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re: Marilyn vos Savant and the history of the Montel Hall question
Posted on 2/24/15 at 10:51 pm to TigerstuckinMS
Posted on 2/24/15 at 10:51 pm to TigerstuckinMS
What I love about this is that it gets brought up every now and then and there are somehow people alive that still don't get it
1
Posted on 2/24/15 at 10:53 pm to rockchlkjayhku11
I love that this thread is over 48 hours old and it is still referencing "Montel Hall".
Posted on 2/24/15 at 10:59 pm to TigerstuckinMS
quote:
What I love about this is that ANYONE can prove out for themselves that switching wins 2/3 of the time by using nothing more than three pieces of paper to represent the three doors and following the rules of the problem and playing the game themselves. No math involved, just play the game.
Yet, people STILL insist it's 50/50. Bless their hearts.
You need two people. One that knows which piece of paper has the car, and which has the goats.
Otherwise this won't work.
Posted on 2/24/15 at 11:22 pm to KosmoCramer
quote:
You need two people. One that knows which piece of paper has the car, and which has the goats.
Otherwise this won't work.
Nah, Monty Hall had rules. As long as you follow the rules of the game, you can do this yourself. It doesn't matter because you're not making decisions based on the information you have, you're just performing defined actions based on what you see.
Take three playing cards. Two jacks and an ace. Ace wins. Put them facedown and shuffle. Pick one at random. Flip the other two over. They will either be two jacks or a jack and an ace. Throw away a faceup jack and swap your facedown card for the remaining face up card. If you get a faceup ace, you win. Keep track of how many times you win and lose.
Remember, you're not making decisions. You are ALWAYS going to switch here. The only decision you can possibly make is which of the two jacks to toss out if they both get turned over when you set your originally picked card aside. Tally it up and pretty quickly, you'll see the 33:66 split between losses and wins if you switch every time.
This post was edited on 2/24/15 at 11:24 pm
Posted on 2/24/15 at 11:39 pm to KosmoCramer
quote:You could easily do it by yourself. Just choose a card (or cup or whatever), say you will never switch (or always switch), and see how many times you win. It would be easier to simulate the show with two people though.
KosmoKramer
Posted on 2/24/15 at 11:43 pm to TigerstuckinMS
quote:
Take three playing cards. Two jacks and an ace. Ace wins. Put them facedown and shuffle. Pick one at random. Flip the other two over. They will either be two jacks or a jack and an ace. Throw away a faceup jack and swap your facedown card for the remaining face up card. If you get a faceup ace, you win. Keep track of how many times you win and lose.
Remember, you're not making decisions. You are ALWAYS going to switch here. The only decision you can possibly make is which of the two jacks to toss out if they both get turned over when you set your originally picked card aside. Tally it up and pretty quickly, you'll see the 33:66 split between losses and wins if you switch every time.
I suppose, but it will be a little bit harder to make the connection for those that aren't believers.
Posted on 2/24/15 at 11:44 pm to buckeye_vol
quote:
You could easily do it by yourself. Just choose a card (or cup or whatever), say you will never switch (or always switch), and see how many times you win. It would be easier to simulate the show with two people though.
What happens if you when you remove one of the cards (or cup or whatever) and it's the one that contains the car(which never happens in the game)?
Posted on 2/24/15 at 11:53 pm to KosmoCramer
I guess you have to find one that isn't. This is why you would have to make a decision to always stay or always switch. It would be more practical with two people though.
Posted on 2/24/15 at 11:55 pm to buckeye_vol
I think another great way to look at it is:
You win if you pick one of the two doors that DOESN'T have the car behind it 100% of the time if you switch.
2/3 of the doors don't have the car.
You win if you pick one of the two doors that DOESN'T have the car behind it 100% of the time if you switch.
2/3 of the doors don't have the car.
Posted on 2/25/15 at 6:07 am to Big Scrub TX
The issue that is problematic is that most view this as a show of probability when it is more a show of profitability. Analytic thought process requires one to consider all alternatives including those outside the norm.
In some ways you need to understand this game show is like being at a casino. You are playing against the house and the game is rigged to ensure the house stays profitable.
There is a final component in this show - it is called profitability- Though the show is designed to be entertaining and on some level challenges the math component of our intelligence - ultimately this show is about marketing. IF the show is not successful in viewer ratings then securing sponsors would falter and the show would not be a business success.
The fact that this show has endured time indicates how strong the "house" component is integrated into the show.
A solid contestant could increase his odds of winning by never forgetting the show is designed for the house to not win but be profitable.
You gotta know when to holdem and when to foldem!
In some ways you need to understand this game show is like being at a casino. You are playing against the house and the game is rigged to ensure the house stays profitable.
There is a final component in this show - it is called profitability- Though the show is designed to be entertaining and on some level challenges the math component of our intelligence - ultimately this show is about marketing. IF the show is not successful in viewer ratings then securing sponsors would falter and the show would not be a business success.
The fact that this show has endured time indicates how strong the "house" component is integrated into the show.
A solid contestant could increase his odds of winning by never forgetting the show is designed for the house to not win but be profitable.
You gotta know when to holdem and when to foldem!
Posted on 2/25/15 at 8:36 am to KosmoCramer
quote:
What happens if you when you remove one of the cards (or cup or whatever) and it's the one that contains the car(which never happens in the game)?
In this case you would correctly assume that Monty would have removed the other card instead, so that you would still have the choice to switch or stay. This actually helps reinforce the 2/1 odds proof.
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