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

[2020hindsight]

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 ..)

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

 

[korrupt_1]

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...

 

[J.B.Nimble]

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...

 

[2020hindsight]

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

 

[brty]

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

 

[2020hindsight]

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

 

[slackjaw]

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.

 

[2020hindsight]

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).

 

[bryaneasy]

: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)

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...

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.

 

[slackjaw]

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