galanter wrote:

Arson Smith wrote:This reminds me of a coin-flipping problem. The odds of flipping 'heads' 100 times in a row are verrry slim. Let's say you flip 99 heads in a row, and now you're freaking out, about to take that 100th coinflip.

You think and then say to yourself "WTF? This is a damn coin, and it either does 'heads' or 'tails', and I have a 50/50 shot at either on this 100th flip, regardless of my flippin' history!"

But that is incorrect - although it's true that you have defied some wicked odds to get 99 'heads' in a row, the probability that your 100th coinflip will be 'heads' are still pretty fucking slim, because then we're still talking about you satisfying the one case out of 1267650600228229401496703205376 where all 'heads' come up.

Uh Oh...

Assuming you are using an unbiased coin, after flipping 99 heads in a row the probability of flipping heads again is 1/2.

I agree with this. Each pick is independant.

galanter wrote:Uh Oh...

Assuming you are using an unbiased coin, after flipping 99 heads in a row the probability of flipping heads again is 1/2.

Positively. That is a true reset of the odds to 50/50 before every choice scenario. There were never two heads and one tails on the coin at any point, so there has never existed a 2/3 and 1/3 pair of options. A heads flip is a win on the 50/50 gamble *and* a win on the astronomical 'flip heads 100 times in a row' gamble. It is both.

galanter wrote:

Arson Smith wrote:This reminds me of a coin-flipping problem. The odds of flipping 'heads' 100 times in a row are verrry slim. Let's say you flip 99 heads in a row, and now you're freaking out, about to take that 100th coinflip.

You think and then say to yourself "WTF? This is a damn coin, and it either does 'heads' or 'tails', and I have a 50/50 shot at either on this 100th flip, regardless of my flippin' history!"

But that is incorrect - although it's true that you have defied some wicked odds to get 99 'heads' in a row, the probability that your 100th coinflip will be 'heads' are still pretty fucking slim, because then we're still talking about you satisfying the one case out of 1267650600228229401496703205376 where all 'heads' come up.

Uh Oh...

Assuming you are using an unbiased coin, after flipping 99 heads in a row the probability of flipping heads again is 1/2.

No, and this is where I think clocker bob got off on the wrong foot with the Monty problem. He was saying that once Monty opens the goat door, that the probability is now "reset" to 50/50. So you're saying the coin, after 99 flips of 'heads' is still "reset" - whereas I still say that the odds of flipping that 100th coin 'heads' still has to be 1/2^100

Part of it is in the phrasing of the original set, I think.

Like in the Monty problem - pretend you didn't know how many doors there were... if someone said "Well Monty may or may not have opened an unknown bunch of doors, and there's two left, now pick the non-goat", then you have no framework from which to operate. But we know there are 3, therefore we can talk about odds in 1/3 and 2/3 probabilities.

With the coins - if I just said I'd flipped a bunch of coins an unknown number of times and didn't really track how many 'heads' & 'tails' then told you I was about to flip again - then you would be correct to say 1/2 chance.

But given that I set up the framework of 100 flips, and the past 99 were 'heads' then it becomes way less likely that I should expect 'heads' again... because even if getting 99 out of 100 is astronomical luck, geting 100 out of 100 is even more astronomical luck, by a power of 2.

So... yeah - I think it just has to do with the given 'frame of reference' when you discuss probability.

I'll say it again: probability is weird.

Arson Smith wrote:But given that I set up the framework of 100 flips, and the past 99 were 'heads' then it becomes way less likely that I should expect 'heads' again... because even if getting 99 out of 100 is astronomical luck, geting 100 out of 100 is even more astronomical luck, by a power of 2.

I edited my previous post to add some details...sorry if you missed that...maybe I should have posted again.

Anyway...using this "reset" kind of thinking is error prone. But to work within your frame of reference...

Yes, flipping 99 heads and then another head is extremely unlikely.

But flipping 99 heads and then a tails is also extremely unlikely.

In fact both are equally unlikely. And that's why you can calculate the odds of that 100th flip to be even...1/2 heads, 1/2 tails.

Man, people on here are really and truly high.

One of my pet peeves is people who refuse to use the information available to them to find an answer to their problem.

I'll admit that I am guilty of this on many accounts, but when I see other people doing it, it drives me nuts.

It takes about 10 seconds to look up Monty Hall on Wikipedia. It takes about another 3 minutes to read about the problem and then read the solution.

That's it. It's done. Over. You know the answer. You can report it to your friends or whatever. You can even explain your answer. You can back it up with references. The whole shebang.

