Computer Simulations from yesterday's Let's Make a Deal problem. | Page 3 | TigerDroppings.com
Page 1 2 3 4 5 ... 10
Posted by
Message
castorinho
Oklahoma Fan
13623 posts
Member since Nov 2010
29990 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.

I actually started a thread about that episode of deal or no deal a while back. I'll try to find it in a second.

Dude had 3 cases left (two \$1 million dollar cases and one \$1 case). The offer at that point was \$600K plus, which he naturally declined, as most would (and should). Then he opens one of the million dollar cases. He's now down to a milli and a dollar case. Offer was just below \$500K, and he declined....he had the dollar case .

I think the thread I started was asking how much would the offer have to be for you to personally decline. This would obviously vary from person to person, and there were people saying it'd have to be at least \$500K .

That would only make sense if you had the chance to play the game multiple times as you'd eventually break even but a once in a lifetime chance like that I'm sure a lot of people would take low offers....some even 100K

Wavefan
Tulane Fan
St. Tammany
Member since Mar 2005
65 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
You have exactly the same odds whether you switch or not. You have exactly the same odds whether Monty knows what is behind the doors or does not. Those odds are 1/3 you get the car, 1/3 you get one hamburger, 1/3 you get the other hamburger. Period. Switching cannot increase those odds. Let me explain why:

First scenario: Monty knows what is behind each door. You pick first. We can all agree that when you do this you have one out of three odds to get the car, two out of three to pick the burger. Now follow this closely, this is the key: you can't pick both burgers; you only get one pick. If you picked the car, there are two burgers left. If you picked a burger, there is a car and a burger left. So regardless, there is still going to be a burger behind a door you did not pick. Remember, Monty knows this. Just for the sake of completeness, let's say that Monty, through some innate meanness, opens a door you didn't pick and the car is behind. Obviously you don't increase your odds by switching. You have a burger either way. But remember, Monty knows what is there, and to make the show more "exciting," he will probably open a door that hides a burger. If you picked a burger, he will open THE only door left that hides a burger. If you picked the car, he can open either door, and voila, there is a burger. But neither of these maneuvers increases the 1/3 odds that you picked right in the first place, it simply leaves you with 50/50 odds that EITHER THE DOOR YOU PICKED OR THE OTHER DOOR NOT OPENED BY MONTY HIDES THE CAR! You have 50/50 odds ON EITHER DOOR whether you switch or not. Think of it like this: the odds that car was behind one of the three doors is 1/3, and the odds that it is behind ONE of the other two is 2/3, but once you pick, you don't shift yourself to 2/3 on the car by switching your pick. You simply shift that initial 1/3 chance from one door to another. Got it?

Suppose Monty doesn't know. Like you, he is guessing. The odds still stay the same. You have a 1/3 of picking the car. When Monty picks, because like you he doesn't know what is where, he also has a 1/3 chance of picking the door hiding the car. The fact that you already picked one door doesn't change Monty's odds, it simply eliminates one of his choices. Again, if Monty picks a door and opens it and there sits the car, switching does you no good. If Monty picks a door and there sits a burger, you know there are equal odds that the car is behind either the door you initially picked or the door neither you nor Monty picked, but switching does NOT IMPROVE YOUR ODDS. By the process of elimination, your odds are now better (1/2 instead of 1/3) but switching picks does not change those odds.

The crucial point on either scenario, either the "he knows" or "he doesn't," is that the initial odds are 1/3 for you to pick the car regardless, and if the door opened by Monty reveals a burger, by process of elimination your chances on the car become 1/2 regardless. Switching does not affect those odds.

Dooshay
LSU Fan
CEBA
Member since Jun 2011
29318 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
the simulations disagree with you.

Replies (0)
Replies (0)
castorinho
Oklahoma Fan
13623 posts
Member since Nov 2010
29990 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

You have exactly the same odds whether you switch or not. You have exactly the same odds whether Monty knows what is behind the doors or does not. Those odds are 1/3 you get the car, 1/3 you get one hamburger, 1/3 you get the other hamburger. Period. Switching cannot increase those odds.

Replies (0)
Replies (0)
castorinho
Oklahoma Fan
13623 posts
Member since Nov 2010
29990 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.

Replies (0)
Replies (0)
Nuts4LSU
LSU Fan
Washington, DC
Member since Oct 2003
18919 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

Are you familiar? Monty gives you a choice of three doors. Behind one is a new car. Behind the other two are hamburgers. You pick Door #1. Before the contents are revealed to you Monty says "hold on, Jack. Let's look behind door number 3!". The door opens to reveal a hamburger. Then Monty says "OK, do you want to keep your first choice or would you like to switch to door number 2?" What should you do?

Obviously, the thing to do is switch.

There is a 2/3rds chance you picked the wrong door. IF you picked the wrong door, switching will win for you. The only way switching will lose for you is if you picked the right door at the beginning. Therefore, if you switch every time, you will win 2/3rds of the time.

It's easier to picture if you change the total number of choices, but keep the principle the same.

Suppose you have a deck of 52 cards and you have to pick the ace of spades. You select a card, leaving 51 others. Then, 50 cards from that group are revealed, none of which are the ace of spades. Now, which card is more likely to be the ace of spades, the one in your hand or the one that's left?

slackster
LSU Fan
Houston
Member since Mar 2009
6943 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

this only holds water if the announcer knows which one and always opens a goat door.

This.

