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Fair coins tend to land on the same side they started

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Re: Fair coins tend to land on the same side they started

#271

Earlier quoted context omitted.

>models under which "it doesn't matter if you switch or not". Could you provide an example? It seems obvious that a switcher wins exactly when a non switcher looses, which is 2 / 3 ?

Take a game show host who lets you choose a door, randomly reveals what is behind one other door, and then gives you an opportunity to change your choice. This game show host CAN (randomly) reveal the prize; he has equal probability of revealing ANY of the unchosen doors. Say you are playing the Monty Hall game with this host. You choose your door, he opens another door, and it happens (purely by chance) that there i…

Isn't that a different problem entirely? The original is that the host reveals a door without the prize.

Aren't you modeling an entirely different problem as opposed to modeling the same problem with a different model, since the problem states the parameters and you are changing those?

Re: Fair coins tend to land on the same side they started

#272

Earlier quoted context omitted.

Take a game show host who lets you choose a door, randomly reveals what is behind one other door, and then gives you an opportunity to change your choice. This game show host CAN (randomly) reveal the prize; he has equal probability of revealing ANY of the unchosen doors. Say you are playing the Monty Hall game with this host. You choose your door, he opens another door, and it happens (purely by chance) that there i…

Isn't that a different problem entirely? The original is that the host reveals a door without the prize. Aren't you modeling an entirely different problem as opposed to modeling the same problem with a different model, since the problem states the parameters and you are changing those?

> Aren't you modeling an entirely different problem...

Not really, but read on:

You correctly state that in the Monty Hall problem, the host reveals a door without the prize. That's the same situation which I described in my previous comment.

Try thinking about it this way: Say you are the contestant on that show. You have never played the game before, and you will never play it again. So you don't know how the host behaves. You pick your door, he reveals another door, there is no prize behind it. You would have to ask yourself: did he deliberately open that door because it had no prize? Or did he just happen to open a door that had no prize?

Your best estimation of your odds of winning changes completely depending on how you model the behavior of the host.

However, with any type of host, the situation whereby "contestant opens door with no prize, host reveals another door with no prize" can still occur, and regardless of whether you deem that the 'original' Monty Hall problem or not, it is the most interesting way to define the Monty Hall problem. Call it the extended Monty Hall problem if you want: the situation described above has occurred, and you have to both define a model for the behavior of the host (and game) and calculate your odds under that model.

Here's a challenge for you: Can you find a model under which the contestant has 100% chance of winning by not switching to the unopened door?

Re: Fair coins tend to land on the same side they started

#273

Earlier quoted context omitted.

When one door was opened it revealed information about the other two doors.

But it didn't. Before, we knew that one of those two doors could contain either a prize or a goat. After, we know the same exact thing. No information was gained there.

As a result of how the doors were selected one of them is more likely to contain the prize. That is information.

Re: Fair coins tend to land on the same side they started

#274

Von Neumann described a very elegant way to get fair results from a biased coin. 1. Flip the coin twice 2. If you get the same result both times, goto 1 3. Now that you have different results for your pair of flips, use the first element of the pair of flips as your result. https://en.wikipedia.org/wiki/Fair_coin#Fair_results_from_a_...

Couldn't you still game that by choosing the starting side? If you want heads, and you start on heads and get a tails for the first round, you could start on tails on the second round to try to get another tails and reset to step 1.

Re: Fair coins tend to land on the same side they started

#275

I'm still looking for an intuitive or ELI5 explanation of the mechanism for this bias. The original paper says: The standard model of coin flipping was extended by Persi Diaconis who proposed that when people flip a ordinary coin, they introduce a small degree of precession’ or wobble—a change in the direction of the axis of rotation throughout the coin’s trajectory. According to the Diaconis model, precession causes…

Perhaps it helps to imagine someone had a "screwy thumb" and the coin only precesses when they "flip" it (in fact people can train themselves to do this, and its very difficult for you, the sucker, to see in the air that the coin is not rotating but just precessing!). Hopefully its obvious that whatever side is initially facing up will be the same one facing up when its caught? The next step is not at all intuitive t…

Video of how to train yourself to precess the coin and cheat:

https://m.youtube.com/watch?v=A-L7KOjyDrE

Re: Fair coins tend to land on the same side they started

#276
post #145

How do I bias a coin flip? Based on the paper it looks like 55% chance that it will land on the same side it started is possible. This was the most extreme subject. The bias is caused by procession so I want my flip to process as much as possible. Maybe I offset my finger as far away from the center of the coin as possible. Also putting as much force into it as possible is probably a good idea. Finally I have to catc…

@robocat linked a great video showing how to do this: https://m.youtube.com/watch?v=A-L7KOjyDrE

Re: Fair coins tend to land on the same side they started

#277
post #84

Earlier quoted context omitted.

If anyone wants to test it, someone wrote a short code that simulates doing that 100,000 times: https://www.techiedelight.com/generate-fair-results-biased-c... The coin is biased to come up TAILS 80% of the time, but using Von Neumann's method in the program I got HEADS 50.035%, TAILS 49.965%.

Why would you test it? Probability of two heads: p*p Probability of two tails: (1-p)*(1-p) Probability of head followed by tails: p*(1-p) Probability of tails followed by heads: (1-p)*p It's not difficult to notice that if you remove the first two, the last two form a 50/50 distribution

Put another way p*(1-p) = (1-p)*p

So the probability is the same. Whereas p doesn't equal (1-p) unless p=0.5

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