# 50/50/90 rule - do I hear people snicker?



## 123enen (3 May 2006)

Did you know that the 50/50/90 rule is real? 
This is how it works, but first I need to explain the 50/50/75 rule.

50/50/75 rule

Four playing cards are laid face down on a table.
One of them is the Ace of diamonds.
Your job is to nominate the card that you believe is the ace of diamonds.

After you have made your choice the moderator turns one of the three other cards face up.
It is not the Ace of diamonds.
The game continues.

Another card is turned face up
It is not the Ace of diamonds.
The game continues with two card left face down on the table.
One of them is the Ace of diamonds, the other is not.

You are now given an opportunity to change your selection.
A 50/50 choice. 
Do you stay with the original selection or do you change cards.
Think about it, make your choice then continue on.

















Remember, the original card you chose had a 75% chance (1 in 4) of being the wrong card. There was a 75% chance that one of the other three cards was the ace of diamonds. 
The game has done you a favour. It has exposed two of those other cards, and now tells you that out of TWO remaining cards, ( 50/50 chance)  one has a 75% chance of being the Ace of diamonds and the other, your original selection, only has a 25% chance.

That is the 50/50/75 rule.


The 50/50/90 rule; start with 10 playing cards. The one you choose has a 90% chance of being wrong. What do you do when you get down to two cards.


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## ctp6360 (3 May 2006)

wow that just blew my mind, that's a great post thanks!


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## GreatPig (3 May 2006)

123enen said:
			
		

> out of TWO remaining cards, ( 50/50 chance) one has a 75% chance of being the Ace of diamonds and the other, your original selection, only has a 25% chance.



That's crap. Both cards have a 50% chance of being the ace, so it makes no difference whether you change your selection or not (in terms of the long term average).

Your selection only has a 25% chance of being the ace when all four cards are present. The moment one is removed, the probability changes.

GP


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## WaySolid (3 May 2006)

Looks like a variant of the 3 door gameshow puzzle.

It should be stated that the person turning over the cards knows the location of the Ace of Diamonds and deliberately does not turn it over.

Yes you should switch as it definitely improves your odds although intuitively it's very difficult to believe this.


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## slackjaw (11 August 2008)

WaySolid said:


> Yes you should switch as it definitely improves your odds although intuitively it's very difficult to believe this.




The dealer deliberately didnt turn over the Ace because he knew where it was, or because he couldn't, because it was the card you chose. Both are of equal probability. This aint even high school maths. You're down to 2 cards, regardless of which of these you chose to begin with each has the same probability of being the Ace.

If intuition played any part, then you would have chosen the correct card in the first place, so intuitively, your odds might be better if you dont change


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## WaySolid (11 August 2008)

slackjaw said:


> The dealer deliberately didnt turn over the Ace because he knew where it was, or because he couldn't, because it was the card you chose. Both are of equal probability. This aint even high school maths. You're down to 2 cards, regardless of which of these you chose to begin with each has the same probability of being the Ace.
> 
> If intuition played any part, then you would have chosen the correct card in the first place, so intuitively, your odds might be better if you dont change



Intuition doesn't imply magical forecasting ability.


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## ghotib (11 August 2008)

Umm. Why didn't each of the first 4 cards have a 1 in 52 chance of being the ace of diamonds? The setup didn't say it was definitely there at all. 

Think I'll stick to Lotto.


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## 2020hindsight (11 August 2008)

Lol
It's like the old blokes at the roulette wheel - 
you get 5 reds in a row, they'll swear that the chance of a black next is improved by this "sequence-of-past-events" , i.e.  is better than 50-50  (ignoring the green for a moment)  

(PS or ignoring  the *two* greens !!, if you're stupid enough to play roulette in Vegas


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## pepperoni (11 August 2008)

*snickers*


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## Buster (11 August 2008)

Hmmm..

I always thought that it was to explain why it is that whenever I am given a 50/50 opportunity, I'll pick incorrectly and get it wrong 90% of the time.. 

