Poll: A little math problem

[Saskwach]

Nice try. All you need to recognize is that having two dogs, and knowing one is male, doesn't have the same probabilities as just having one unknown dog. There are two unknowns. One of the dogs has an unknown gender, and which dog is a male is unknown. You must see that there is some difference between the situation where you know Dog 1 is male, and where you know one of the dogs is a male. On the problem that Alex_P proposed, which is an exact replica of this one, but in simpler terms, do you think the chances of the result being 2 are equal to the chances of a 1?

This is what we were actually talking about: we can't be sure which dog is male, so both are still unknown quantities, so it's not as simple as flipping a coin for the other dog's gender - that dog might very well be the male we referred to.

 

[Samirat]

How do you respond to the number generator case? It is a pretty straightforward example of this. It makes the solution relatively clear. If you are looking for one, look at that.

I'm not sure why you think the XOR has any bearing on the probability. Saying that one or the other is true indicates two different cases. All of these are exclusive of each other. 2 males is exclusive of Dog 1:M, Dog 2:F. Likewise, Dog 1:M, Dog 2:F is exclusive of Dog 2:M, Dog 1:F. But all are equally likely.

 

[dMighty]

50% right? Since you've ruled out GG and GB, BG and BB remain.

The real question would be, is eating the beagle barbaric?

 

[Saskwach]

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.

 

[Samirat]

Unhappy with theory, I've decided to experiment. I got two $1 coins in change from a nice dish of Kueh Teo. 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.

For some reason, Cheeze doesn't believe that it's okay to reflip. He thinks it is like changing the sex of the dogs or something, and that unless you're getting at least one heads 100 percent of the time, your experiment is obviously flawed.

For the last time, Cheeze, this is identical to the situation in the OP's problem. The two dogs have random gender. One happens to be male. This is the same as flipping two coins, where one happens to be heads. If both happen to be tails, it's not the same problem anymore, and can safely be discarded. Effectively, you are narrowing down your case set to correspond to what occurred in the problem.

 

[Samirat]

50% right? Since you've ruled out GG and GB, BG and BB remain.

The real question would be, is eating the beagle barbaric?

How have you ruled out GB?

 

[Lukeje]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

May I point out that a literal interpretation of this problem leads as follows:
There are two dogs. You're told that both can't be female. This rules out FF.
You are then told to calculate the probability of the 'Other' dog being male. If we think about this completely logically, if both dogs are male, then there is no 'Other' dog. This means that the probability of the 'Other' being male is 0.

 

[dMighty]

How have you ruled out GB?

By reading it wrong. I had thought the questioned implied the first dog the washer inspected was male, which it didn't. So there are the three possibilities with a 33% chance of the dog being male.

 

[Ancalagon]

What about my post no. 363? Do you disagree that the situation is as I describe in the first of my statements, i.e. "you could mean that "I've won either bet 1 or bet 2" and maintain that the real situation is one of the other two?

Glancing back on it quickly, I guess--what's your point?

That if you follow the logic that follows from situation A, i.e. that "I've won either bet 1 or bet 2"

bet1/bet2
M/M
M/F
F/M

And we go back [http://www.escapistmagazine.com/forums/jump/18.73797.811084] to what I was saying like, seven pages ago, that that kind of matrix is inapplicable in light of the new information, and I say that we now have to use this matrix:

Bet I Won/Other Bet
Even/Even
Even/Odd

+++++

Sorry, but, we're just going around in circles now.

Why is my matrix inapplicable? Because we've received new information? But I've modelled that new information perfectly in the matrix. Remember, you agreed that 'you've won one of the bets' could mean different things, you agreed that the proper interpretation was "I've won either bet 1 or bet 2". That leaves you with three equally likely outcomes. How do I know they're equally likely? Because there were four equally likely outcomes to begin with, and we ruled out one.

Now you're saying that the information that "I've won either bet 1 or bet 2" entirely invalidates the matrix. Why? Because you know you've won one of the bets? But remember, you've agreed that means "I've won either bet 1 or bet 2", not that you've won a particular bet.

 

[Samirat]

How have you ruled out GB?

By reading it wrong. I had thought the questioned implied the first dog the washer inspected was male, which it didn't. So there are the three possibilities with a 33% chance of the dog being male.

Hah. Sweet. This guy gets it, why can't everyone.

I think most people would, if they took a second to think about it. Only a few think about it and still remain unconvinced.

 

[Jimmydanger]

All you know when the man says you won one bet is that you didn't lose them both, removing odd odd.

When someone tells me I've won one of my bets, all I know is that I know I didn't lose both of them? Really? When someone tells me I've won one of my bets, I don't know that I've one at least one of my bets?

Think about it...

Ok I know you gave up last night on this but this morning I finally found out how to isolate the flaw in your logic.

As far as I can tell you would agree with us if the man had stated not that "there is a male" but rather "there are not two females." You believe that these two statements are not the exact same and that the first statement gives you the right to remove one of the M/F F/M choices. The problem with this argument is that these two statements are equivalent.

"there is a male" tells you that there is at least one male and you can infer that there cannot be two females

"there are not two females" tells you that there is not two females and you can infer that there is at least one male.

