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probability question / brain teaser!


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#1 of 17 EricW

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Posted March 05 2012 - 02:46 PM


hey guys, i need some input as i've been going back and forth with my brother-in-law with this all night.  it a probabilities question involving three coins.


here is the scenario.


1) you have three coins in hand

2) you throw all three in the air

3) the three coins fall to the ground.

4) before knowing the orientation of any coins, you cover one at random.

5) *IF* the 2 exposed coins are both heads, what is the probability that the hidden coin will be a Head?


my answer was that it is 50/50 whether the hidden coin is H or T.


here's the proof:  below is the outcome table.
-along the left are the possible outcomes of a coin toss.

-along the top are the possible positions of the "hiding hand"

-there are only 6 possible outcomes which we would consider (red boxes - where both exposed coins are H's)

-of those 6 outcomes, 3 have a hidden Head and 3 have a hidden Tail.


is this sound logic?  if i'm missing something please tell me.


http://static.hometh...um.com/imgrepo/


oh by the way, the loser has to buy Boardwalk Empire on Blu-ray :)



"now, if that's a fact, tell me... am i lying?"

#2 of 17 Mike Frezon

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Posted March 05 2012 - 03:28 PM

Eric:


I'm no math whiz (and I know the HTF is full of them!)...but isn't the orientation of the hidden coin independent of whatever the other outcome of the other two coins?

That would also mean it's a simple 50/50 proposition for the hidden coin..

There's Jessie the yodeling cowgirl. Bullseye, he's Woody's horse. Pete the old prospector. And, Woody, the man himself.Of course, it's time for Woody's RoundUp. He's the very best! He's the rootinest, tootinest cowboy in the wild, wild west!


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#3 of 17 EricW

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Posted March 05 2012 - 04:46 PM



yeah but i'd rather have a mathematical, explained argument :P it's more convincing that way to people who would believe it's not 50/50


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#4 of 17 David Norman

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Posted March 05 2012 - 05:52 PM

As Mike says since the events are 100% independent the answer is always 50/50 for heads or tails whether it's coin 1, 2, or 3 and whether you tossed all 3 coins simultaneously or consecutively. 50% There are only 2 outcomes -- it's either Heads/Heads/Heads or Heads/Heads/Tails. 50/50 Different version of same argument and removes any . Coin 1 has Black and White sides Coin 2 has Blue and Green Sides Coin 3 has Red and Yellow sides. Throw all 3 coins into the air and cover 1 randomly. 1st coin in Blue 2nd coin is Black What's the odds that 3rd coin is Yellow? 50% Otherwise just go back to your chart. As an aside I did a coin flip years ago and the coin spun and spun and actually ended up standing on its edge.
 

 


#5 of 17 EricW

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Posted March 06 2012 - 12:45 AM

i agree with all your agruments.  this whole thing came about as a tangent to the Monty Hall Problem.


for those who don't know, this is the MHP:

you're on a game show, there's 3 doors.  behind 1 is the Prize, behind the other 2, nothing.


-you pick one door.

-the game show host, who knows where the prize is, opens one of the remaining doors to show nothing.

-host gives you the option to stick with your door, or switch to the other closed door.


what do you do?

one might intuitively say it doesn't matter, but mathematically, you double your chances by switching doors


if you need the proof, just google "monty hall problem"


our current argument was similar (at least to him) in that "when you only reveal when 2 Heads are showing, it's conditional probability"

since i wanted to be open minded, i listen but still don't agree.




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#6 of 17 EricW

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Posted March 06 2012 - 01:13 AM


to add, i think that in the current scenario, if the coin flipper sees the coins land and can choose which coin to cover, THEN the hidden coin is 75% likely to be T, and 25% likely to be H.  is this correct?


if so, i think he was just getting confused with the variable of [randomly covering a coin] vs [knowledgeably covering a coin]


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#7 of 17 Jason Charlton

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Posted March 06 2012 - 02:40 AM

The Mythbusters tackled the Monty Hall problem last season.  It was a fascinating episode.  They had pretty conclusive evidence that it's to one's benefit to change their answer when given the option.


The coin flip issue is fundamentally different than the monty hall problem.  As was discussed prior, each coin has it's own probability (50/50) of being in either state.  With the MH issue, there was only ONE door that had a prize (heads?).


With the MH problem, the three options are related - meaning if you know what two doors are, you automatically know the state of the third door.  Not so with the coins.


A coin flip is always a 50/50 proposition.  One flip has no bearing on the next.


