Last edited by Decourcy; 06-12-2022 at 06:14 PM.
But also, if you intend to pull the trigger 6 times without spinning the cylinder, the the odd are 6 in 6 (100%) that you will shoot yourself by the time you are you reach the 6th pull if not sooner. Your example looks at the probability for a single pull while counting how many many pulls have already beet taken.
Going back to Monty. The contestant makes one choice and has a 33% chance Monty had two doors - 66% chance. The contestant does not open the door until Monty has opened both of his doors. That means there is a 66% chance that one of his doors is the car. If he does not trade either door with the contestant, then he still has 66% chance. His chances haven't changed, because he has not changed either of the two doors in his possession. But, his odds don't change after he has opened the first door assuming it is a goat because we are saying it is a 66% chance that one the doors has a car. It could be either one and ther ren=mains a 33% chance that neither door has the car: i.e. the contestant is the winner.
It really is quite difficult to build an ugly wooden boat.
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I just played the game 20 more times using the "always switch" strategy. The contestant scored 14 wins and 6 losses. A winning percentage of 70%.
I have now played the game 80 times with the contestant scoring 53 wins and 27 losses for a winning percentage of 66.25%.
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
...
Simpler is better, except when complicated looks really cool.
I have ignored all Russian Roulette posts, sorry, that's my quirk. When I was in high school the boyfriend of a girl in my class killed himself by actually playing Russian roulette when drunk one night. I can't hardly stand to read the term in a small town, that caused a huge amount of trama.
Yeah. But I rather enjoyed playing.Originally Posted by Paul Pless
I just completed 100 plays of the game employing the "always switch" strategy.
The first twenty plays resulted in 10 contestant wins and 10 losses for a winning percentage of 50%.
The second twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.
The third twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
The fourth twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
The fifth twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.
The final total was 100 plays resulting in 68 contestant wins and 32 losses for a winning percentage of 68%.
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
I may have misunderstood your earlier post. If you find the ace in the two cards held by Monty, then I will agree that's a win for switch.
I just played it 45 times and it came out closer to 50:50, but in the process I realized that it's actually difficult to shuffle 3 cards well. I started seeing patterns in the outcome that I'm not sure are random.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
Suppose you had only two choices to begin with. You'd have a 50/50 chance. Now a third choice is added before you choose. Now you've a 1 in 3 chance, no?
Once the 3rd choice is eliminated, it's no longer a factor as to odds. The host may have some sort of established pattern. If the prize is behind the even numbered curtain, he eliminates one of the odd numbers, and if that's known by the contestant, then the contestant is more likely to choose the even numbered curtain. That would show up in long term results.
"alternative facts (lies)" are a cancer eating through a democracy, and will kill it. 1st amendment is not absolute.
I’m seeing the ‘problem’ as a paradox of time, with both arguments being correct.
Looked at as a singular event/game sample, the odds at the second choice – to switch, or not – are 50-50. However, with increased frequency of the events/samples, over time, say, playing 100 games, it would seem to be logical to choose the switch, because statistically the first choice would have a 66% probability of being wrong. Therefore, the ‘correct answer’ is to switch if playing multiple games, but in a singular instance it’s 50-50.
But, there is randomness, intuitiveness, woo-woo, in the first selection – how do you put numbers to that?
Prolly not.
When I shared the puzzle with the mob at dinner, Son1 said - quickly and with no hesitation - 'switch, every time'. He said he visualized the 2/3 chance being compressed into the 'switch' door in terms of quantum physics. He tried to explain it... but it was a leap too far, for me. I suspect he was wrong also. Or correct for the wrong reasons.
David G
Harbor Woodworks
https://www.facebook.com/HarborWoodworks/
"It was a Sunday morning and Goddard gave thanks that there were still places where one could worship in temples not made by human hands." -- L. F. Herreshoff (The Compleat Cruiser)
Wikipedia has a good page on the problem. https://en.wikipedia.org/wiki/Monty_Hall_problem
When you first choose a door, you have a 2 in 3 chance of being wrong. Even after seeing in the second door, you still have the 2 in 3 chance of being wrong. Opening the door doesn't improve your chances if you stick with your first choice..
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Twelve pages over the course of 32 hours!
When I started this thread yesterday I had no idea….
