Haha, man this is so frustrating that Flip it will not accept the correct solution. TEST IT YOURSELF! You can easily gather the evidence and find that it is NOT 50/50!! It is an interesting problem because most people want to say 50/50 at first.
Haha, man this is so frustrating that Flip it will not accept the correct solution. TEST IT YOURSELF! You can easily gather the evidence and find that it is NOT 50/50!! It is an interesting problem because most people want to say 50/50 at first.
Because they are thinking about the red/green ball problem as
trial 1: three possibilities of three
trial 2: two possiblities of two
When the "right" answer is:
trial 1: three possibilities, two possibilities of three.
Some math is full of stupid tricks like that.
the 100 door scenario is not the same as the 3 door scenario although it looks the same.
put it this way. 100 doors, you choose one, monty opens one door with a goat. do you switch or stay the same?
poker face wrote:
the 100 door scenario is not the same as the 3 door scenario although it looks the same.
put it this way. 100 doors, you choose one, monty opens one door with a goat. do you switch or stay the same?
I switch because my chances just doubled from 1/100 to 2/100.
DiscoGary wrote:
poker face wrote:the 100 door scenario is not the same as the 3 door scenario although it looks the same.
put it this way. 100 doors, you choose one, monty opens one door with a goat. do you switch or stay the same?
I switch because my chances just doubled from 1/100 to 2/100.
Doubled? Really?
DiscoGary wrote:
poker face wrote:the 100 door scenario is not the same as the 3 door scenario although it looks the same.
put it this way. 100 doors, you choose one, monty opens one door with a goat. do you switch or stay the same?
I switch because my chances just doubled from 1/100 to 2/100.
How did you get 2 chances to win with only one car?
Flip it wrote:
DiscoGary wrote:I switch because my chances just doubled from 1/100 to 2/100.
How did you get 2 chances to win with only one car?
The same way the original chances went from 1/3 to 2/3 when Monty opened one of the bad choices. Every door taken out of the game increases the chances that the remaining doors have the car behind them.
By the time Monty had opened 98 bad doors the chances that the last door has the car are 99/100. Still, only one car.
DiscoGary wrote:
poker face wrote:the 100 door scenario is not the same as the 3 door scenario although it looks the same.
put it this way. 100 doors, you choose one, monty opens one door with a goat. do you switch or stay the same?
I switch because my chances just doubled from 1/100 to 2/100.
You can have a seat at my poker table anytime. Bring lots of money. :)
DiscoGary's numbers are correct. Some of you need to take an intro prob and stats course and then get back to us.
Are you serious, Disco Gary?One door out of 100 was opened and you doubled your chances of winning?You went from 1/100 to 1/99 (99 doors are still closed). And you think that is doubling your chances?
Of course it is wrote:
DiscoGary's numbers are correct. Some of you need to take an intro prob and stats course and then get back to us.
You pick door 1. Monty has doors 2 and 3. There are 3 possibilities for Monty's doors:
1) Door 2 has a goat. Door 3 has a goat.
- Monty opens either door. Switch and get the other door with a goat, you lose
2) Door 2 has a goat. Door 3 has a car.
- Monty must open door 2. Switch and you get door 3 with the car, you win
3) Door 2 has a car. Door 3 has a goat.
- Monty must open door 3. Switch and you get door 2 with the car, you win.
Oh no! Another thread on Powerball hijacked into Monty Hall. Not the first time.
http://www.letsrun.com/forum/flat_read.php?board=1&thread=567711&id=606604#606604
Mathwise, I think the percentage odds of success by switching are calculated as follows:
( (n-1) / (n*x) )*100%
where n is the number of possibilities in a 1 in n chance, x is the number of doors Monty has left to open.
Monty always starts out with 99/100 chances of having the car, because he has 99 of the doors. Any particular door has the above odds of being a winner. If Monty opens no doors, thus having 99 doors to open, it should be 1/100, or 1% chance of any particular one of his doors holding a car.
((100-1)/(100*99))*100% = 1% chance.
If Monty opens one goat door, he has 98 doors left. Your odds of picking the car from Monty now are:
((100-1)/(100*98))*100% = 1.01%, slightly better than if you stay with your door. You can only have a 1% chance of winning if you stay.
If Monty opens 2 doors:
((100-1)/(100*97))*100% = 1.02%. A slight increase, but not double. You don't get double the odds when Monty opens another door.
If Monty opens 50 doors and has 49 left, the odds of picking the car by switching are now:
((100-1)/(100*49)*100% = 2.02%. Now you've ~doubled your chances of success by switching, but Monty had to open ~50 doors before you get there. In other words, Monty has to reveal ~half of his available doors to double your chance of success. This makes sense.
If Monty opens 98 doors and leaves 1, the odds of winning by switching are:
((100-1)/(100*1)) * 100% = 99%. You have a 99% chance of success by switching.
