|
Post by haxemon on Feb 9, 2018 16:48:03 GMT
Since beren44 has me thinking of math puzzles/curiosities, I thought I'd post the Monty Hall problem. The situation is this: You're a contestant on Let's Make A Deal. Monty Hall offers you three giant boxes on the stage, one of which holds a prize, the other two are empty (or given the show, maybe one has a prize and the other two have jokes such as a stack of old newspapers or a goat). Monty knows which box holds the prize of course. After you pick a box (let's say box 1) he first shows you one of the boxes you didn't chose and the joke non-prize inside. Then you're offered the option to stick with the box you chose or switch. Should you switch? There is a right or wrong answer (spoiler tagged below). You actually double your chances if you switch!
If it's just a straight pick and reveal, clearly the odds of winning are 1/3. If you don't switch then you maintain those odds. So let's look at if you always switch.
When you chose box 1, there was a 2/3 chance you were wrong and that box was empty. When Monty removes one of the empty boxes from the game that means there's still a 2/3 chance box 1 is empty by the other remaining box now has a 1/3 chance of being empty (the total has to remain 1/1). Which of course means there's a 2/3 chance the other box has the prize. Switching means you double your odds from 1/3 to 2/3.
The key is in the fact that Monty is removing a wrong answer.
Looking at all possible scenarios if you always pick box number 1 and always switch:
If box #1 has the prize you always lose. If box #2 has the prize Monty will reveal box #3 to be empty and you always win. If box #3 has the prize Monty will reveal box #2 to be empty and you always win.
Switching means you win 2/3 times - ironically only losing when your first guess was correct.
|
|
|
Post by beren44 on Feb 9, 2018 18:07:59 GMT
Since beren44 has me thinking of math puzzles/curiosities, I thought I'd post the Monty Hall problem. The situation is this: You're a contestant on Let's Make A Deal. Monty Hall offers you three giant boxes on the stage, one of which holds a prize, the other two are empty (or given the show, maybe one has a prize and the other two have jokes such as a stack of old newspapers or a goat). Monty knows which box holds the prize of course. After you pick a box (let's say box 1) he first shows you one of the boxes you didn't chose and the joke non-prize inside. Then you're offered the option to stick with the box you chose or switch. Should you switch? There is a right or wrong answer (spoiler tagged below). You actually double your chances if you switch!
If it's just a straight pick and reveal, clearly the odds of winning are 1/3. If you don't switch then you maintain those odds. So let's look at if you always switch.
When you chose box 1, there was a 2/3 chance you were wrong and that box was empty. When Monty removes one of the empty boxes from the game that means there's still a 2/3 chance box 1 is empty by the other remaining box now has a 1/3 chance of being empty (the total has to remain 1/1). Which of course means there's a 2/3 chance the other box has the prize. Switching means you double your odds from 1/3 to 2/3.
The key is in the fact that Monty is removing a wrong answer.
Looking at all possible scenarios if you always pick box number 1 and always switch:
If box #1 has the prize you always lose. If box #2 has the prize Monty will reveal box #3 to be empty and you always win. If box #3 has the prize Monty will reveal box #2 to be empty and you always win.
Switching means you win 2/3 times - ironically only losing when your first guess was correct.
My answer, without looking, is keep it, or trade it..it doesn't matter. The first round, you have a 1/3 chance of winning. The second time, there is a 50/50 chance you win, no matter which one you choose. But I want to think about it a bit more before I commit to that.
|
|
|
Post by backroadjunkie on Feb 9, 2018 18:43:54 GMT
I haven't looked at the answer, but I have seen empirical evidence you should always switch, because you have to figure your odds based on the original three cans, not two.
|
|
|
Post by haxemon on Feb 9, 2018 18:47:01 GMT
My answer, without looking, is keep it, or trade it..it doesn't matter. The first round, you have a 1/3 chance of winning. The second time, there is a 50/50 chance you win, no matter which one you choose. But I want to think about it a bit more before I commit to that. I don't think I can edit the poll to add an option for "it doesn't matter" but (spoiler alert) that would be the wrong answer to the poll anyway.