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
About 10 or 12 pages in the middle, i didn't bother to read.
But some seem to still not believe that.
Don’t expect much, and you won’t be disappointed…
This is amazing to me, that people don't understand this.
Here's a simulator I wrote in the Go programming language: https://goplay.tools/snippet/l-0dNVMAhlq
It's in the [a] Go Playground, so you can compile the code and run it. Currently set up to do runs of 1,000,000 plays of the game.
The rules of the game are
- There are 3 doors. One of them a valuable prize behind it (a car!). The other two has worthless prizes (goats!).
- The contestant selects a door. It remains closed.
- Monty Hall, the preceptor of the game show, opens one of the two remaining doors. It cannot be the door containing the valuable prize, as that would ruin/end the game. It's always a worthless prize.
- The contestant is offered the chance to switch to the remaining door, and can either switch, or hold.
The question is, which strategy is more effective? Switch to the last door?, or holding on to your initial choice?
When the contestant first makes their selection, they have a 1/3 chance (1:2 odds) of selecting the door with the car behind it, and a 2/3 chance (2:1 odds) of selecting a door with a booby prize behind it.
After Monty Hall has opened the door, and revealed the booby prize behind it, he has taken 1 booby prize out of contention, because his pick was not random.
That makes the remaining door coalesce into a 1/2 chance (even odds) that it is the valuable prize, because it's either behind that door, or it's behind the door selected by the contestant. And the door selected by the contestant is still sitting there with a a 1/3 chance of it being the valuable prize.
It's not so much a probability & stats problem as it is
Here's output from a sample run:
This is the guts of the thing:Code:Strategy: Switching +---------+-----------+---------+ | OUTCOME | FREQUENCY | PERCENT | +---------+-----------+---------+ | Win | 666164 | 66.616 | | Lose | 333836 | 33.384 | +---------+-----------+---------+ | 1000000 | 100.000 | +---------+-----------+---------+ Strategy: Holding +---------+-----------+---------+ | OUTCOME | FREQUENCY | PERCENT | +---------+-----------+---------+ | Win | 332717 | 33.272 | | Lose | 667283 | 66.728 | +---------+-----------+---------+ | 1000000 | 100.000 | +---------+-----------+---------+
Code:func main() { // Our sample size is 1,000,000 plays of the game const sample = 1_000_000 // for each possible strategy (switching or holding) for _, strategy := range []string{SWITCHING, HOLDING} { // zero the counters win, lose, total := 0, 0, 0 // play the game, over and over... for n := 0; n < sample; n++ { // play the game didWin := play(strategy) // record the outcome as either a win or a loss if didWin { win++ } else { lose++ } // record the total number of games played (wins+losses) total++ } // report the outcome. This just formats it as a nice table. tbl := tablewriter.NewWriter(os.Stdout) tbl.SetAutoMergeCells(true) tbl.SetHeaderLine(true) tbl.SetHeader([]string{"Outcome", "Frequency", "Percent"}) tbl.SetColumnAlignment([]int{tablewriter.ALIGN_LEFT, tablewriter.ALIGN_RIGHT, tablewriter.ALIGN_RIGHT}) tbl.SetFooterAlignment(tablewriter.ALIGN_RIGHT) tbl.Append(row("Win", win, total)) tbl.Append(row("Lose", lose, total)) tbl.SetFooter(row("", total, total)) // render the table with a nice header to tell us which strategy was in use. fmt.Println() fmt.Printf("Strategy: %s\n", strategy) tbl.Render() } return } func play(strategy string) (didWin bool) { // populate the doors. The car is placed behind door at random and // every other door has a goat behind it. doors := []string{GOAT, GOAT, GOAT} doors[randomDoor()] = CAR // The contestant selects a door at random from the 3 available doors. contestantSelection := randomDoor() // Monty Hall opens a door that is neither the winning door // or the door selected by the contestant. // // We don't really care what door he opened, as that is guaranteed // not to be of any interest (he's not going to open the winning door), // but we do care about the single door that is neither selected by the // contestant or opened by Monty Hall. _, remaining := montyHallOpensDoor(doors, contestantSelection) // if our strategy is switching, we switch to the remaining door, // otherwise we stand pat. if strategy == SWITCHING { contestantSelection = remaining } // The contestant wins if his selected door has the car behind it. didWin = doors[contestantSelection] == CAR // return the win/lose indicator to the caller return didWin }
You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)
This diagram show why this is true.
