KasinoKing's SlotBeater Strategy

[gloria460]

In the case at hand he not only required constant surveillance he actually went around provoking us by knowingly and repeatedly violating the Forum Rules. If that doesn't qualify that person as someone deserving scrutiny and being disliked for what they get up to I don't know what does or would.

At the risk of belaboring the point they're called 'troublemakers' because they make trouble.

I can't argue with you there and I won't even try but there is more to this guy than meets the eye and I'll leave it alone.

Hiya guys,


Now I said I had told myself I wasn't gonna get involved in this thread at all - and I stepped in only to clear the maths - but I would feel sorry for JHV if what he was trying to question now completely gets ignored. I'll be biting my fingernails and fingers not to respond to this again, but I hope for JHV that someone notices what he was really trying to say.
Bingo and it's not me agreeing with that statement it's pretty much what he said in his email.
Also, JHV would be dissapointed if he wouldn't be banned over this. All rude people should be banned. I'm confident he feels the same way.

Cheers,

Enzo

He didn't think he was going to get banned and ok I said I would leave it alone ) "Maybe when people agree that I occasionally have a point " Ooops I do have one more thing to add..for those of you that wanted to know what his website looked like or whatever it's actually very nice and professional... There's a forum but I didn't see any post from him and actually I don't know if that's his forum or not..I think he just contributes there as a blogger..Rob you're right...he's grown but he looks so young.

[KasinoKing]

Sad I know, but I've been thinking about this puzzle most of the weekend.
Previously I finally agreed with Enzo, but now I'm doing a complete U-turn and will prove him wrong!

It wasn't a trick question - it was a trick answer!

Say I have a closed bag with in it two colored chips. One is red on two sides, and the other is green on one side and red on the other. I take one chip out of the bag and show you one side. You see red. What are the odds that I picked the chip with the green ?

I'm sorry KK. The pick is totally blind and random.
yet the answer is NOT 1/2.
and its NOT 1/4.

How is

"what are the odds that I picked the chip with the green"

different from

"what are the odds that the backside of this chip I picked, which I have not yet shown you the backside of is the one green side from the remaining three sides you didn't see ?"

Jasmine ?

The answer is 1/3, I was very careful in how I posed the question. There is no room for discussion. The answer is 1/3.

OK, so the answer was that there were 4 "sides" in the bag; 3 Red and 1 Green.
You pick 2 sides from the bag (1 chip) and show 1 side is Red.
The answer therefore states that of the 3 sides you haven't seen, one is Green, so the odds of any of those sides being Green is 1 in 3. That is true.

But what Enzo seems to have missed (deliberately?) is that it was not 4 individual sides in the bag, but 2 pairs - joined together by being on one chip.
So although you have only seen 1 Red side, you also know without seeing it that there is another Red side still in the bag, since at least one side of both chips is red.
So now you know there is one Red in Enzo's hand and one Red in the bag.
That only leaves 2 unknown sides - one Red and one Green. One in the bag & one in the hand.
Therefore the odds that Enzo picked the Green-sided chip is 1/2.

There is no room for discussion. The answer is 1/2

KK

[gloria460]

KK it is 1-in-3...you are thinking after the fact...what are the odds that I picked the chip with the green You don't count the red side if he ask what are the odds of picking the green side. sides is the key word not amount.

[Jufo]

So now you know there is one Red in Enzo's hand and one Red in the bag.
That only leaves 2 unknown sides - one Red and one Green. One in the bag & one in the hand.
Therefore the odds that Enzo picked the Green-sided chip is 1/2.

There is no room for discussion. The answer is 1/2

Even though there are only two possibilities they are not equally likely. If you see a red side, it is more likely that it belongs to a RED-RED chip than RED-GREEN chip.

If you still don't believe that the answer is 2/3, why don't you do an experiment? Take two pieces or paper and write RED-RED to both sides of first paper and RED-GREEN to both sides of second paper. Then pick one paper at random, if you see GREEN then pick again. When you see RED, guess what reads on the other side of paper and you will eventually see that 2/3 of the time the other side reads RED.

Here's another one:

"I have two children. One of them is called John. What are the odds that I have two boys?"

[RobWin]

Even though there are only two possibilities they are not equally likely. If you see a red side, it is more likely that it belongs to a RED-RED chip than RED-GREEN chip.

How about this one:

"I have two children. One of them is a boy. What are the odds that the other one is a boy as well?"

0% Chance since you already stated that one of them is a boy but the odds that the other one is a girl are 100%!
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[Jufo]

0% Chance since you already stated that one of them is a boy but the odds that the other one is a girl are 100%!

Hmph I rephrased the question now...

[RobWin]

Hmph I rephrased the question now...

LOL..., good move..the other way was too easy, I won't play no more, I'll let KK answer..
____
____

[3Dice]

But what Enzo seems to have missed (deliberately?) is that it was not 4 individual sides in the bag, but 2 pairs - joined together by being on one chip. So although you have only seen 1 Red side, you also know without seeing it that there is another Red side still in the bag, since at least one side of both chips is red. So now you know there is one Red in Enzo's hand and one Red in the bag. That only leaves 2 unknown sides - one Red and one Green. One in the bag & one in the hand. Therefore the odds that Enzo picked the Green-sided chip is 1/2.

There is no room for discussion. The answer is 1/2

KK

{Edit} Actually, the bit about them being 2 sides on one chip is irrelevant - even if it were 4 separate items and he pulled out any 2 you would still know for sure that at least one of the 2 in the bag were Red, so the answer is still the same.

KK,

I think you need another shave . Lets rephrase the question. Instead of using chips and sides lets just use 4 marbles, 3 red ones, 1 green. Now surely you are not to claim there's a difference between pulling out two marbles at random in one time, or pulling out two marbles, at random, one after another.

In the chip example you pull out two marbles at once and only show one (= pick random chip, show side ..). But imagine you just picking out one marble, its red. Now guess the odds the next pick will be green.

There's three marbles left, so obviously the odds are now 1/3. Ok, we pick out a new random marble but don't look at it yet. Now we know at least one of the marbles left is red .. that doesn't exactly make the odds we picked the green one out 1/2 ... (which is what you claim now ..)

There's no trick KK, not in the question and not in the answer.

Cheers,

Enzo

[Zoozie]

I have another explanation for KK.
Let R1,R2 denote the two red sides on coin #1, and let R3,G1 denote the red and green side on coin #2.
Taking a coin at random and showing a side at random can result in 4 outcomes each with equal probability:

R1
R2
R3
G1

But the G1 possiblity is excluded since it was given that this event had not happened!

So we now only have 3 outcomes left with equal probability (R1,R2 and R3). And only in 1 of them (R3) will the other side be green! So probability is 1/3.

If you still doubt it simply find 2 coins and mark one of the sides. Then draw a coin and side at random and write down the outcome. Repeat 50 times

Learning Bayers Theorem will come in handy in many situations and it is not hard to use.

[4 of a kind]

What is the point everyone is trying to make here with all these explanations for the same answer?

Are you trying to prove someone is slow?