Simple gambling problem

Originally posted by brobluto
Yes, and for MULTIPLE tries you are right, we need to consider sides and I agree. But for the one time event, it's 50/50 SS vs SG.

So are you saying that for a one-time bet, the probability of the other side being silver P(other side is silver) = 1/2, or are you denying the existence of P(other side is silver) altogether?

Originally posted by PBE6
So are you saying that for a one-time bet, the probability of the other side being silver P(other side is silver) = 1/2, or are you denying the existence of P(other side is silver) altogether?

I am saying that the P(SG)=P(SS) since P(GG) does not exist.

This guy had a 1 in 3 chance of pulling out either SS, SG, or GG. So that P(GG)+P(SS)+P(SG)=1 correct?

We KNOW that P(GG)=0. so P(SS)+P(SG)=1. After ONE try, either P(SS)=1 or P(SG)=1 that's it.

After 3 tries, I agree, it's probably P(SG)=.33 and P(SS)=.66, but that's assuming that the 3 tries:

(a) always had the silver side show AND
(b) an even money bet for the other side showing silver.

if a/b changed then it would be different odds, would it not? The original question does not specify this which is why the bet is fair for the ONE time deal.

Originally posted by brobluto
Yes, and for MULTIPLE tries you are right, we need to consider sides and I agree. But for the one time event, it's 50/50 SS vs SG.

Why? We don't know which silver side we see? Why do you even take into consideration the number of times we try this? It's a matter of probability of something to happen, independently of the number of times repeated.

Originally posted by brobluto
I am saying that the P(SG)=P(SS) since P(GG) does not exist.

This guy had a 1 in 3 chance of pulling out either SS, SG, or GG. So that P(GG)+P(SS)+P(SG)=1 correct?

We KNOW that P(GG)=0. so P(SS)+P(SG)=1. After ONE try, either P(SS)=1 or P(SG)=1 that's it.

After 3 tries, I agree, it's probably P(SG)=.33 and P(SS)=.66, but that's assuming that the 3 ...[text shortened]... e original question does not specify this which is why the bet is fair for the ONE time deal.

Yes P(SS) + P(SG) = 1, but P(SS) =/= P(SG). These values don't change after looking at the other side. If you want to repeat the experiment, the probability remains the same.

Originally posted by kbaumen
Why? We don't know which silver side we see? Why do you even take into consideration the number of times we try this? It's a matter of probability of something to happen, independently of the number of times repeated.

Exactly, we don't know which side is showing, but we know it's either SS SG. It's a 50/50 shot for the one time deal.

Originally posted by brobluto
Exactly, we don't know which side is showing, but we know it's either SS SG. It's a 50/50 shot for the one time deal.

But SS is more probable than SG because it has 2 silver sides of whom one we might see, while SG has only one silver side.

Originally posted by kbaumen
Yes P(SS) + P(SG) = 1, but P(SS) =/= P(SG). These values don't change after looking at the other side. If you want to repeat the experiment, the probability remains the same.

P(SS) does = P(SG) they have the same probability of being pulled from the hat, do they not?

Originally posted by brobluto
P(SS) does = P(SG) they have the same probability of being pulled from the hat, do they not?

It's the same probability if we talk about drawing cards from the hat without any other information. Then, yes, it's either SG or SS but now we have additional information that one of the sides is S. One of the possible cards has both sides S from which one random side can be shown to us, the other has only one S at which we might be staring. So P(SS) = 2P(SG), hence, the solution.

Originally posted by kbaumen
It's the same probability if we talk about drawing cards from the hat without any other information. Then, yes, it's either SG or SS but now we have additional information that one of the sides is S. One of the possible cards has both sides S from which one random side can be shown to us, the other has only one S at which we might be staring. So P(SS) = [b]2P(SG), hence, the solution.[/b]

Are you saying that the method in which we acquire the information changes the odds?

Example 1: We are shown a side thereby negating one of the 3 card. You are suggesting that the odds are 1/3.

Example 2: Neither side is revealed from the card chosen, but after choosing the card, the dealer removes another card from the hat which happens to be the GG card. Thereby saying that the odds that he has the SS card is 1/2.

Originally posted by brobluto
Are you saying that the method in which we acquire the information changes the odds?

Example 1: We are shown a side thereby negating one of the 3 card. You are suggesting that the odds are 1/3.

Example 2: Neither side is revealed from the card chosen, but after choosing the card, the dealer removes another card from the hat which happens to be the GG card. Thereby saying that the odds that he has the SS card is 1/2.

No, I'm saying that the odds depend on the information given, not the way we acquire it. If we know ONLY that there are two cards, then yes, it's 50/50. But if we know that one card has a characteristic that is stronger for one card than the other, that it's more likely that it isn't the latter.

We know that one side of the chosen card is S, so it's more likely that it's SS, not SG, because it has two S's, while SG, has one S.

The beans' analogy was alright. If we have 3 beans, 1 red, 2 white, then we are more likely to choose (randomly) one of the white beans. So we can say about the cards. 2 white beans refer to SS, while 1 red refer to SG.