You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)
From a link to a Reddit page in the link that Tom first posted - it sort of flips the problem on it head, and made more sense of it to me than all the charts and tables .
Quote:
To me, the clearest explanation is that the only way to get it wrong by switching is to have picked the correct door in the first place. The odds of picking the correct door first are 1 in 3.
Pete
The Ignore feature, lowering blood pressure since 1862. Ahhhhhhh.
I have laid it out in the form of a white board, as in Toms video above.
The error that the statistician with his sharpie made is that he did not model all the facts in his working. He ignored that we know that Monty's door ALWAYS has O probability of revealing the car. The probability that the car is behind the other two doors is equal, we simply don't know which door.
This is how it should be worked, without the waffle.
White board.jpg
I hope that is clear enough.
Last edited by Peerie Maa; 06-13-2022 at 04:21 AM.
It really is quite difficult to build an ugly wooden boat.
The power of the web: Anyone can post anything on the web
The weakness of the web: Anyone can post anything on the web.
I will change my answer to 'switch'.
Contestant's chances to win are 1 in 3
Monty's chances are 2 in 3.
If you switch,you get Monty's chances ,minus the door that he opened.
Not unlike Monty allowing you to pick 2 doors.
R
Sleep with one eye open.
It really is quite difficult to build an ugly wooden boat.
The power of the web: Anyone can post anything on the web
The weakness of the web: Anyone can post anything on the web.
Nick. How does Monty's removal of on of the doors that you did not pick retroactively change the chance of you choosing the right door first, from 1/3 to 1/2 ?
I'll admit not having read through all 14 pages, so sorry if I'm just repeating an already stated point (which I probably do)
What is wrong is the assumption that all options are equally plausible. Corrected version:
Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Scenario probability Stays Switches Weight 1 1 2 1/6th Win Lose .5 1 1 3 1/6th Win Lose .5 1 2 3 1/3rd Lose Win 1 1 3 2 1/3rd Lose Win 1 cut for brevity other scenarios analytically redundant Weighted win ratio 2/3rds 1/3rd 3
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There are lots of ways to fiddle the math to fit the desired outcome.
But you need to justify why you chose those numbers. Why is the probability of any pick a variable. The probability of picking 1 from 3 is always the same.
It really is quite difficult to build an ugly wooden boat.
The power of the web: Anyone can post anything on the web
The weakness of the web: Anyone can post anything on the web.
So if your original pick remains at 1/3, the probability of the prize being one of the other two choices must be 2/3 and you now know not to pick one of those.
If, as you assert, that it was not an advantage to change now the probability of either staying or changing must be 1/2. Which means that the probability of you originally making the correct choice must have changed, which is obviously nonsense
I was in the 50/50 camp when I first saw the problem but now would switch.
Putting aside all the trees and statistics, the nub of the issue is what do you know after Monty opens the door that you did not know before. Answer nothing, Monty will always have the option to open a door with a goat therefore nothing has changed.
There are 2 sets, my first choice 33% the other 2 doors 66%. Monty selecting a goat tells us nothing about what is in the second set but there is still a 66% chance that the second set had the car.
Were Monty to open an other door at random and picked a goat then it would be 50/50.
Apologies to all posters who have already made the point, I do not have time to read all 470 replies.
Good point, I should've explicitly written that:
Only when a contestant picks correct gate, which will happen in 1 of every 3 games, Monty has to choose a goat to be revealed. Simply put, the scenarios 1-1-2 and 1-1-3 have to "share" the same 1/3rd chance of happening. They are, naturally, equivalent, meaning 50% chance of happening *within* the initial 1/3rd thus being half as likely as the other two scenarios where Monty has no choice in which goat to reveal.
If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.
This is the key: he's not. There is 0% chance he'll show you the car. This is the introduced bias.
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But but but. . . Monty always has to open a door with a goat, whatever the contestant picks. So the probabilities are the same whichever the contestant picked, car or goat.
Which is what I meant, choosing numbers to agree with a desired outcome.If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.
It really is quite difficult to build an ugly wooden boat.
The power of the web: Anyone can post anything on the web
The weakness of the web: Anyone can post anything on the web.