Regards,

Buster.


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## slackjaw (11 August 2008)

WaySolid said:


> Intuition doesn't imply magical forecasting ability.




http://dictionary.reference.com/search?q=intuition


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## 2020hindsight (11 August 2008)

Buster said:


> I always thought that it was to explain why it is that whenever I am given a 50/50 opportunity, I'll pick incorrectly and get it wrong 90% of the time.. .



buster, 

something like murphy's law ?
drop the toast, it always lands butter side down 

PS Murphy was an optimist


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## Buster (11 August 2008)

Hey 2020..



2020hindsight said:


> something like murphy's law ?
> drop the toast, it always lands butter side down




Ha ha.. Well _ALMOST_ always.. 

Regards,

Buster


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## Lubenc (11 August 2008)

Sorry slackjaw but you are wrong.

Say there were 10 cards and one is an ace.

When you get down to the last two cards there are two possiblities either:

1. the card you chose (when there were 10 to choose from) is the ace 
2. the other card is ace

The odds that the card you chose initially was the ace are 1 in 10

The odds you chose a card other than the ace are 9 in 10

So, some may say counter-intuitively, the odds are significantly greater
that you will end up with the ace if you switch.


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## 2020hindsight (11 August 2008)

anyone know the formula(e) they use on the "deal or no deal" show?
I've only seen it a few times (and certainly haven't cracked the code). 

Maybe something like - "the mean of the remaining suitcases less 10% ??"


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## white_goodman (11 August 2008)

2020hindsight said:


> anyone know the formula(e) they use on the "deal or no deal" show?
> I've only seen it a few times (and certainly haven't cracked the code).
> 
> Maybe something like - "the mean of the remaining suitcases less 10% ??"




im sorry i cant watch it andrew okeefe is a douche


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## korrupt_1 (11 August 2008)

2020hindsight said:


> anyone know the formula(e) they use on the "deal or no deal" show?
> I've only seen it a few times (and certainly haven't cracked the code).
> 
> Maybe something like - "the mean of the remaining suitcases less 10% ??"




:topic

so... after all but two cases remain... the 50c and $200k...

your chance of getting the $200K is 50/50...

what will the bank offer you? $99,995? i dont think so..... im guessing it will be $80,000 or something like that....

i think the bank takes human greed and emotion into it...

what would u do... take the unfair offer of $80K or take a chance at 50/50... knowing you could go home with 50c... 

remember, the bank KNOWS what you have... much like the card dealer knew where the Ace of Diamond was...


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## 2020hindsight (12 August 2008)

white_goodman said:
			
		

> im sorry i cant watch it andrew okeefe is a douche



that sounds like a variation of ad hominem mate 



korrupt_1 said:


> so... after all but two cases remain... the 50c and $200k...
> your chance of getting the $200K is 50/50...
> 
> im guessing it will be $80,000 or something like that....
> ...



ahhh - now we're getting down to the short strokes here korrupt  . 

I'd reason as follows ...
You have the option to go on ... with a negative casino advantage !!
There is $200,000.50 out there -  On average (if you did it 100 times whatever) you will go home with $100,000.25.  

And they are only offering $80,000.

When next in your life will the odds of a positive outcome be in YOUR favour rather than the bank's ? 

PS lol, then again - perhaps you're right - one bird in a penthouse or two birds in the bush, whatever that saying is


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## 2020hindsight (12 August 2008)

It's like the odds bet at the craps table ..

