The reason why the probabilities of your door are still 1/3 is not because we are counting the original cases, but because both the cases in which you could have failed and in which you could have succeeded were reduced by half (in the example, success cases were reduced from 300 to 150 and failure cases were reduced from 600 to 300) and to reduce both by half results in their respective probabilities being maintained.But that’s not what happened. Thanks for clearly showing how the code in this comment thread does not work correctly.Why is this the accepted solution when it’s easily debunked. A row in the table represents one combination of the sequence of events and the outcome for whether you switch or stay. Shares . The question itself is not asking about first or second events, or anything like that. If the host only sometimes reveals what’s behind a door, and other times not, then the best bet is to stay put with your initial pick, since the host is probably only going to reveal a door when he wants you to switch away from the door you selected, because the big prize is behind it. He can’t open 1 since you picked it and can’t open 3 since it has the prize.If you choose door 2 he will be FORCED to open door 1. It’s called “The Riddler,” and it’s Important small print: Please wait until Monday to publicly share your answers. Look at #2 in your list. This leaves us with 2 possible wins to 2 possible losses out of 4 probable outcomes. In the table below, I show the six scenarios and their outcomes. This scenario might be slightly different than the various ways he presented the information in the game show. No offense to Jim Frost, but I think that the table is a little bit misleading, and somewhat obscures the correct solution. Just run through this scenario a number of times, record the results, and see how you fare when you switch and don’t switch. The goat appears. For example, in scenario 1, you choose door one and the prize is behind door one. I am happy to admit that when I first came across the Monty Hall problem the answer seemed unbelievable. But if no other doors have goats behind them, he will tell you that is the case.It just so happens that when you play, Monty is able to open another door, revealing a goat behind it. There are only three combinations of two doors that the host inherits after the contestant’s selection.

Monty knows all about this contest. You pick one door and Monty has the other 99 doors. Since it includes explicitly the other steps, it is more complicated, but that way nobody can make any excuse.By the way, with Monty “Fall”, I was referring to the variant of the game in which the host randomly chooses a door. A game theory problem of determining the best strategy with poor knowledge. When you pick door 1 and the prize is behind door 1, Monty can pick either door 2 or 3. I think the example would be clearer with the “Monty Opens” column removed entirely.I sort of see what you’re saying. I guess what I would do is to accompany the table with a probability tree, which is a good way to visually illustrate the conditional probabilities.By the way, I love the optical illusion in the beginning. However, if I list both options on separate lines, that artificially inflates that outcome because it is listed twice. This meant your everyday blinker simply couldnt exist on our modified grid, and you had to hunt for more complicated oscillators.The smallest value of N that supported an oscillator turned out to be four, and it had a period of two (i.e., it returned to its original arrangement every two ticks).

You just need to look into it more deeply to obtain all the information you need.I think it’s important to realize that for this to make sense, you need to understand that Monty cannot open the door you originally chose. It’s possible but unlikely.The problem with 3 doors follows a similar logic–just fewer doors.