Poll: A little math problem

[Jimmydanger]

Win Both Bets = .50 (1.00 win bet A x .5 win bet notA)

Win One Bet, Lose the Other = .50 (1.00 win bet A x .5 lose bet notA + .5 win bet notA x 0.00 lose bet A)

Lose Both Bets = 0.00 (0.00 lose bet A x .5 win bet notA)

this is the part that is not correct. You have put Lose both bet to zero and are re flipping the coins. this is not a continuation of a previous scenario but a whole new one.

I assure you no coins were reflipped, injured, or harmed in the making of this solution.

What I am saying is your two scenarios have nothing to do with each other the first one is the correct chances of winning or losing bets.

The second probability is the chances of getting a second win when the first one is already predetermined. But you have started a new probability matrix instead of continuing working with the previous one. This second probability matrix does not take into account the chances of getting two tails. Even though it did not happen the chance that it could have happened changes the probabilities. I agree with all of your math here I just do not think that the second half applies to our situation.

All you have done is recreated someone elses idea from pages and pages ago about placing one coin and flipping the other. In order to have accurate results coins must be flipped simultaneously and then evaluated from there.

I wish you knew something about experimental design because my first experiment I asked you to do is perfectly valid. Simply flip two coins at least 20 times and records the results. then record out of all the times at least one is heads count the times they are both heads. the result will be close to 33%.

edit: this is probably what saskwatch is doing now

 

[Saskwach]

The results from my second trial (this time with two 10c pieces) are in, and they are: 33 tails, 67 heads.

So is that a .67 for two male puppies?

Whoops, I mixed them up. In fact it was 33 heads and 67 tails (I swear! Doesn't anyone trust me?).

Heh--I'll trust you if you'll trust that I'm not disregarding that result just because it's sour grapes, but rather because, well, runs of luck happen. I think we'd need a longer trial, but, I don't know enough about Monte Carlo simulations to know what how long that trial would have to be.

Oh, I understood that totally, which is why I wasn't happy with the first 100 - and still I'm not happy! I intend to keep testing until I reach at least a thousand! I'm stubborn that way.

 

[Jimmydanger]

The results from my second trial (this time with two 10c pieces) are in, and they are: 33 tails, 67 heads.

So is that a .67 for two male puppies?

Whoops, I mixed them up. In fact it was 33 heads and 67 tails (I swear! Doesn't anyone trust me?).

Heh--I'll trust you if you'll trust that I'm not disregarding that result just because it's sour grapes, but rather because, well, runs of luck happen. I think we'd need a longer trial, but, I don't know enough about Monte Carlo simulations to know what how long that trial would have to be.

I do know about that in my second semester statistics textbook they give about 30 tries before the chances of error are acceptable <10% at 100 tries the chances are less than 1%

Edit: this is also dependant on what you are trying to prove I used numbers on proving it is not 50% it is slightly harder to statistically prove that it is 33% but not much more.

 

[Saskwach]

All you have done is recreated someone elses idea from pages and pages ago about placing one coin and flipping the other. In order to have accurate results coins must be flipped simultaneously and then evaluated from there.

I wish you knew something about experimental design because my first experiment I asked you to do is perfectly valid. Simply flip two coins at least 20 times and records the results. then record out of all the times at least one is heads count the times they are both heads. the result will be close to 33%.

edit: this is probably what saskwatch is doing now

In fact it is:

Unhappy with theory, I've decided to experiment. I got two $1 coins in change from a nice dish of Kueh Teow. I then proceeded to flip them.
The rules were as follows:
1)Flip both coins at once or in sequence - just so long as nothing else is done until both coins hit the table.
2)Check if at least one is a heads. If not, re-flip without going further.
3)Take out one of the heads coins.
4)What is the other coin? If tails, tally it under your "tails" column (you do have one right?) and if heads, tally under the "heads" column.
5)Repeat steps 1-4 until you get a total of 100 flips (or more if you have the time).

My first tally for this test was 32 heads and 68 tails.

Of course this is not conclusive - perhaps one or both coins were weighted badly. Besides, 100 flips, though not bad, doesn't mean undeniable proof. because of these concerns I will get two new coins and repeat the experiment 100 times with those two. I'll then do the experiment 100 times with another pair of coins and so on. I also encourage anyone who's curious to try the experiment themselves.

Edit: To keep a running tally, we have 65 heads and 135 tails, which gives us a total heads percentage of 32.5%.

 

[Alex_P]

Could you quote me and put where I did this in bold?

Et voila!