When you play craps, the casino advantage for bets on the pass line is 1.4% i.e. for every $100 bet, the casino keeps $1.40 in the long run.  If a number comes up first roll ( other than a natural or a crap out), you can then place an odds bet behind your first bet of at least the same value - and the casino gives you perfect odds.  i.e. suppose the first roll is a four, and the chances of you winning (getting a 4 before you get a seven) is 1:2, then the casino  pays you 2:1.   It is a GIFT!! - one of the few in the casino , other than the cup of lousy coffee that the girl in the skimpy dress brings you 

(i.e. 3 ways for two dice to come up with 4 , = 1+3, 2+2, 3+1
6 ways for them to come up with 7, = 1+6, 2+5, 3+3, 4+3, 5+2, 6+1
casino pays you 2:1 = perfect odds, ok)

So my question then is ... *if you can't afford to put the odds bet down,* - because you claim that's like putting all your eggs into one basket sort of thing - *should you pack up and go home*? - or hang around for the next chance to bet with 1.4% casino advantage?. 

I'd answer that you'd be mad to hang around.   *If you can't afford to bet when the casino is effectively GIVING money away, (- correction - make that GIVING you free entertainment lol)  then you can't afford to bet period.* 

PS Compare that to roulette wheel with 1/37 = 2.7% casino advantage (in Aus)
or 2/38 = 5.3% in Vegas where there is a "00" as well as an "0" - sheesh. now that is crazy.



> Albert Einstein is reputed to have stated, "You cannot beat a roulette table unless you steal money from it."




https://www.aussiestockforums.com/forums/showthread.php?p=175398



> Say Vegas permits odds bet of 3 times your first bet, (although Reno is better, up to 10)
> 
> and you have choice of either .......
> 
> ...


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## 2020hindsight (12 August 2008)

2020 said:
			
		

> Say Vegas permits odds bet of 3 times your first bet, (although Reno is better, up to 10)THEN
> a) craps would cost you $14.40 per hour,



oops, but Aus is more expensive of course  ( not permitting such large odds bets.)


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## slackjaw (12 August 2008)

Lubenc said:


> Sorry slackjaw but you are wrong.
> 
> Say there were 10 cards and one is an ace.
> 
> ...




Yes but had you chosen the other card to begin with, those odds would be the same, therefore equal probability


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## WaySolid (12 August 2008)

slackjaw said:


> http://dictionary.reference.com/search?q=intuition



When you can use that to make money from forecasting visit the casino.


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## pepperoni (12 August 2008)

Deal or no deal makes high or low offers to make the show interesting. No forumula.

The question contestants should ask themselves are "is the offer better than the odds I could get at the casino" ... if it is take offer everytime.

You can take the money to the casino down the track.

IE 2 cases 1c and 200k ... offer 150k ... TAKE IT ... then take it to the casino and bet it all on black for a better potential return.  Or bet 100k on black and pocket the 50k. :

But the rule in this thread is embarrasing .... the odds when you picked from 4 are no longer relevant in a 50 50 pick


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## theasxgorilla (12 August 2008)

GreatPig said:


> That's crap. Both cards have a 50% chance of being the ace, so it makes no difference whether you change your selection or not (in terms of the long term average).
> 
> Your selection only has a 25% chance of being the ace when all four cards are present. The moment one is removed, the probability changes.




Exactly to all of the above.


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## Lubenc (12 August 2008)

Both cards have a 50% chance of being the ace when the there are 2 remaining but the process leading to the selection of the cards is important. 

If you switch cards the only way you can lose is if you chose the ace as your card.

If you chose any of the other cards you will end up with the ace when you switch.

This is why the odds when there are 4 cards are relevent. And the more cards there are the more likely you will not chose the ace and therefore end up with it when you switch.


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## 2020hindsight (12 August 2008)

pepperoni said:


> Deal or no deal makes high or low offers to make the show interesting. No forumula....
> 
> IE 2 cases 1c and 200k ... offer 150k ... TAKE IT ...



I sorta agree pepperoni 
but not sure they offer high offers - or that they'd offer you 150K  
heck if they did you'd have to take it.. like you say.  And you could do what you like with the profit lol.  

But the crime (to my mind) would be if they offered you $80K - when the fair amount was $100K - and you took it - only for you then to go to the local club say 8000 times before you die, spending $10 each time - and it ending up as $50K or so - when you had an excellent chance of turning it into $100K on "one turn of pitch and toss" or "one turn of the deal-or-no-deal challenge" etc. (imo)


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## WaySolid (12 August 2008)

theasxgorilla said:


> Exactly to all of the above.