Here's the Mythbuster episode video.  Skip to about 14 minutes to see them start their final "test" and discuss the paradox:





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#8 of 17 Stan

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Posted March 06 2012 - 02:45 AM

Las Vegas casinos do this all the time, especially with roulette. Six red numbers hit in row, everybody thinks the seventh number will also be red. Not necessarily wrong, but as everyone has pointed out, it's at 50/50 chance it will be black. Every spin of the roulette ball is a fresh spin, has absolutely nothing to do with past results. Never heard of the Monty Hall story, will do a little checking on that.
Stan

#9 of 17 Stan

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Posted March 06 2012 - 02:45 AM

Las Vegas casinos do this all the time, especially with roulette. Six red numbers hit in row, everybody thinks the seventh number will also be red. Not necessarily wrong, but as everyone has pointed out, it's at 50/50 chance it will be black. Every spin of the roulette ball is a fresh spin, has absolutely nothing to do with past results. Never heard of the Monty Hall story, will do a little checking on that.
Stan

#10 of 17 Stan

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Posted March 06 2012 - 02:48 AM

Sorry for the double post, got an error saying there was a problem submitting the post to the server, tried again and now I have two.
Stan

#11 of 17 cafink

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Posted March 06 2012 - 04:49 AM

The important difference between your coin-flipping problem and the Monty Hall problem is that you cover one of the coins randomly, but Monty Hall does not act randomly when he opens one of the doors; he specifically opens a door he knows to be empty. If your chosen door is empty (which happens two-thirds of the time), then Monty is forced to open the single remaining empty door. There is a variation of the Monty Hall problem in which Monty does not know the location of the car, and opens his door randomly. In this variation, there is no advantage to switching--you have a 50/50 shot whether or not you stick with your original door. But there is also a 1/3 chance that Monty will open his door to reveal the prize, ending the game prematurely!
 

 


#12 of 17 EricW

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Posted March 06 2012 - 06:02 AM


do you guys agree with this part? :

if the coin flipper sees the coins land and can choose which coin to cover (hide 3rd coin when the other 2 are H+H), THEN the hidden coin is 75% likely to be T, and 25% likely to be H.

"now, if that's a fact, tell me... am i lying?"

#13 of 17 Jason Charlton

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Posted March 06 2012 - 06:22 AM


Originally Posted by EricW 

do you guys agree with this part? :

if the coin flipper sees the coins land and can choose which coin to cover (hide 3rd coin when the other 2 are H+H), THEN the hidden coin is 75% likely to be T, and 25% likely to be H.


I can see the thinking there - according to your outcome table, there are a total of 4 possible outcomes that include at least 2 heads.  Of those four, 3 of the situations involve the final coin being tails, while only 1 scenario has all three coins being heads.


However, I can't reconcile the fact that choosing which coin to cover in order to ensure that the remaining two coins are heads screws up the whole shebang.  It's no longer random and no longer "objective", if that makes any sense...


So I guess I would agree with that part of the scenario, but I think somewhere along the lines the validity of the "experiment" was lost.



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#14 of 17 EricW

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Posted March 06 2012 - 06:28 AM


clarification - the question of "what happens if the coin flipper sees the coin" is a twist on the original post, and should be considered seprate of my original scenario / post.  i was only trying to illustrate what might have been going on in my brother-in-law's head when i was discussing this with him :P


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#15 of 17 betooz

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Posted March 06 2012 - 02:30 PM

From Stan: Las Vegas casinos do this all the time, especially with roulette. Six red numbers hit in row, everybody thinks the seventh number will also be red. Not necessarily wrong, but as everyone has pointed out, it's at 50/50 chance it will be black. Every spin of the roulette ball is a fresh spin, has absolutely nothing to do with past results. Never heard of the Monty Hall story, will do a little checking on that. ****Nice, messed up the quote feature.**** You're forgetting that the roulette wheel in Vegas also has two green spaces, zero and double zero. This makes the odds less than 50/50 that red or black will come up. To answer the original question, the odds are still 50/50 on the third coin coming up heads or tails.

#16 of 17 EricW

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Posted March 06 2012 - 05:18 PM


yeah, my brother-in-law conceeded, looks like i'm getting Boardwalk Empire on Blu-ray :)


"now, if that's a fact, tell me... am i lying?"

#17 of 17 Mike Frezon

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Posted March 07 2012 - 01:28 AM



Originally Posted by EricW 


yeah, my brother-in-law conceeded, looks like i'm getting Boardwalk Empire on Blu-ray :)



THAT'S what I'm talkin' 'bout!  Posted Image


There's Jessie the yodeling cowgirl. Bullseye, he's Woody's horse. Pete the old prospector. And, Woody, the man himself.Of course, it's time for Woody's RoundUp. He's the very best! He's the rootinest, tootinest cowboy in the wild, wild west!


HTF Rules | HTF Mission Statement | Father of the Bride

Dieting with my Dog & Heart to Heart/Hand in Paw by Peggy Frezon





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