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re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by threeputt on 3/6/13 at 3:31 pm to NaturalBeam)

quote:

t does matter. If you started with a $1M case and 25 $1 cases, you have a 3.8% chance of picking the $1M case. If (big if) you then eliminated 24 other cases and were still left with the $1M, you still only have a 3.8% chance of having picked the $1M initially. You had a 96.2% chance of picking a $1 case.

If you have 26 unique cases, eliminated 24 of them, and are left with 2 amounts ($1 and $1M), it doesn't matter if you switch b/c you had only a 3.8% chance of picking $1M AND you only had a 3.8% chance of picking $1.

no just no

This post was edited on 3/6 at 3:32 pm

quote:

. So even if I've narrowed it down to where I have a 50% chance that I chose correctly I'm switching just based on the fact that when I originally chose there was a small chance I picked the right one. In reality it doesn't matter I guess

quote:

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?

No, wrong. If you choose a burger at first (2 out of 3 games), then Monty WILL choose the other burger, so the car WILL be behind the unopened door 2 out of 3 games. We are not shifting the initial 1/3 chance to the other door, we are leaving the 1/3 chance for a 2/3 chance since EVERY time we are wrong, Monty will essentially show us the right door.

Now, if Monty doesn't know where the car is and there is a 1/3 chance that he reveals it, then your odds depend on how the game is played at that point. If you just lose when this happens, then you will lose 1/3 by switching, lose 1/3 by Monty revealing the car, leaving an overall 1/3 win rate by switching. The same as if you don't switch.

If, instead, the game is void and you get to play again, THEN your overall chance of victory becomes 50/50. Here's how:

You play 36 games

12 games you choose the car at first, Monty reveals a burger, and you switch to the other burger and lose

24 games you choose a burger at first, 12 of those he reveals the other burger and you switch to the car, and the other 12 he reveals the car and the games are replayed

At this point your record is 12 wins, 12 losses. We play the 12 voided games again

4 of these you choose the car at first, he reveals a burger you switch and lose

8 of them you choose a burger first, 4 he reveals the other burger you switch and win, the other 4 he reveals a car and play again

Now your record is 16 wins, 16 losses, and we have 4 games to replay.

It should now be easy to see that the only way your overall odds of winning are 50% is if the game is replayed when Monty reveals the car, which would generally be a LOSS because it's as if he doesn't give you the option to switch at all. I think we can agree that a game will never be void and replayed, and if Monty doesn't know where the car is then your overall chance of winning will be 33%. 1/3 games Monty will reveal the car and you lose, and of the remaining 2/3 your odds will be 50/50, which of course 50% of 2/3 is 1/3, switch or no switch.

So, finally, if Monty is clueless you will win 1 out of 3 games whether you switch or not. If Monty knows where the car is, then you will win 1/3 by staying pat, and 2/3 by switching. The ONLY way to get your odds to 50/50 would be if games are replayed when Monty screws up, which would never happen.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by NaturalBeam on 3/6/13 at 3:33 pm to threeputt)

I'm right bro

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by OneMoreTime on 3/6/13 at 3:34 pm to threeputt)

Actually I just ran the numbers because I'm procrastinating, and it's exactly the same odds

ETA: frick math

ETA: frick math

This post was edited on 3/6 at 3:36 pm

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by NaturalBeam on 3/6/13 at 3:35 pm to Korkstand)

We've moved on from Monty. We're on to Howie.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by Tiger1242 on 3/6/13 at 3:36 pm to OneMoreTime)

quote:

Actually I just ran the numbers because I'm procrastinating, and it's nearly the same odds

How nearly? (and which "gameshow" principal are you running numbers with?)

quote:

No I know it's 50/50, but when I first chose there was a 96% chance I chose wrong. So even if I've narrowed it down to where I have a 50% chance that I chose correctly I'm switching just based on the fact that when I originally chose there was a small chance I picked the right one. In reality it doesn't matter I guess

But, there's like a 4% chance you picked 24 straight losers. Either way, you displaying a lot of fail.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by OneMoreTime on 3/6/13 at 3:38 pm to Tiger1242)

quote:I went back and checked and saw I forgot a number. It's exactly the same odds of you picking the case right off the bat and of you eliminating 24 $1 cases with the million out there.

How nearly? (and which "gameshow" principal are you running numbers with?)

Deal or no deal

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by Korkstand on 3/6/13 at 3:43 pm to NaturalBeam)

quote:

We've moved on from Monty. We're on to Howie.

Sorry, took a while to type all that.

Howie is easy. The odds of choosing right out of 26 from the start is 1/26 =

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 =

Switch cases or not, doesn't matter.

quote:

But, there's like a 4% chance you picked 24 straight losers. Either way, you displaying a lot of fail

Not sure how I demonstrated any fail, I was just saying I would switch and trying to back up why with my theory. I recognized it's a 50/50 chance so really it doesn't matter but I would still switch, I dk how that's failing

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by OneMoreTime on 3/6/13 at 3:44 pm to Korkstand)

quote:whew. looks like i did it right

Howie is easy. The odds of choosing right out of 26 from the start is 1/26 = 0.03846

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

quote:

Yes there are two left and they have the same odds. But a large factor of the case you picked originally still being there is because it's the one you picked originally. It is very unlikely you picked correctly originally, therefore switching is the smart decision

Er, no.

The odds are the same with the final two suitcases whether you switch or not. That's easy to see if you just imagine that you had picked the OTHER one, instead. You'd be in the exact same boat.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by lsutgrfan10 on 3/6/13 at 3:58 pm to OneMoreTime)

Yes, you have the same odds of picking the million dollar case first out of 26 cases as you do for elimininating 25 cases without picking the million dollar case.

(1/26)=(25!/26!)

MATH

(1/26)=(25!/26!)

MATH

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by LSUBoo on 3/6/13 at 4:19 pm to OneMoreTime)

quote:

Actually I just ran the numbers because I'm procrastinating, and it's exactly the same odds ETA: frick math

I was going to say that, but didn't feel like running the numbers.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by Rex on 3/6/13 at 4:22 pm to lsutgrfan10)

quote:

Yes, you have the same odds of picking the million dollar case first out of 26 cases as you do for elimininating 25 cases without picking the million dollar case.

That's easy to see by just offering somebody two choices:

"Hey, Joe, pick the million dollar case out of these 26 cases/"

or

"Hey, Joe, eliminate all the 25 cases out of these 26 that doesn't contain the million dollars."

It's the exact same task worded two different ways.

quote:

The odds of choosing right out of 26 from the start is 1/26 = 0.03846 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 Switch cases or not, doesn't matter.

Exactly.

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