I think you will find that switching is the way to go!

The thing is that the probability does actually change.

There's quite a bit on the web about the three door game.
http://www.theproblemsite.com/games/monty_hall_game.asp


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## brty (12 August 2008)

Hi,

I can't believe this thread has gone on for so long.

In a 'fair' game you switch every time, because the odds are in your favour to do so.

If 10 cards are placed face down, and you chose 1 of them, you have a 10% probability of guessing which one is the ace.

The other 9 cards represent a 90% probability of the ace being there.

If the other person, in charge of the game knows where the ace is, and turns up 8 non ace cards, then the odds are still 90% that their last card is the ace, and your odds are still 10% of having the ace. 

The odds never changed in that scenario.

If you think this is wrong, then I would really like to play some card games with you.

brty


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## pepperoni (12 August 2008)

brty said:


> If the other person, in charge of the game knows where the ace is,




Is this element in the original ... there is a big diff between this and random flips by a dealer IMO.

If the dealer knows the cards he will just flip the bs cards so 2 or 2000000 cards it was always gonna come to 50 50


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## pepperoni (12 August 2008)

WaySolid said:


> Looks like a variant of the 3 door gameshow puzzle.
> 
> It should be stated that the person turning over the cards knows the location of the Ace of Diamonds and deliberately does not turn it over.




I see someone else clarified  ... this makes it misleading bs ... better to say "you can pick this one card or these 3"!  .


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## brty (13 August 2008)

Pepper,

It does not come to 50/50.

If the odds started at 90/10, then that is what they remain, whether you see 'most' of the cards face up or face down makes no difference, the cards as delt are the same.

In the games mentioned, the dealer knows where the ace is, therefore all he is showing face up are the irrelevant cards. He has to have at least 8 irrelevant ones. If he shows them to you face up or face down makes no difference to the odds, he had a 90% probability of having the ace you had a 10% probability of having the ace.

Are you, per chance, interested in a game of cards???:

brty


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## 2020hindsight (13 August 2008)

brty said:


> It does not come to 50/50.
> 
> If the odds started at 90/10, then that is what they remain, whether you see 'most' of the cards face up or face down makes no difference, the cards as delt are the same.
> 
> ...




I must be misunderstanding you brty. (or you're having a lend surely?) 

There are two cards (at the end) - one is the ace , one isn't 
and you say that there 's 90% chance one is, and 10% the other? 

So when you play cards, if we could increase our bets at that point, you'd give me 9-1 if I added to my bet on my card?  

..........

PS Rethink - Strange rules - Since when does the dealer know where all the cards are , and which ones to show you? -  Are you saying  all the cards are effectively face up for him (after being dealt), and face down for you? - and he chooses which ones to show you ?    I guess with enough rules like that you are right .


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## 2020hindsight (13 August 2008)

korrupt_1 said:


> what would u do... take the unfair offer of $80K or take a chance at 50/50... knowing you could go home with 50c...




On second thoughts, you might have talked me round to this philosophy korrupt lol.  i.e. take the $80K  (at least in my present predicament).   

like trading - it depends whether you're talking about money you can spare, or money you need to stay afloat 
https://www.aussiestockforums.com/forums/search.php?searchid=1456777


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## 2020hindsight (13 August 2008)

The maths said go for the motsa
but the dice came down the wrong way
now I've got this bill - in fact lotsa
and no money to mathematically pay


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## pepperoni (13 August 2008)

brty said:


> Pepper,
> 
> It does not come to 50/50.
> 
> ...




I know but as I said as he knows where the cards are its all a charade.

He should ask "find the ace .... do you want to pick one card or 9 cards" 

If he doesnt know where the cards are, as per the original post, there is no point in changing as they are all 1 in 4.