Originally posted by brobluto
I am saying that the P(SG)=P(SS) since P(GG) does not exist.

This guy had a 1 in 3 chance of pulling out either SS, SG, or GG. So that P(GG)+P(SS)+P(SG)=1 correct?

We KNOW that P(GG)=0. so P(SS)+P(SG)=1. After ONE try, either P(SS)=1 or P(SG)=1 that's it.

After 3 tries, I agree, it's probably P(SG)=.33 and P(SS)=.66, but that's assuming that the 3 ...[text shortened]... e original question does not specify this which is why the bet is fair for the ONE time deal.

You said the following:

(1)

I am saying that the P(SG)=P(SS) since P(GG) does not exist.

(2)

We KNOW that P(GG)=0. so P(SS)+P(SG)=1.

If this is the case, then we have a system of equations:

(1) P(SG) = P(SS)
(2) P(SS) + P(SG) = 1

Rearranging, we get:

(1*) P(SG) - P(SS) = 0
(2*) P(SG) + P(SS) = 1

Adding the equations, we get:

2*P(SG) = 1

Therefore:

P(SG) = 1/2

And subbing into (1) above we get:

P(SS) = 1/2

However, you also said the following:

After 3 tries, I agree, it's probably P(SG)=.33 and P(SS)=.66, but that's assuming that the 3 tries:

(a) always had the silver side show AND
(b) an even money bet for the other side showing silver.

This means you repeat the game exactly as it was run the first time. If that is the case, and we have already established that you are arguing that P(SG) = P(SS) = 1/2, then how does running the game 3 times change P(SG) and P(SS) from 1/2 to .33 and .66, respectively?

Originally posted by kbaumen
No, I'm saying that the odds depend on the information given, not the way we acquire it. If we know ONLY that there are two cards, then yes, it's 50/50. But if we know that one card has a characteristic that is stronger for one card than the other, that it's more likely that it isn't the latter.

We know that one side of the chosen card is S, so it's more ...[text shortened]... eans. So we can say about the cards. 2 white beans refer to SS, while 1 red refer to SG.

But we are not choosing between 3 independent beans, we are choosing between whether or not we believe it is the SS or SG card.

The original question does not say that if the gold side showed that the question would be different.

That's what I'm trying to say. For the one time event in the scenario posed, it's 50/50.

You are making assumptions, however logical or illogical they may be, that the bet would change if the side that came up was gold instead of silver. Also, that sides would not change for future bets.

For example, I take that bet, I win or lose either way, doesn't matter. If now he puts that card back into the hat, takes out another one and:

1. says this side is gold, and even money that the other side is gold, I walk away because he changed the question with the scenario to match the previous bet that I either won or lost on, so I'm either up money and wlak away, or I lost money and don't want to lose any more. (not in my favor)
2. says this side is gold, and even money that the other side is silver, I stay and make the bet, because he didn't change the question even though the scenario changed (in my favor)
3. says this side is silver, and even money that the other side is gold, I stay and make the bet, because he changed the question without changing the scenario. (in my favor)
4. says this side is silver, and even money that the other side is silver, I walk away for the same reason as (1) (not in my favor)


But on the FIRST offer, I stay and make the bet because he either has the SS card or the SG card.

Originally posted by eldragonfly
Irrelevant. geepamoogle once again your explanation is worthless. You can jump through all the hoops that you want, your answers and shallow justifications are still in error.

read these two questions again.

What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?

Both questions a ...[text shortened]... r coin analogy is bogus and erroneous, it is a combinations problem, not a permutations problem.

I tried it 100 times, never got less than 60%. Shall I try it 1000 times? The good thing is that this discussion over boys and girls has helped me produce a few resources for when I next teach conditional probability

Originally posted by brobluto
But we are not choosing between 3 independent beans, we are choosing between whether or not we believe it is the SS or SG card.

The original question does not say that if the gold side showed that the question would be different.

That's what I'm trying to say. For the one time event in the scenario posed, it's 50/50.

You are making assumptions, ...[text shortened]... he FIRST offer, I stay and make the bet because he either has the SS card or the SG card.

Oh yes we are choosing between 3 independent beans or between 3 S's. Actually any further playing would be irrelevant, because we are talking about this one time - you said that yourself. And you also said that there are only two possible outcomes. But that doesn't make it 50/50. One outcome is more probable because it's more likely that we see one of the two SS S's, than the single S from SG. The question "is it a fair bet?" means "Are the chance of this card being SS or SG are even if you take into consideration the info you've been given (you see one side which is silver)?"

If we want to talk about further bets, than the man should say "I bet ya money that the other side is the same color you see, no matter what you see." Then it would always be 2/3 that he wins the bet.

Originally posted by kbaumen
We have three possible combinations then

1)older brother, younger sister
2)older sister, younger brother
3)two brothers

so 66% that they're different gender.

And, please, don't say it's a worthless solution. Instead, explain why so, if you think it's worthless.

Wrong, those are pemutations, not combinations. This is just a simple 2 coin toss, disguised as a poorly worded problem.

http://www.knowyourluck.com/coins2u.html

Your answers are ridiculous, and your tired solutions are simplistic.