There is no difference in these two statements.

 

[tocj]

We should also consider the fact she is lying, why whould she react so enthousiastic with a smile. I'd say they're both female (so 0%) and she wants to sell you a male..for odd reasons.

 

[Jimmydanger]

Why is my matrix inapplicable? Because we've received new information? But I've modelled that new information perfectly in the matrix. Remember, you agreed that 'you've won one of the bets' could mean different things, you agreed that the proper interpretation was "I've won either bet 1 or bet 2". That leaves you with three equally likely outcomes. How do I know they're equally likely? Because there were four equally likely outcomes to begin with, and we ruled out one.

Apples and oranges. Try this:

1) There are two bets: A and notA, that's the principle of the excluded middle. When there are only two possible states a thing can be, it is either A or it is notA. This way you won't get hung up on the 'order' anymore.

2) The bets were equal, which means winning bet A and losing bet notA yields the same payoff--breaking even--as losing bet A and winning bet notA.

3) There are three possible outcomes: win both, win one and lose one, lose two, so let's see the various possibilities:


Win Both Bets = .25 (.5 win bet A x .5 win bet notA)

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

Lose Both Bets = .25 (.5 lose bet A x .5 lose bet notA)

+++++

Fine so far, right? Now we get some new information: we find out we've won one of our bets. This means we've one either bet A or bet notA, and we know that by virtue of (1), the principle of the excluded middle. So let's imagine two hypothetical scenarios that cover all the possibilites. One where the bet we've won is A, and one where we've won bet notA:

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)

+++++

See what happened?

--Under either possibility (won A or won notA) the odds of all the outcomes--and especially important because it's the question we're asked, the odds of winning BOTH bets--come out the same;

--We know both possibilities cover the range of all possible outcomes thanks to (1) the principle of the excluded middle;

--We know that if we win one bet and lose the other we break even no matter which specific bet was lost and which specific bet was won thanks to (2) because the payout on each bet was equal (EXTRA CREDIT: although technically this is irrelevant--can you guess why?)

Whether we win bet A or bet notA, the odds come out the same under either of the two possible scenarios, so if winning a bet is like finding a male puppy, the answer to the two male puppies question is the same as the answer to the two won bets question: 50/50

you are re flipping the coins in this scenario. If it were coins.

what I mean by that is you are treating the chance of there being two losses like it was never an option. Your experiment is the exact same as taking one coin placing it as heads and then flipping the other coin and therefore it is 50-50. Even though in our scenarios it did not occur you still have to take into account the chance that it would be two losses not just make that chance a 0%.

 

[Alex_P]

Fine so far, right? Now we get some new information: we find out we've won one of our bets. This means we've one either bet A or bet notA, and we know that by virtue of (1), the principle of the excluded middle. So let's imagine two hypothetical scenarios that cover all the possibilites. One where the bet we've won is A, and one where we've won bet notA:

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)

+++++

See what happened?

--Under either possibility (won A or won notA) the odds of all the outcomes--and especially important because it's the question we're asked, the odds of winning BOTH bets--come out the same;

--We know both possibilities cover the range of all possible outcomes thanks to (1) the principle of the excluded middle;

--We know that if we win one bet and lose the other we break even no matter which specific bet was lost and which specific bet was won thanks to (2) because the payout on each bet was equal (EXTRA CREDIT: although technically this is irrelevant--can you guess why?)

Whether we win bet A or bet notA, the odds come out the same under either of the two possible scenarios, so if winning a bet is like finding a male puppy, the answer to the two male puppies question is the same as the answer to the two won bets question: 50/50

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

There is only one set called "Win Both Bets." In "Win Both Bets" is special because it allows either thing to be the referent at the same time.

...

Imagine that we play this "game":

I generate a set that contains at least one male. I then silently pick a "referent," which is any dog that is male.

Then I show you the set and ask "Which of these things was my referent?"

Based on your response, we bin the sets into "Jesse is the referent" and "OSAN is the referent."

Right now what you're doing is putting any set where either of them could be the referent into both bins.

-- Alex

 

[Jimmydanger]

All you know when the man says you won one bet is that you didn't lose them both, removing odd odd.

When someone tells me I've won one of my bets, all I know is that I know I didn't lose both of them? Really? When someone tells me I've won one of my bets, I don't know that I've one at least one of my bets?

Think about it...

Ok I know you gave up last night on this

No--I solved this last night: check out posts #409 and #426

As far as I can tell you would agree with us if the man had stated not that "there is a male" but rather "there are not two females." You believe that these two statements are not the exact same and that the first statement gives you the right to remove one of the M/F F/M choices.

That's not what I believe. I believe they are equivalent statements and that both require you to remove not only F/F, but also one and only one of M/F and F/M, just like "you won (at least) one of your bets" and "you didn't lose both bets" are equivalent statements.

Then how would you separate these two statements. E.G. removing only the FF scenario without removing any other?

If you cannot ask yourself again are you inferring something from the statement that is not actually there

 

[Fraught]

Does it matter what gender the other dog is?

Maybe it doesn't to you, but I know many people care from what gender someone is.

 

[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.

 

[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