Page First 7 8 9 10
Posted by
Message
Dooshay
LSU Fan
CEBA
Member since Jun 2011
29879 posts

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

The odds of eliminating 24 out of 25 losers in a row is 24/25 * 23/24 * 22/23 * ... * 1/2 = 0.0400 * 0.9615 chance of actually starting with the winner in the group = 0.03846

Unfortunately, that's incorrect.

If they were independent choice, that would be accurate, but the next choice is dependent on the fact that you chose correctly previously, so it's a conditional probability.

Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

If they were independent choice, that would be accurate, but the next choice is dependent on the fact that you chose correctly previously, so it's a conditional probability.

I know, that's why I said "in a row" (implies the condition that all previous choices were "right"), and that's the reason why the math I did was correct.

Dooshay
LSU Fan
CEBA
Member since Jun 2011
29879 posts

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

Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

It is incorrect.

You've said that twice now without offering the "correct" math. My math works out to the right answer, and has been verified by multiple posters.

It is correct.

Dooshay
LSU Fan
CEBA
Member since Jun 2011
29879 posts

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

Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

Nope

I'll take this as a troll and ignore you until you explain your flawed thinking.

Cap Crunch
LSU Fan
Member since Dec 2010
50775 posts
Online

re: Computer Simulations from yesterday's Let's Make a Deal problem.
I'm bumping this thread because I need something explained to me

I understand the Monty Hall problem and why it gives you a better chance if you switch. For the most part, I'm also grasping why it doesn't matter in the deal or no deal scenario.

My question is why does it matter if the host knows or not? If hypothetically he opens a door and its a goat, how would that be any different than if he opened the door knowingly and its a goat?

Now in the real world, it makes sense that he'd only open another door if he knew what was behind it. I'm asking in purely a hypothetical sense.

Though this also pertains to the Deal or No Deal question. So far the only difference that I've seen is that it isn't a host removing cases knowing what's behind them, instead its randomly by the person playing

WDE24
Wright St. Fan
Member since Oct 2010
45226 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Because the host knows where the prize is/was when you made your original selection. His targeted elimination of the options means that the original choice odds matter. It is your pick vs. the field.

He intentionally left two choices (whether you started with 3 or 100) a right one and a wrong one. Your original choice still only had the odds of being right 1 out of how ever many options there were. The other one has the original odds of being right every time except the 1 you picked.

Poorly worded, but I hope that makes sense.

Cap Crunch
LSU Fan
Member since Dec 2010
50775 posts
Online

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Hmm. I kinda get it. I guess I'm just saying it seems like its possible for everything to fall into place the same way even if he didn't know what was behind each door

Replies (0)
Replies (0)
00
TheWalrus
LSU Fan
Memphis
Member since Dec 2012
15838 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Kind of how I explained it.

Replies (0)
Replies (0)
00
PepaSpray
Alabama Fan
Member since Aug 2012
10754 posts

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

In that scenario, you will only get the goat 1/3 of the time when you pick the car first. But 2/3 of the time, you will pick the goat first and will automatically end up with the car if you switch

damn world of conundrums, huh?

Replies (0)
Replies (0)
00
Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

My question is why does it matter if the host knows or not? If hypothetically he opens a door and its a goat, how would that be any different than if he opened the door knowingly and its a goat?

The only reason the odds change is because there is a chance he reveals the car. If you count it as a win if Monty reveals the car, then your odds are still 2/3 when switching. If you count it as a loss, your odds are 1/3 whether you switch or not.

quote:

Though this also pertains to the Deal or No Deal question. So far the only difference that I've seen is that it isn't a host removing cases knowing what's behind them, instead its randomly by the person playing

The choosing of the case, the opening of cases one by one, and the switch at the end are all just theatrics. In the end, it's just a roundabout way to ultimately choose 1 case out of 26. Your odds stay the same just like if Monty opens a random door with the chance of revealing the car, and counting that as a loss.

Replies (0)
Replies (0)
00
foshizzle
LSU Fan
Washington DC metro
Member since Mar 2008
36304 posts

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

why does it matter if the host knows or not?

Suppose host Monty has no clue. This introduces a new possibility, namely that Monty unwittingly picks the car himself. Now it doesn't matter whether you switch or not because you can't win.

Scenario 1 - your first choice really does have the car
Variation a - Monty shows you a goat. You should not switch.
Variation b - Monty shows you a goat. You should not switch.
Scenario 2 - your first choice was one of the bad doors
Variation a - Monty has no clue which is which and flings open the door to the car. You lose since you can't pick a winner.
Variation b - Monty shows you the goat. Now you should switch.
Variation a - Monty has no clue which is which and flings open the door to the car. You lose since you can't pick a winner.
Variation b - Monty shows you the goat. Now you should switch.

