Please ,please Help Me

[xyz]

Hello I have had an idea for over 6 years , this is probably my last forum left to prove my idea, I am running out of forums.  The world is not understanding something very simple. 

 

Please answer this.

 

I am pulling my hair out and ready to bang my head against a wall, this is baby maths and easy maths yet people are denying it.

Take 2 sets of 2 sweets, in each set there is a blue sweet and a red sweet, 


So we have two people with two sweets each a red and a blue sweet, ok so far?


Each person swaps shuffles the sweets between their hands behind their back and then both persons hold out their left closed hand and ask you to pick a one of the peoples left hands

the odds of a blue sweet are?........

 

 

Now the world is saying and keeps saying it is 1/2, that is wrong.

 

 

I get p=0  p=.5  p=1 = P(A)=0_1

Edited August 30, 2015 by xyz

[HydrogenBond]

You are saying each person has two sweets, red and blue. Both people shuffle two sweets, and then both people pick a sweet from the other. This means there are two chances to get a blue sweet.

 

Another way to look at this is it is analogous to two people, each flipping a coin. Since we only need to find one blue sweet among the two coins, we are throwing two coins and need only 1 head. For two coins we can get HH, HT, TH, TT. So three out of four times will be a winner; 0.75. 

[xyz]

You are saying each person has two sweets, red and blue. Both people shuffle two sweets, and then both people pick a sweet from the other. This means there are two chances to get a blue sweet.

 

Another way to look at this is it is analogous to two people, each flipping a coin. Since we only need to find one blue sweet among the two coins, we are throwing two coins and need only 1 head. For two coins we can get HH, HT, TH, TT. So three out of four times will be a winner; 0.75. 

Thank you for that, a slightly different scenario, 2 people have 2 sweets each, a blue one and a red one, both people randomly throw away 1 of the 2 sweets leaving one sweet concealed in their left fists, I ask you to choose one of the people who will give you their remaining sweet, what is the chance you will receive a blue sweet?

[pgrmdave]

50%.

There are four scenarios of the initial order of the sweets:

1. RB RB
2. RB BR
3. BR RB
4. BR BR

After throwing away the sweet on the right, that leave us with:

1. R R
2. R B
3. B R
4. B B

That leaves us with four equally likely scenarios (so 25% chance of each), of which we need to choose one sweet (so a 50% chance of each) which leaves us with:

1. 25% chance overall, 50% chance of R, 50% chance of R
2. 25% chance overall, 50% chance of R, 50% chance of B
3. 25% chance overall, 50% chance of B, 50% chance of R
4. 25% chance overall, 50% chance of B, 50% chance of B

Mathematically, that can be reduced to the following:

0.25 * (0.5R + 0.5R) + 0.25 * (0.5R + 0.5B) + 0.25 * (0.5B + 0.5R) + 0.25 * (0.5B + 0.5B) =
0.25 * R + 0.125 * R + 0.125 * B + 0.125 * B + 0.125 * R + 0.25 *B =
0.25 * R + 0.125 * R + 0.125 * R + 0.125 * B + 0.125 * B + 0.25 *B = 
0.25 * R + 0.25 * R + 0.25 * B + 0.25 * B =
0.5 * R + 0.5 * B

So, a 50% chance of a blue sweet.