|
|
|
Post by Jemma Simmons on Feb 10, 2018 0:33:24 GMT
Hi haxemon I just wanted to let you know that since I just made this Puzzles and Brain Teasers board, I shuffled this thread on over there, too.
|
|
|
Post by DoTheMath on Feb 10, 2018 0:37:47 GMT
Since beren44 has me thinking of math puzzles/curiosities, I thought I'd post the Monty Hall problem. The situation is this: You're a contestant on Let's Make A Deal. Monty Hall offers you three giant boxes on the stage, one of which holds a prize, the other two are empty (or given the show, maybe one has a prize and the other two have jokes such as a stack of old newspapers or a goat). Monty knows which box holds the prize of course. After you pick a box (let's say box 1) he first shows you one of the boxes you didn't chose and the joke non-prize inside. Then you're offered the option to stick with the box you chose or switch. Should you switch? There is a right or wrong answer (spoiler tagged below). You actually double your chances if you switch!
If it's just a straight pick and reveal, clearly the odds of winning are 1/3. If you don't switch then you maintain those odds. So let's look at if you always switch.
When you chose box 1, there was a 2/3 chance you were wrong and that box was empty. When Monty removes one of the empty boxes from the game that means there's still a 2/3 chance box 1 is empty by the other remaining box now has a 1/3 chance of being empty (the total has to remain 1/1). Which of course means there's a 2/3 chance the other box has the prize. Switching means you double your odds from 1/3 to 2/3.
The key is in the fact that Monty is removing a wrong answer.
Looking at all possible scenarios if you always pick box number 1 and always switch:
If box #1 has the prize you always lose. If box #2 has the prize Monty will reveal box #3 to be empty and you always win. If box #3 has the prize Monty will reveal box #2 to be empty and you always win.
Switching means you win 2/3 times - ironically only losing when your first guess was correct.
|
|
|
Post by beren44 on Feb 10, 2018 15:33:00 GMT
Since beren44 has me thinking of math puzzles/curiosities, I thought I'd post the Monty Hall problem. The situation is this: You're a contestant on Let's Make A Deal. Monty Hall offers you three giant boxes on the stage, one of which holds a prize, the other two are empty (or given the show, maybe one has a prize and the other two have jokes such as a stack of old newspapers or a goat). Monty knows which box holds the prize of course. After you pick a box (let's say box 1) he first shows you one of the boxes you didn't chose and the joke non-prize inside. Then you're offered the option to stick with the box you chose or switch. Should you switch? There is a right or wrong answer (spoiler tagged below). You actually double your chances if you switch!
If it's just a straight pick and reveal, clearly the odds of winning are 1/3. If you don't switch then you maintain those odds. So let's look at if you always switch.
When you chose box 1, there was a 2/3 chance you were wrong and that box was empty. When Monty removes one of the empty boxes from the game that means there's still a 2/3 chance box 1 is empty by the other remaining box now has a 1/3 chance of being empty (the total has to remain 1/1). Which of course means there's a 2/3 chance the other box has the prize. Switching means you double your odds from 1/3 to 2/3.
The key is in the fact that Monty is removing a wrong answer.
Looking at all possible scenarios if you always pick box number 1 and always switch:
If box #1 has the prize you always lose. If box #2 has the prize Monty will reveal box #3 to be empty and you always win. If box #3 has the prize Monty will reveal box #2 to be empty and you always win.
Switching means you win 2/3 times - ironically only losing when your first guess was correct.
I knew there was a reason I hesitated on my answer . I had a feeling it had to do with the number of initial choices, but could not get my head around it yesterday. Funny thing is, I solved this one day while driving, which I did a LOT of for years, on a 3 lane freeway...which lane is the fastest, and should you switch lanes....
|
|
|
Post by haxemon on Feb 10, 2018 17:43:46 GMT
I had a feeling it had to do with the number of initial choices, but could not get my head around it yesterday. Funny thing is, I solved this one day while driving, which I did a LOT of for years, on a 3 lane freeway...which lane is the fastest, and should you switch lanes.... If only the freeway had a Monty Hall there to tell you which of the lanes you are not in is definitely *not* the fastest.
|
|