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## 2020hindsight (13 August 2008)

Good point pepperoni - we need to define the game..
I'll repost the definition of the 50/50/75 rule:-




123enen said:


> Four playing cards are laid face down on a table.
> One of them is the Ace of diamonds.
> Your job is to nominate the card that you believe is the ace of diamonds.
> 
> ...




ok , I think I 've worked it out.

a) If the "moderator" originally knew where the ace was, and has a good memory , and is the one choosing which cards to turn over - you should change your choice  (the odds are 75/25)

b) If the "moderator" originally knew where the ace was, and has a bad memory, and is the one choosing which cards to turn over , you might as well change your choice (as it's somewhere between 50/50 and 75/25)

c) If the "moderator" never knew where the ace was, and was above cheating or dealing from the bottom of the pack etc. - and is the one choosing which cards to turn over - You might as well change your choice just in case your assumption that he wasn't dealing from the bottom of the pack was wrong, lol. 

d) If YOU were the one choosing which cards, it's 50/50.  (irrespective of whether the moderator knew where the ace was or not).  

Summary.  You might as well change your bet when it gets down to 2 cards - nothing to lose. (imo)   .


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## cuttlefish (13 August 2008)

If the moderator does know where the ace is then you should change because its 75/25 in favour of it being the one the mod has left.

If the moderator doesn't know, and is randomly turning cards over - then if the cards were in his lot there is a 2/3 chance he won't turn it over first go, a 1/2 chance he won't turn it over second go.  Therefore if the card is in his lot of three and he hasn't turned it over through random selection then there is only a 1 in 3 chance he wouldn't have turned it over.  So that makes me think the odds are in favour of not swapping because if it was the moderators lot of three there's a 2/3 chance he would have turned it over already.

i.e. extrapolate this out to a million - if the ace was in the moderators set of 999,999 cards and he is truning them over randomly, then odds on the moderator turning over 999999 cards without turning over the ace is very slim - thus if he manages to do this its likely to be because the ace is your card and not amongst his cards.

I could probably have explained that better but am feeling too lazy to rewrite it.


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## pepperoni (13 August 2008)

cuttlefish said:


> If the moderator does know where the ace is then you should change because its 75/25 in favour of it being the one the mod has left.
> 
> If the moderator doesn't know, and is randomly turning cards over - then if the cards were in his lot there is a 2/3 chance he won't turn it over first go, a 1/2 chance he won't turn it over second go.  Therefore if the card is in his lot of three and he hasn't turned it over through random selection then there is only a 1 in 3 chance he wouldn't have turned it over.  So that makes me think the odds are in favour of not swapping because if it was the moderators lot of three there's a 2/3 chance he would have turned it over already.
> 
> ...





I struggle with this ... they were all 1/4 when they started and 50/50 on last pick. 

But yes nothing lose by changing cards.

The other one is obvious ... just a roundabout way of saying "which is more likely, the ace is *that* card, or one of the other 3!


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## 2020hindsight (13 August 2008)

cuttlefish said:


> i.e. extrapolate this out to a million - if the ace was in the moderators set of 999,999 cards and he is truning them over randomly, then odds on the moderator turning over 999999 cards without turning over the ace is very slim - thus if he manages to do this its likely to be because the ace is your card and not amongst his cards.
> 
> I could probably have explained that better but am feeling too lazy to rewrite it.




lol
I struggle with this as well.
you're making it real complicated there cuttlefish - got my eyebrows in a knot here lol 

It could also be argued (erroniously / mischievously) that there are two cards left (of the 1,000,000) - the one you chose when you had odds of 1 in 1,000,000 of getting it right,   ...

and the one the moderator chose when he dismissed all the others - and when he had 1 in 999,999 chance of picking it if you missed it.    Therefore the one he chose has the slightly better odds of being right    (PS you could drive a truck through that logic but after a couple of beers it starts to make more sense lol)

probably exhausted this one surely


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## bryaneasy (13 August 2008)

If you disagree with the OP's theory then you are quite ignorant.