I realize that it's "fun" and "educational" to try to determine an answer oneself, but there are plenty of other situations where you'll be under fire, have to come up with a workable solution, and be forced to go with your results to not have to waste your time debating an already solved problem when you don't even possess the required knowledge to start to solve said problem!! And if that's the case, you should be researching the root of the problem (in this case, probability). Not the problem itself!

Here is yet another way to think about it. Say you bet you can flip 100 heads in a row. What are the odds you will win?

1/2^100

Now after having flipped 50 heads in a row someone else wants in on the action. Well, you're already halfway to your goal, so the odds on the bet should reflect that. (Sort of (*sort of*) like betting on a football game at halftime)). Well what are the odds you will be able to complete the 100 heads at this point? There are 50 flips to go, so the odds are:

1/2^50

Now you've gotten 99 heads in a row. And at the last minute someone wants in on the action. You've already gotten 99 heads in a row. The chances that you will reach your goal are now much better. You only have one head to go. What are the odds you will get it?

1/2^1

i.e. the odds on that last flip are 50/50.

In other words if you bet someone that you could flip 100 heads in a row you might expect them to put down some huge amount of money, and you should only have to put down a dollar. But if you only have one flip to go, and someone wants to get in on the bet, you don't really expect them to put down some huge amount of money against your dollar at that point do you?

russ wrote:One of my pet peeves is people who refuse to use the information available to them to find an answer to their problem.

But in this specific case, speaking for myself, it was not a refusal to use information available. I wanted to have fun with the puzzle, didn't know what the answer was supposed to be, and respected galanter's request in his beginning post to not go googling the answer.

And I did have fun trying to work through the problem, confined to the thread. If I had an immediate need for the answer, then I would just look it up as usual.

clocker bob wrote:

russ wrote:One of my pet peeves is people who refuse to use the information available to them to find an answer to their problem.

But in this specific case, speaking for myself, it was not a refusal to use information available. I wanted to have fun with the puzzle, didn't know what the answer was supposed to be, and respected galanter's request in his beginning post to not go googling the answer.

And I did have fun trying to work through the problem, confined to the thread. If I had an immediate need for the answer, then I would just look it up as usual.

It's good mental exercise.

I looked it up in short order, but I had to work it out on my own for a while before I really got it.

The coin flip is 1/2 chance of heads, always. Assuming perfectly balanced coin, no ability to finesse the flip on the part of flipper, etc. It's truly random.

In the M. Hall problem, if his act is random, the results of switching are also random--doesn't matter if you switch. It's the fact that he _had_ to show you the other goat that tips the scales.

I keep saying that b/c it still blows my mind.

russ wrote:It takes about 10 seconds to look up Monty Hall on Wikipedia. It takes about another 3 minutes to read about the problem and then read the solution.

That's it. It's done. Over. You know the answer. You can report it to your friends or whatever. You can even explain your answer. You can back it up with references. The whole shebang.

Things like this aren't deadline work. They're mental exercise. Reference materials are good, but becoming beholden to them is lazy and prevents you form solving novel problems.

Here's one from the pool room:

I offer to play you Nine-ball for $100 a set (10 games -- first to 10 wins $100), plus $10 a game on the side (every game, winner takes $10). The first session, we both win 10 sets. What's the most I could have beaten you out of?

Second session, same offer. This session, we both win 100 games. What's the most I could have beaten you out of?

This is a common hustle in pool rooms, to make a contest seem fair when it actually favors the hustler.

galanter wrote:Here is yet another way to think about it. Say you bet you can flip 100 heads in a row. What are the odds you will win?

1/2^100

Now after having flipped 50 heads in a row someone else wants in on the action. Well, you're already halfway to your goal, so the odds on the bet should reflect that. (Sort of (*sort of*) like betting on a football game at halftime)). Well what are the odds you will be able to complete the 100 heads at this point? There are 50 flips to go, so the odds are:

1/2^50

Now you've gotten 99 heads in a row. And at the last minute someone wants in on the action. You've already gotten 99 heads in a row. The chances that you will reach your goal are now much better. You only have one head to go. What are the odds you will get it?

1/2^1

i.e. the odds on that last flip are 50/50.

In other words if you bet someone that you could flip 100 heads in a row you might expect them to put down some huge amount of money, and you should only have to put down a dollar. But if you only have one flip to go, and someone wants to get in on the bet, you don't really expect them to put down some huge amount of money against your dollar at that point do you?

Aha - yes, I totally agree with all of what you said. When someone new walks in at the 100th flip, they are in a sense not tied to the history of the established framework. To them it IS just a 50/50 bet. As was said previously each unlinked event (a coin flip) is 50/50. But right - when you start linking them: "I bet I can flip 100 'heads' out of 100 flips." Then we're talking about the probability of the totality of the 100 events. Which makes me go back and say I probably screwed up the wording of what I said the first time.