If Monty only opened 97 doors and leaves 2, the odds of picking the winning door by switching are:
((100-1) / (100*2)) * 100% = %49.5.
think folks think wrote:
You pick door 1. Monty has doors 2 and 3. There are 3 possibilities for Monty's doors:
1) Door 2 has a goat. Door 3 has a goat.
- Monty opens either door. Switch and get the other door with a goat, you lose
2) Door 2 has a goat. Door 3 has a car.
- Monty must open door 2. Switch and you get door 3 with the car, you win
3) Door 2 has a car. Door 3 has a goat.
- Monty must open door 3. Switch and you get door 2 with the car, you win.
How could anyone argue against this logic? This was also a CarTalk Puzzler years ago.
No. No. No. I doubled my chances when open one door.
think folks think wrote:
Mathwise, I think the percentage odds of success by switching are calculated as follows:
( (n-1) / (n*x) )*100%
where n is the number of possibilities in a 1 in n chance, x is the number of doors Monty has left to open.
Monty always starts out with 99/100 chances of having the car, because he has 99 of the doors. Any particular door has the above odds of being a winner. If Monty opens no doors, thus having 99 doors to open, it should be 1/100, or 1% chance of any particular one of his doors holding a car.
((100-1)/(100*99))*100% = 1% chance.
If Monty opens one goat door, he has 98 doors left. Your odds of picking the car from Monty now are:
((100-1)/(100*98))*100% = 1.01%, slightly better than if you stay with your door. You can only have a 1% chance of winning if you stay.
If Monty opens 2 doors:
((100-1)/(100*97))*100% = 1.02%. A slight increase, but not double. You don't get double the odds when Monty opens another door.
If Monty opens 50 doors and has 49 left, the odds of picking the car by switching are now:
((100-1)/(100*49)*100% = 2.02%. Now you've ~doubled your chances of success by switching, but Monty had to open ~50 doors before you get there. In other words, Monty has to reveal ~half of his available doors to double your chance of success. This makes sense.
If Monty opens 98 doors and leaves 1, the odds of winning by switching are:
((100-1)/(100*1)) * 100% = 99%. You have a 99% chance of success by switching.
If Monty only opened 97 doors and leaves 2, the odds of picking the winning door by switching are:
((100-1) / (100*2)) * 100% = %49.5.
I think that all of us who understand the problem need to accept that no matter how clearly we explain it some people simply will not accept the answer.
Disco Gerry wrote:
No. No. No. I doubled my chances when open one door.
he was talking about 100 doors. The Monty Hall problem has 3 doors.
logic and reading wrote:
Disco Gerry wrote:No. No. No. I doubled my chances when open one door.
he was talking about 100 doors. The Monty Hall problem has 3 doors.
While I don't think he understands this solution, I did say that the odds jumped from 1/100 to 2/100 when one door opened. That was wrong. I knew the chances were 99/100 with only one of Monty's doors left. I failed to see how the odds would not increase linearly, but would increase hugely right as the last few doors were opened.
It still makes sense to switch.
Flip it wrote:
Wrong. 2 Doors left. One has it, one doesn't. 50/50.
Flip it, I am not a math/stats guy, but this really is more a logic than fancy statistics category. You may be missing the implied scenario that the host is never going to open the door with the prize behind it.
2/3 of the time you will initially pick a loser door. That means 2/3 of the time, Monty only has one door to open, meaning in those situations (which are 2/3 of the scenarios) you win by switching doors. The problem hinges on the fact that Monty never chooses the door with the prize behind it.
Door A is the winner
Scenario 1: You pick Door A, you switch your pick and lose.
Scenario 2: You pick Door B, Monty can only open Door C. You switch and win.
Scenario 3: You pick Door C. Monty can only open Door B. You switch and win.
Those are the only 3 scenarios, and in 2/3 of them you win by switching.
logic and reading wrote:
Disco Gerry wrote:No. No. No. I doubled my chances when open one door.
he was talking about 100 doors. The Monty Hall problem has 3 doors.
The 'double your odds' by opening 1 door only works in the 3 door scenario. The formula I gave is general so just make n = 3. If Monty opens no doors, he has 2 doors closed:
((3-1)/(3*2))*100% = %33.333...
If Monty opens a door:
((3-1)/(3*1))*100% = %66.666...
When you add more doors, it no longer doubles the odds. Make n=4, Monty has 3 doors to your 1. Your starting odds are %25. No open doors:
((4-1)/(4*3))*100% = %25
Monty opens a door:
((4-1)/(4*2))*100% = %37.5
You didn't double when Monty opened 1 door
You do double the odds when Monty opens the last door:
((4-1)/(4*1))*100% = %75
but it only applies on the last door
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