The maths behind it is simple and it involves the player playing from the VERY BEGINNING.  Now you are playing from the beginning you have picked one card and randomly (I mean of course if the dealer knows the ace then the game has absolutely no meaning to it) comes down to the last 2 of 10 cards, then that makes switching the card for you a 90% favourite from the beginning, of course the luck on the way is now disregarded simply because of the fact that you are now on the last 2 cards and you have played all the way through.

It is like the simple maths of if you have 2 different pairs of socks in the drawer, what is the chance of picking out a pair simultaneously - well 25%, however if you pick out one sock first then what are the odds? well it changes to 33% doesn't it?

Roulette too, the chances of you starting from spin number 1 and spinning 20 blacks in a row (the highest I have seen at the casino is 9 spins in a row - I won $200 from my original $5 ) is almost impossible, however if it happens and then you start playing from the spin thereafter then yes your chances are 50/50 of black coming up next spin. 

It doesn't matter what game you play - maths is maths think what you like but our world revolves around these magic figures and formulas - it is the reason we are so technology advanced to date.


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## 2020hindsight (13 August 2008)

well I wouldn't have used the word ignorant, and it requires a bit of concentration surely (even if only for a coupla minutes) - depends on the definition of the problem etc...   variations of the three-door-problem etc.

Waysolid is spot on  (and that website ..)



WaySolid said:


> I think you will find that switching is the way to go!....  the three door game.....
> http://www.theproblemsite.com/games/monty_hall_game.asp




Likewise this explains it (a t least one version thereof):-

http://mathforum.org/dr.math/faq/faq.monty.hall.html


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## korrupt_1 (13 August 2008)

bryaneasy said:


> Roulette too, the chances of you starting from spin number 1 and spinning 20 blacks in a row (the highest I have seen at the casino is 9 spins in a row - I won $200 from my original $5 ) is almost impossible, however if it happens and then you start playing from the spin thereafter then yes your chances are 50/50 of black coming up next spin.




:topic

Does probability have memory? Does it know what was tossed/spinned/picked in the past? Each and every turn will still result in a 50/50 chance of getting it right... no matter if you spun in 5, 10, 100, 1000000000 times... every single spin is 50/50...

I pulled this from a PDF that was posted recently on another thread about setting stoplosses... (Jan Bylov Stop techniques)

It's interesting that the Gambler will always think he has increased probability of a correct pick if it continues to turn up with the same result consequtively in a row...


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## J.B.Nimble (13 August 2008)

Anyone still struggling with this, please prove it to yourself. 

Take 10 cards and sit yourself down somewhere no one will catch your looks of bewilderment and cries of WTF... You are going to lay the cards out in a row on the table but before you deal them decide on one position in that row that is going to be your pick. Now deal them face up and observe your pick - have you picked the Ace? Now, remove 8 other cards (not including the ace). So do you swap or not? 

Now repeat this about a hundred times and report back here... Here endeth the snickers...


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## 2020hindsight (13 August 2008)

J.B.Nimble said:


> Anyone still struggling with this, please prove it to yourself.
> 
> Take 10 cards and sit yourself down somewhere no one will catch your looks of bewilderment and cries of WTF... You are going to lay the cards out in a row on the table but before you deal them decide on one position in that row that is going to be your pick. Now deal them face up and observe your pick - have you picked the Ace? Now, remove 8 other cards (not including the ace). So do you swap or not?
> 
> Now repeat this about a hundred times and report back here... Here endeth the snickers...



nimble, yeah but  ..
are you talking post #1 ?
or post #4 ?
different ballgame m8


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## brty (13 August 2008)

2020 and others,

The point you are still missing, which nimble points out, is that the odds are the same, whether you see the cards or not.

If you have 10 cards, select 1, then you have a 10% chance/probability of having the ace. It doesn't matter if it is face up, face down or sideways 
The person who has the 9 cards has a 90% chance/probability of having the ace.

