Pizza brainteaser - here is another

[jeffereynelson]

I'm just going to claim to be dense, because I don't see what difference it makes that you had already chosen.  The end result is still the same; you have 2 choices, the one in hand or the other one.  Convince me otherwise.

Did you read the explanation I typed up to the problem just a few posts up?

[TXCraig1]

I'm just going to claim to be dense, because I don't see what difference it makes that you had already chosen.  The end result is still the same; you have 2 choices, the one in hand or the other one.  Convince me otherwise.

Because the only way you wouldn't switch is if you believed you picked the ace the first time when it was 1:100.

Say Jeff started with 1 million cards and only one ace. You pice one, and he throws all the others away except one - now you each have one card. One is the ace. Would you switch? If you say no, it means you believe you hit the 1:1,000,000 choice off the bat. The odds of Jeff having the ace are 999,999:1,000,000.  That's pretty close to 1:1. I don't know about you, but I'm not lucky enough to hit the 1:1,000,000, so I'll take Jeff's card.

[Chicago Bob]

The odds are 2:3, you are correct chicago bob.

Could somebody please tell me when and where I can pick up my new car that I've won? I'm really stoked about it possibly being just like one of these here custom jobs man. If it is I think I will be so happy I will want to give all my fellow contestants here a great big 'ol smoooch!

Wooo hoo...thank you Monty!

[Don K]

No, here's the car you win Bob...

[Chicago Bob]

Hey man... that's not the Curtain/door I chose last night...is it?   
Daaang!   

Craig...can ya help a brother out here?

[Chicago Bob]

No, here's the car you win Bob...

That's funny Don...at least you didn't yoke a jackass up to that rig. I was wondering why you asked me to send you a pick of myself......

Please tell me that is not you tucked in the grill there!

[jeffereynelson]

Here's a similar question to those posted-

All of the women in the world who have two children are rounded up, and one is randomly selected from the group. When asked if she has a girl, she replies "yes." What are the odds the other child is a boy?

(assume the likelihood of birthing boys and girls is exactly 50/50)

[Chicago Bob]

50/50

[pizzablogger]

50/50

+1

There are three child combinations for each woman (boy-boy, boy-girl, girl-girl). The lady replies one of her children is a girl, which eliminates the boy-boy combination. That leaves only the boy of boy-girl and girl of girl-girl a possibility. 50-50.

--K

[jeffereynelson]

+1

There are three child combinations for each woman (boy-boy, boy-girl, girl-girl). The lady replies one of her children is a girl, which eliminates the boy-boy combination. That leaves only the boy of boy-girl and girl of girl-girl a possibility. 50-50.

--K

Pretty close

[Chicago Bob]

Forget about the "all the women in the world, one randomly chosen part"

Take ANY woman....each of her children have a 50/50 Chance of being a boy or a girl(that is given). When she was asked if she has a girl...that was a 50/50 chance also.

[jeffereynelson]

Forget about the "all the women in the world, one randomly chosen part"

Take ANY woman....each of her children have a 50/50 Chance of being a boy or a girl(that is given). When she was asked if she has a girl...that was a 50/50 chance also.

If the woman only has one child, then yes it is 50/50 that it is either a girl or a boy. However if you want to have a girl, and you are going to have two kids, the odds are not 50/50 that one of them will be a girl.

[pizzablogger]

If the woman only has one child, then yes it is 50/50 that it is either a girl or a boy. However if you want to have a girl, and you are going to have two kids, the odds are not 50/50 that one of them will be a girl.

If you plan on having two kids, there are four possible orders of the children being born (first born, second born), boy-girl, boy-boy, girl-girl, girl-boy. That's four combinations of first born-second born. Only one of those combinations excludes a girl, leaving three combinations that do, which would be a 75% chance that if you are planning on having two kids, one will be a girl. We don't know which combination would sequence as there is no first born kid yet....you are only planning to have kids.

I change my answer to the original question: Since we already know one kid is a girl, the chances of the other kid being a boy are 4 in 6, or 66.66%

Am I close?

[jeffereynelson]

Ya 66% is right, if one is a girl, then the parent has girl/boy, boy/girl, or girl/girl. 2 out 3 are the other child being a boy.

[Chicago Bob]

(assume the likelihood of birthing boys and girls is exactly 50/50)

If this is true...then I don't see that it matters how many children she already has. Even if she already has 66 children....the likelihood of her next birthing will result in a 50/50 chance that it will be a boy (or a girl) no?

[jeffereynelson]

If this is true...then I don't see that it matters how many children she already has. Even if she already has 66 children....the likelihood of her next birthing will result in a 50/50 chance that it will be a boy (or a girl) no?

Yes this is correct. But my question assumes the mother already has given birth to two children, not what are the odds of the next child be a girl or boy.  I didn't say we know the mothers first born child was a girl, what is the sex of the second. I said one of her two children is a girl, which may be the older or the younger. That is the key to understanding why it is not 50/50.

[Chicago Bob]

Where an when do I get my car man.....somebody owes me.....

[TXCraig1]

Here is an easy one:

Say there is a very long piece of string (131,479,658 feet) that is wrapped exactly around the Earth's equator and a much shorter piece of string (5.2 inches) that is wrapped exactly around a golf ball. Assume both the earth and the golf ball are perfect shperes; the strings are both equidistant from the center of their respective spheres at all points.

If we increase the length of either string, it must now be lifted off the surface of the earth of golf ball to be equidistant from the center at all points. Say we add 1 foot of length to each string. The string around the earth is now 131,479,659 feet (+0.0000008%) and the string around the golf ball is now 17.2 inches (+230.8%).

Which longer string (the one around the earth or the one around the golf ball) must be lifted higher off the surface to maintain a perfect circle that is equidistant from the center of the sphere at all points?

[pizzaneer]

golf ball. seems too easy, so i'm likely wrong...

[TXCraig1]

golf ball. seems too easy, so i'm likely wrong...

I said it was easy. Anyone disagree?