As far as the "Deal or No Deal" question, I guess you can turn it into the goat/goat/car scenario in the video. 26 suitcases, 1 contains \$1M, 25 contain goats. After you pick your suitcase, there is a 3.85% chance you have picked the \$1M. There is a 96.15% chance the \$1M is in the lot of suitcases still remaining. If, at the end of the game, the lot is dwindled down to 1 suitcase, it doesn't change the fact that there was a 96.15% chance that the \$1M was in the lot, so I say you switch.

This is all much different than a roulette wheel with 10 red turns in a row. The ENTIRE lot is always in play and is independent of the last turn. With "Deal or No Deal" or the Monty Hall game, the next turn is dependent on the previous one. 96.15% of the time, you did not pick the \$1M suitcase. That is true even when it is down to the \$.01 and the \$1M suitcase left.

Replies (0)
Replies (0)
Nuts4LSU
LSU Fan
Washington, DC
Member since Oct 2003
18919 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

Wavefan

I take it the math department at Tulane isn't their strongest?

threeputt
Notre Dame Fan
God's Country
Member since Sep 2008
21743 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

The one in your hand or the ones that left

Its a 50/50 chance at that point. A ton of people here are giving the right answer but have no clue why they are correct

Replies (0)
Replies (0)
Catman88
LSU Fan
Baton Rouge, LA
Member since Dec 2004
40845 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
How was this computer "simulation" set up? I cant find that in this thread. I very rarely trust things like this done via excel.

Wavefan
Tulane Fan
St. Tammany
Member since Mar 2005
65 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
There was 2/3 chance you were wrong to begin with. That means that if you picked door number 1 there is a 2/3 chance that the car is actually behind door 2 OR door 3. When Monty opens door 3 to show you the burger, that doesn't make the odds 2/3 that the car is behind door number 2. Rather, by eliminating a choice, Monty CHANGES the odds to 50/50 on the two unopened doors. But the odds are still evenly distributed among the remaining doors. You do not increase your odds by switching.

bigblake
Member since Jun 2011
1120 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
(no message)
This post was edited on 5/4 at 11:03 am

threeputt
Notre Dame Fan
God's Country
Member since Sep 2008
21743 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Your odds are the same unless the host knew where the prize was located with three boxes or a million.

The Easter Bunny
Georgia Tech Fan
Member since Jan 2005
42556 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

There was 2/3 chance you were wrong to begin with. That means that if you picked door number 1 there is a 2/3 chance that the car is actually behind door 2 OR door 3. When Monty opens door 3 to show you the burger, that doesn't make the odds 2/3 that the car is behind door number 2. Rather, by eliminating a choice, Monty CHANGES the odds to 50/50 on the two unopened doors. But the odds are still evenly distributed among the remaining doors. You do not increase your odds by switching.

You're wrong. Just have a random number generator select 1000 integers from 1-3 inclusive and see how many times you would win if you always picked Door #1. I'll give you a hint, it isn't 50%

LSU Fan
Baton Rouge
Member since Dec 2005
16353 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

It's easier to picture if you change the total number of choices, but keep the principle the same.

Suppose you have a deck of 52 cards and you have to pick the ace of spades. You select a card, leaving 51 others. Then, 50 cards from that group are revealed, none of which are the ace of spades. Now, which card is more likely to be the ace of spades, the one in your hand or the one that's left?

FOr the first time, I understand why switching gives you the best odds.

lsutgrfan10
LSU Fan
Member since Jun 2011
212 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Think of it like this: 3 Doors

Behind Door 1 is a goat
Behind Door 2 is a goat
Behind Door 3 is a car

Switching Situations:
- You pick Door 1, host removes Door 2, you switch, you win.
- You pick Door 2, host removes Door 1, you switch, you win.
- You pick Door 3, host removes Door 1 or 2, you switch to the other, you lose.
SUCCESS RATE - 66.7%

OR

Don't Switch Situations:
- You pick Door 1, host removes Door 2, you don't switch, you lose.
- You pick Door 2, host removes Door 1, you don't switch, you lose.
- You pick Door 3, host removes Door 1 or 2, you don't switch, you win.
SUCCESS RATE - 33.3%

Once the host removes one choice as the non-car that gives you the ability to double your chance of winning the car by switching.

Replies (0)
Replies (0)
Catman88
LSU Fan
Baton Rouge, LA
Member since Dec 2004
40845 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
quote:

Just have a random number generator select 1000

Is this how the simulation was done? Computers have a real problem really creating random numbers. RAND in general has a history of doing a shite job at it even MSFT admits it. Just wanted to point out that computers random number generators often suffer from patterns.

Replies (0)
Replies (0)
Wavefan
Tulane Fan
St. Tammany
Member since Mar 2005
65 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Yes. In that case you do. The difference is that the choices were more than three and each time one is eliminated by the host KNOWING it is not THE winner the chances that the winner is not what you picked increased. In your example, by "talking" your choice down to two, there is a higher percentage that the one remaining option you didn't pick is the right option. But with three choices, what I said above works. The given initial options are one car and two burgers. When you pick, if you pick the car, there are two burgers left and if you pick a burger there is one burger left. Monty opens one door to show you the burger he knows is behind it. There is 50/50 chance that you cannot improve by gamemanship that the car is behind either the door you initially picked or the door Monty did not open. Switching only improves your odds when the process of elimination involves larger numbers of choices and the elimination of incorrect choices is a stated rule.

Replies (0)
Replies (0)
iheartlsu
USA Fan
70820
Member since Sep 2005
25787 posts
Online

re: Computer Simulations from yesterday's Let's Make a Deal problem.

Replies (0)
Replies (0)
threeputt
Notre Dame Fan
God's Country
Member since Sep 2008
21743 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
But hes not correct
This post was edited on 3/6 at 1:57 pm