That's it there is no more. If you don't understand it, don't play that type of game. There is no point argueing about any other so called odds. In a random shuffle over a couple of thousand hands you will eventually get it.

Comparing it to a casino and a roulette wheel is irrelevant. Each spin of the wheel is independant. With the card game the ace has to come up. They are different and so are the probabilities

brty


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## 2020hindsight (14 August 2008)

brty said:


> 1. Comparing it to a casino and a roulette wheel is irrelevant.
> 
> 2. With the card game the ace has to come up.



brty
1. agreed  (apologies if I was labouring the discussion)
I'll stick with my thoughts on #36. 
btw, I don't think we were arguing - just trying to define the problem lol. (the first post is a bit ambiguous)
As Plato sad, "Until you define your terms, I refuse to argue" 

2.  :topic Speaking of card games, isn't there a new movie  "21" - about some kids who combine together at blackjack to rip off the casino? - counting the colour cards, and mathematically deciding when to double - can tip the casino advantage in your favour)     Happened in real life in Macau about 1980.  THe Chinese called em "The Magnificent 7"  Canadians I think.  They were banned from the casino.  Didn't the local population yell and scream


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## slackjaw (14 August 2008)

2020hindsight said:


> http://mathforum.org/dr.math/faq/faq.monty.hall.html




 After following this link, I became almost convinced, but not entirely grasping the concept I decided to simulate the thing in MATLAB, simulating with 1000 samples at a time and varying numbers of cards. Sure as hell, the theory withstood the test, leaving me with no other option but to cross the floor on this issue.

So I found that for n cards originally on the table the odds when you make the switch are:

1/n of probability losing and (n-1)/n probability of winning

My mistake was in thinking that the probability lay in the cards themselves, when in fact the probability lies in the person, whether he or she switches or not.


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## 2020hindsight (14 August 2008)

slackjaw said:


> I decided to simulate the thing in MATLAB, ...



slackjaw - lol - Matlab  - well done, 
talk about using a sledgehammer to crack a walnut   

I think the problem is that it runs against the Irish grain in all of us lol - once you bet on a horse, stick with it !  (not).


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## bryaneasy (15 August 2008)

korrupt_1 said:


> :topic
> 
> Does probability have memory? Does it know what was tossed/spinned/picked in the past? Each and every turn will still result in a 50/50 chance of getting it right... no matter if you spun in 5, 10, 100, 1000000000 times...
> I pulled this from a PDF that was posted recently on another thread about setting stoplosses... (Jan Bylov Stop techniques)
> ...




Off-topic? I am just trying stipulate that the maths involved in this topic is derived from the fact that the person is playing from the very beginning!

*"every single spin is 50/50"*... .  Yes you just proved that you are missing the bigger picture - every single spin is a 50/50 because from your mathematical point of view that is when you have just started playing/participating.

Now some high school maths will tell us that if you play for 2 spins then your chances of getting tails 2 times is 25% not 50%..... the fact is we are playing from the very beginning.


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## slackjaw (15 August 2008)

Bryan, that was my thought as well, but if you write down all the possible events that can occur in this example (the three door problem), then 2/3 of the times that you can win happen when you make the switch, 1/3 of the times you can win when you dont.

To think of the same problem in a different way which is easier to see...

How do I steal a number out of your head?

I ask you to think of a number between 1 and 1000.

I tell you I am thinking of the number 6

then you have to eliminate all but 2 numbers, making sure you don't eliminate the number I chose or the number you chose. If I happened to guess right, you can choose one at random not to eliminate. You have to tell me what two you have left.

I can be almost certain that I didn't choose right, so the number you were thinking of was almost definately the number you said that was not number 6, hence I now switch my choice and get the answer right.

Now apply this to the card game, eliminating cards one by one until we have two left, including the card I chose and the prize card. Its the same system.

It took me a while to understand it too, follow the link that 2020 posted or simulate it yourself as I did.

In a game of heads or tails, every toss is still 50/50. The difference is that you are not eliminating options that lead to a loss, so there is no value in switching


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