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Author Topic: Another Probability Problem  (Read 5470 times)

TLEberle

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Another Probability Problem
« on: February 21, 2005, 09:04:45 PM »
I'm trying to figure out what the odds are of winning the bonus round of "Shop 'til You Drop."  (Some people collect bottle caps, I mess around with my graphing calculator.  Go figure.)  I'm working based on the following assumptions: the team gets all six boxes, and one box will guarantee a victory, no matter what is inside the other five.

Using my trusted TI-86, I know that there are 3,003 different ways to pick six boxes out of a possible 14.  Unfortunately, that's about all I've got.  I think the next step is to figure out how many different ways you can combine that one box with five others, but I don't know how.  

If you can help me out, I sure would appreciate it.  Even a nudge in the right direction would be good.
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Desperado

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Another Probability Problem
« Reply #1 on: February 21, 2005, 09:17:57 PM »
The odds of my being interested in this thread:  0.
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zachhoran

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Another Probability Problem
« Reply #2 on: February 21, 2005, 09:38:20 PM »
[quote name=\'TLEberle\' date=\'Feb 21 2005, 09:04 PM\']I'm trying to figure out what the odds are of winning the bonus round of "Shop 'til You Drop."  (Some people collect bottle caps, I mess around with my graphing calculator.  Go figure.)  I'm working based on the following assumptions: the team gets all six boxes, and one box will guarantee a victory, no matter what is inside the other five.

Using my trusted TI-86, I know that there are 3,003 different ways to pick six boxes out of a possible 14.  Unfortunately, that's about all I've got.  I think the next step is to figure out how many different ways you can combine that one box with five others, but I don't know how.   


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13 choose 5, meaning 13*12*11*10*.......*2*1 divided by(8*7*6*5*4*3*2*1)(5*4*3*2*1). There would be 1287 ways to combine that guaranteed win box with 5 others. If one has to get the box with the prize over $700 to win, there would be a win about 1287 divided by 3003, or 43% of the time(rounding to the nearest whole percent). I've seen an occasional win without getting the $700+ prize box.
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parliboy

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Another Probability Problem
« Reply #3 on: February 21, 2005, 10:58:56 PM »
[quote name=\'TLEberle\' date=\'Feb 21 2005, 09:04 PM\']I'm trying to figure out what the odds are of winning the bonus round of "Shop 'til You Drop."  (Some people collect bottle caps, I mess around with my graphing calculator.  Go figure.)  I'm working based on the following assumptions: the team gets all six boxes, and one box will guarantee a victory, no matter what is inside the other five.

Using my trusted TI-86, I know that there are 3,003 different ways to pick six boxes out of a possible 14.  Unfortunately, that's about all I've got.  I think the next step is to figure out how many different ways you can combine that one box with five others, but I don't know how.   

If you can help me out, I sure would appreciate it.  Even a nudge in the right direction would be good.
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As with other probability questions we've seen here, you're overcomplicating this.  If you pick six boxes of 14, and one of them gives a win, the odds are 6 out of 14, yes?

While it's true that you can lose when getting the big box, and you can win without getting it, these events occur no more than once or twice a season, and tend to cancel each other out.  So they're statistically insignificant.
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jmangin

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Another Probability Problem
« Reply #4 on: February 22, 2005, 01:02:13 AM »
A better experiment would be figuring out the probability of this show continuing for another season with new episodes.
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whewfan

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Another Probability Problem
« Reply #5 on: February 22, 2005, 09:07:04 AM »
Considering the show has had various runs since 1990, I'd say there's a fair chance.

Does anyone know why the format and set change?

STYD was never that great of a show to begin with, but the original format did have some perks that the new format doesn't have

PRO- new format
No more Dee Baker- Dee's cross dressing and antics did get creepier, and more annoying as the series progressed. He even modeled the prizes in drag.

No more worrying about contestants falling down the spiral steps- Ok, maybe some of us were HOPING for a big fall down those steps.

AGAINST-
The stunts were pretty dumb, but at least they had more movement and were slightly more interesting than just standing behind the podium the whole time.

Some of the new games in the new format are kind of easy, and some are way too much like TPIR.

Unfortunately, this format doesn't offer much variety. Each show seems pretty much the same.

JD Roberto... should be JD Roboto. Need I say more?
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jmangin

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Another Probability Problem
« Reply #6 on: February 22, 2005, 09:48:13 AM »
[quote name=\'whewfan\' date=\'Feb 22 2005, 09:07 AM\']...should be JD Roboto.
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HEY!  That's Mister Roboto.
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MikeK

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Another Probability Problem
« Reply #7 on: February 22, 2005, 09:54:27 AM »
[quote name=\'zachhoran\' date=\'Feb 21 2005, 09:38 PM\']13 choose 5, meaning 13*12*11*10*.......*2*1 divided by(8*7*6*5*4*3*2*1)(5*4*3*2*1). There would be 1287 ways to combine that guaranteed win box with 5 others. If one has to get the box with the prize over $700 to win, there would be a win about 1287 divided by 3003, or 43% of the time(rounding to the nearest whole percent). I've seen an occasional win without getting the $700+ prize box.[/quote]
This isn't something said often but it is true in this matter--Zach's right.