There are six scenarios. You should switch in two, shouldn't switch in two, and it doesn't matter in two since Monty keeps the car.

But usually the way this works is that the host does know. That changes 2a and 3a into cases where you should switch. Then you should switch b/c 4 out of 6 times you'll get the car.
This post was edited on 3/6 at 8:59 pm

Cap Crunch
LSU Fan
Member since Dec 2010
50775 posts
Online

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Hmm I guess that makes sense

slackster
Stanford Fan
Houston
Member since Mar 2009
31752 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Ok, this is my question/conundrum:

-26 unique cases
-\$1M case through \$.01, such as Deal or No Deal
-2 cases left, \$1M and \$.01

If you view it as \$1M case against 25 "bad" cases, you should switch. It doesn't matter how unlikely it is that you got to this point, what matters is the decision you make at this point, which should be to switch. Just as in the Monty Hall problem, whether Monty KNOWS where the car is or not is irrelevant, as long as he picks a goat, you should switch. In this case, you were extremely lucky to get to this point, but you're still here nonetheless - SWITCH.

Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

If you view it as \$1M case against 25 "bad" cases, you should switch. It doesn't matter how unlikely it is that you got to this point, what matters is the decision you make at this point, which should be to switch. Just as in the Monty Hall problem, whether Monty KNOWS where the car is or not is irrelevant, as long as he picks a goat, you should switch. In this case, you were extremely lucky to get to this point, but you're still here nonetheless - SWITCH.

Nope. As unlikely as it is to get to that point, it's exactly as likely as choosing the winning case to begin with (3.8%). It doesn't matter how you end up choosing a case, whether you just choose one and open it, or eliminating them one by one to leave your choice, each case is equally likely to be the winner. The odds of winning do increase as cases are eliminated, but only from a given point in the game. From the start, there is no strategy for improving your overall odds.

slackster
Stanford Fan
Houston
Member since Mar 2009
31752 posts

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

Nope. As unlikely as it is to get to that point, it's exactly as likely as choosing the winning case to begin with (3.8%). It doesn't matter how you end up choosing a case, whether you just choose one and open it, or eliminating them one by one to leave your choice, each case is equally likely to be the winner. The odds of winning do increase as cases are eliminated, but only from a given point in the game. From the start, there is no strategy for improving your overall odds.

If you pick a case, then Howie knowingly eliminates 24 "wrong" cases (not \$1M cases), and you are left with the \$1M case and one case that is NOT the \$1M case, you should switch.

If the Monty Hall set up says you should switch, then you should also switch in this set-up. Period. End of story.

slackster
Stanford Fan
Houston
Member since Mar 2009
31752 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Basically, whether Howie/Monty/You "know" or not, as long as Monty picks a goat, you should switch. As long as Howie/you pick 24 cases that are NOT the \$1M case, you should switch.

Replies (0)
Replies (0)
00
Korkstand
LSU Fan
Plaquemine, LA
Member since Nov 2003
14339 posts

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

If you pick a case, then Howie knowingly eliminates 24 "wrong" cases (not \$1M cases), and you are left with the \$1M case and one case that is NOT the \$1M case, you should switch.

If the Monty Hall set up says you should switch, then you should also switch in this set-up. Period. End of story.

Two different setups, not end of story.

If you read MY posts, you will see that I said that if Monty chooses randomly, your odds remain 1/3 whether you switch or not. This is because Monty will end your game half of the 2/3 games that you would win by switching, leaving you only 1/3 odds to win by switching, 1/3 to lose by switching, and 1/3 to lose by Monty revealing the car and preventing you from switching. The difference is whether Monty knows where the car is or not.

This is easier to understand in deal or no deal. You will pick the right case at the start 3.8% of the time, and if you eliminate cases randomly there is no way to improve those odds. When you get down to 10 total cases, sure, AT THAT POINT you have a 10% chance to win. But the odds of getting there randomly without revealing the winning case is 38%, and 10% of that of course shows your overall odds haven't changed. HOWEVER, if someone who knows where the winning case is does the eliminating, then your chance to get down to 10 cases is 100%, so NOW your overall odds can be improved to 10% by choosing a different case. The odds continue to improve each time you switch cases as long as there is zero chance of losing up until that point in the game.

And here's the kicker: if someone with knowledge is eliminating cases, you can keep switching between THE SAME TWO CASES and improve your odds with each switch, provided neither of them gets eliminated.

Replies (0)
Replies (0)
00
When in Rome
USA Fan
Member since Jan 2011
29662 posts

re: Computer Simulations from yesterday's Let's Make a Deal problem.
Bump...Let's Make a Deal is coming on at 2:00