Won bet A, bet notA is just the same as before:

Win Both Bets = .50 (1.00 win bet A x .5 win bet notA)

Win One Bet, Lose the Other = .50 (1.00 win bet A x .5 lose bet notA + .5 win bet notA x 0.00 lose bet A)

Lose Both Bets = 0.00 (0.00 lose bet A x .5 win bet notA)

+++++

Now the other possible scenario, we won bet notA and bet A's chances are just the same as before:

Win Both Bets = .50 (.5 win bet A x 1.00 win bet notA)

Win One Bet, Lose the Other = .50 (.5 win bet A x 0.00 lose bet notA + 1.00 win bet notA x .5 lose bet A)

Lose Both Bets = 0.00 (.5 lose bet A x 0.00 lose bet notA)

-- Alex

 

[Jimmydanger]

If you think flipping a coin 20 times is sufficient, you should either stay away from gambling, or work on Wall Street ;-D

the fact that you are flipping 2 coins at once makes the standard deviation much much smaller I should have said 30 I hadn't looked at the charts in my text in a while.

 

[Alex_P]

I wish you knew something about experimental design because my first experiment I asked you to do is perfectly valid. Simply flip two coins at least 20 times and records the results. then record out of all the times at least one is heads count the times they are both heads. the result will be close to 33%.

The reason this experiment works while a coin-setting one doesn't work is that the coin-setting one breaks a fundamental assumption of the problem: that your have fair coins. If you're only interested in results that include a certain coin being heads, you still have to flip that coin to get the right probability distribution.

-- Alex

 

[Jimmydanger]

Okay, now I really have to get going, so. [http://www.escapistmagazine.com/forums/jump/18.73797.817488]

I've read it and understand it I just think its not applicable.

 

[10BIT]

Okay, now I really have to get going, so. [http://www.escapistmagazine.com/forums/jump/18.73797.817488]

Damn! I've come too late!

To keep a long explanation short since I've got to get going too, Cheeze_Pavilion seems to be under the impression that at least one is the same as saying the first one which, as most of you should know, is incorrect.

 

[Jimmydanger]

Okay, now I really have to get going, so. [http://www.escapistmagazine.com/forums/jump/18.73797.817488]

Damn! I've come too late!

To keep a long explanation short since I've got to get going too, Cheeze_Pavilion seems to be under the impression that at least one is the same as saying the first one which, as most of you should know, is incorrect.

Ya we've explained that too him a lot. He does think that they mean the exact same thing

 

[Saskwach]

Okay, now I really have to get going, so. [http://www.escapistmagazine.com/forums/jump/18.73797.817488]

Damn! I've come too late!

To keep a long explanation short since I've got to get going too, Cheeze_Pavilion seems to be under the impression that at least one is the same as saying the first one which, as most of you should know, is incorrect.

Ya we've explained that too him a lot. He does think that they mean the exact same thing

I know I've said it in so many confused ways that I can't blame anyone for not understanding; when I explain maths I usually tangle myself in knots so bad even Alexander wouldn't know what to do.

 

[Lukeje]

Okay, now I really have to get going, so. [http://www.escapistmagazine.com/forums/jump/18.73797.817488]

Damn! I've come too late!

To keep a long explanation short since I've got to get going too, Cheeze_Pavilion seems to be under the impression that at least one is the same as saying the first one which, as most of you should know, is incorrect.

Ya we've explained that too him a lot. He does think that they mean the exact same thing

Its not the 'first' one its the 'Other' one that is the source of ambiguity. It suggests that you know that a specific puppy is male, and that there is 'another' one still to check.
Which is why I suggested that the probability is actually zero, as if both dogs are male, and the shop-owner has been assured that at least one is male, then there can't be a dog labelled 'Other' as they are both exactly the same! Therefore the other puppy must be female.

 

[fedpayne]

OK, this has been pretty clearly (well) explained with maths, for those who can understand it, but some people still don't seem to be getting it. Here's my understanding, using nice gentle words (literature student). Also, this assumes that the beagles have the same mother.

When mummy beagle has two pups, there can be an older male with a younger female, an older male with a younger male, an older female with a younger male, or an older female with a younger female. Can anyone see any other possible outcomes? Draw a little dog with those couplets coming out if it helps.

Now, we know that one of the pups is a dog. It can either be a younger male or an older male. So for the other dog to be male it could be the younger male of an older male, the older male of a younger male, and nothing else. For it to be a female, it could be the younger female of an older male, or the older female of a younger male.

Wait, what? I was convinced that 1/3 was right, I followed that train of logic and agreed with a third. But I can't see the flaw in my logic above. Unless, when people have been saying that the female male combination can also be written as male female, can male male also be written as male male. Because the other pup could be a younger male or an older male. Can't it? Older female or younger female. Those are all viable outcomes, surely?

God damn it, can someone spot the flaw in that logic? Because I've just confused myself. Surely the boy pup picked out could be either older or younger? So there is the possibility it can have both an older male and a younger male sibling, as well as an older female and younger female sibling?

Like I said, literature student, can someone pick that apart and find out where I made my mistake? Or can I send the gloatiest email in the world to Marilyn vos Savant?



EDIT: Please don't tell me I'm going to wait up for this? I have lectures tomorrow. Someone message me.

 

[Alex_P]

Now the question is what we do with that leftover .25, because all the possible outcomes of an event that is guaranteed to happen must add up to 1.