If you're trying to find out how many combinations there are for selecting the other boxes given you have the $700+ box, it is indeed 13 choose 5.  The probability of getting the $700+ box is (13 choose 5) / (14 choose 6).

Since I haven't seen an episode of STYD for the better part of three years (and thank you WVPX for that blessing!), does getting the $700+ prize guarantee victory?

I had plans to do a presentation on the probabilities and mathematics of Plinko at a math conference in April.  Maybe I should do STYD math instead...:-)
« Last Edit: February 22, 2005, 10:07:54 AM by hmtriplecrown »
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Chelsea Thrasher

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Another Probability Problem
« Reply #8 on: February 22, 2005, 09:58:01 AM »
[quote name=\'jmangin\' date=\'Feb 22 2005, 08:48 AM\'][quote name=\'whewfan\' date=\'Feb 22 2005, 09:07 AM\']...should be JD Roboto.
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[/quote]

HEY!  That's Mister Roboto.
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[/quote]

Domo Arigato...
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zachhoran

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Another Probability Problem
« Reply #9 on: February 22, 2005, 10:37:31 AM »
[quote name=\'hmtriplecrown\' date=\'Feb 22 2005, 09:54 AM\']

Since I haven't seen an episode of STYD for the better part of three years (and thank you WVPX for that blessing!), does getting the $700+ prize guarantee victory?


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The shows I've seen with a win of recent(and I don't follow this show much, especially post-Lifetime) had the $700+ box picked. The probability would be altered a bit if they still did the "two stores have prizes worth over $500 bit" that was done in the Lifetime days with the $2500 rule(Stydfan can be useful and answer if he's listening)
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Dbacksfan12

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Another Probability Problem
« Reply #10 on: February 22, 2005, 04:10:17 PM »
[quote name=\'hmtriplecrown\' date=\'Feb 22 2005, 09:54 AM\']Since I haven't seen an episode of STYD for the better part of three years (and thank you WVPX for that blessing!), does getting the $700+ prize guarantee victory?
[/quote]
Not necessairly....although the probabilty of it happening is zilch; if the team elected to keep 2-3 of the prizes without exchanging; odds are probably against them.
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The odds of my being interested in this thread: 0.
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A better experiment would be figuring out the probability of this show continuing for another season with new episodes
The ignorance is amazing in this thread.  Someone actually poses an intellegent question; and crap like this gets churned out.
« Last Edit: February 22, 2005, 04:37:49 PM by Modor »
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clemon79

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Another Probability Problem
« Reply #11 on: February 22, 2005, 05:18:53 PM »
[quote name=\'Modor\' date=\'Feb 22 2005, 02:10 PM\']The ignorance is amazing in this thread.  Someone actually poses an intellegent question; and crap like this gets churned out.
[/quote]
Indeed. Churn, baby, churn.
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bclark71

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Another Probability Problem
« Reply #12 on: February 22, 2005, 05:38:27 PM »
[quote name=\'jmangin\' date=\'Feb 22 2005, 09:48 AM\'][quote name=\'whewfan\' date=\'Feb 22 2005, 09:07 AM\']...should be JD Roboto.
[snapback]75683[/snapback]
[/quote]

HEY!  That's Mister Roboto.
[snapback]75695[/snapback]
[/quote]

Domo arigato for straightening that out.  :)
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chris319

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Another Probability Problem
« Reply #13 on: February 22, 2005, 06:15:49 PM »
Quote
13 choose 5, meaning 13*12*11*10*.......*2*1 divided by(8*7*6*5*4*3*2*1)(5*4*3*2*1). There would be 1287 ways to combine that guaranteed win box with 5 others. If one has to get the box with the prize over $700 to win, there would be a win about 1287 divided by 3003, or 43% of the time(rounding to the nearest whole percent).

Quote
As with other probability questions we've seen here, you're overcomplicating this. If you pick six boxes of 14, and one of them gives a win, the odds are 6 out of 14, yes?

6 / 14 = 0.428571 or 42.8%
« Last Edit: February 22, 2005, 06:18:14 PM by chris319 »
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GS Warehouse

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Another Probability Problem
« Reply #14 on: February 22, 2005, 07:24:39 PM »
[quote name=\'jmangin\' date=\'Feb 22 2005, 08:48 AM\'][quote name=\'whewfan\' date=\'Feb 22 2005, 09:07 AM\']...should be JD Roboto.
[snapback]75683[/snapback]
[/quote]
HEY!  That's Mister Roboto.
[snapback]75695[/snapback]
[/quote]
[quote name=\'Seth Thrasher\' date=\'Feb 22 2005, 09:58 AM\']Domo Arigato...
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[quote name=\'bclark71\' date=\'Feb 22 2005, 05:38 PM\']Domo arigato for straightening that out.  :)
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[/quote]That's a repeat answer.  [throws parts made in Japan at bclark71]
« Last Edit: February 22, 2005, 07:25:08 PM by GS Warehouse »
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