Poll: A little math problem

[Saskwach]

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

 

[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?). I think this might have been because I tend to find tails more exotic, so mentally fitted them to the least common result. That or I mixed up thirty and tails.

 

[Alex_P]

You're double-counting "Win Both Bets" here.

Not sure what you mean, but look again--I double counted everything.

You did this:

scenario 1 (A is referent):
won A, won notA
won A, lost notA

scenario 2 (notA is referent):
won A, won notA
lost A, won notA

The outcome "won A, won notA" appears twice.

-- Alex

 

[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.