We put it all in Both Male, and here is why--sure having one male puppy makes One of Each seem more probable, but it doesn't. Because it *also* makes Both Male more possible, and to the degree Both Male becomes *more* possible, One of Each becomes *less* possible because they are mutually exclusive outcomes. Remember--One of Each only got to be twice as likely in the beginning because there were two undetermined events and two ways to get One of Each. Once there's only one undetermined event, there's only one way to get One of Each--and there's one way to get Both Male, and each way is equally probable (the math is in posts #409 and #426).

I get why this is so hard for everyone and tricked so many people. We start off thinking about permutations--in fact, we have to in order to figure out the odds--but then we get hit with information that changes the possible combinations AND permutations, and then--quite sneakily--the question asks us about combinations while we're all still thinking about permutations.

What's really counter-intuitive (and made me think a while before posting this) is that getting halfway to One of Each doesn't make One of Each any more possible than it was before. The reason is that One of Each was more possible when there were two permutations that could get us there: M/F and F/M. Like I've been saying from the beginning, though, when the number of indeterminate events goes from 2 to 1, and the determined event fulfills one of two necessary conditions for both remaining possible outcomes--One of Each and Both Male--then Both Male becomes more probable, and any increase in the probability of one exclusive outcome reduces the probability of the other exclusive outcome to the same degree.

No. You're flat-out wrong. I don't want to be an ass about this, but, you're really, really wrong here, and you've been confusing yourself with referents and "undetermined" and "determined" events and weird applications of "stuff adds up to 1" for, like, three days now. All of this part about "the leftover .25" is nonsense.

A fair two-coin generator generates the following probability distribution:
25% two heads
50% one head
25% no heads
You know a single fact about the current result: the final outcome includes at least one heads result.

That's it. That is the entire problem. There is absolutely nothing more to the problem than that.

-- Alex

 

[Alex_P]

Now, we know that one of the pups is a dog. It can either be a younger male or an older male. So for the other dog to be male it could be the younger male of an older male, the older male of a younger male, and nothing else. For it to be a female, it could be the younger female of an older male, or the older female of a younger male.

"The older male of a younger male" and "the younger male of an older male" refer to the same set of dogs.

-- Alex

 

[Unholykrumpet]

I suppose all the people still talking about this are the ones that understand the math problem with the three doors mentioned in the movie "21". Still don't understand that one, and I don't really understand this one, either. Probability takes all the fun out of chance.

 

[Alex_P]

A fair two-coin generator generates the following probability distribution:
25% two heads
50% one head
25% no heads
You know a single fact about the current result: the final outcome includes at least one heads result.

That's it. That is the entire problem. There is absolutely nothing more to the problem than that.

Well, there better be--that 25% for no heads, that's gone down to 0%, because "the final outcome includes at least one heads result" and that no heads result includes, well, no heads.

So what do you do with the 25% and why? Because you can't just leave the sum of the probabilities of all possible outcomes at 75% unless like, one of the coins disappears when you flip it.

1 two heads
2 one head
1 no heads
4 total
probabilties are 1/4, 1/2, 1/4

1 two heads
2 one head
3 total
probabilities are 1/3, 2/3

There's no trick to it.

-- Alex

 

[Alex_P]

Really? It's nonsense? [http://www.escapistmagazine.com/forums/jump/18.73797.809907]

Yes, thinking about it in terms of "Oh now there is a .25 that doesn't go anywhere, what do I do?" is nonsense, even if it gives you the right answer.

Those percentages and decimals represent ratios -- fractions. When you throw out a result, you don't "find the most likely one" or "distribute the percentage equally" or anything like that. You just decrement the total number of possible results (the denominator of all the fractions) because you've thrown out one of the results.

-- Alex

 

[Saskwach]

Okay, here's where you all screwed up. You've confused permutations with combinations. The reason you're all incorrect is *very* subtle, and when someone else explains it to you this way it's very seductive.

The reason is that you've all missed the difference between a *permutation* matrix and a *combination* matrix.

In fact I haven't made this mistake. A few pages back I brought up permutations and combinations myself, and explained why this question was a permutation problem where the first dog (the one that's a male) is uncertain, so we have to factor that in. If it were either a combination problem or a permutation problem where the first male dog was certain, the probability is 50%, but this question is neither.

Maybe if you guys are criticizing each other's math in the process of criticizing mine, you should all, like, get together and get your stories straight before telling me I'm wrong again?

That depends - are we telling the police about a crime we all committed, or are we each in our own search for the right answer? In the first I would support "getting our stories straight". There's nothing wrong with correcting people even on your own side who are being wrong-headed. I'd suggest it's far worse to let their mistakes lie simply because you fear weakening your argument.

 

[MRMIdAS2k]

75.

one is DEFINATELY male, so you've got a 50/50 chance of picking that one. the other has a 50/50 chance of being male, so half of 50 is 25, so it's a 75% chance of getting a male dog.