probability of the Price is Right
« Reply #179 on: January 05, 2012, 02:55:44 AM »
Well, this part didn't need much more work at all...

As the second spinner, if you tie on 65 or less, spin again; tie on 70 or above, stop.