Coin Flip Odds

jimmyg said:
I'm really embarrased, and aside from all the personal insults included in your really nasty post, you are correct.

Mathematically, for someone having two children there is a 50% probability of having a boy and a girl, it is not 33%.

Between the original coin toss odds discussion and this boy/girl scenario..I must have become lost somewhere along the way.

Guess I acted as though I was arguing a weak case before an average jury.

Sorry
Jim
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yeah sorry I reread that post and it was rather vicious. You hit on a couple pet peeves of mine and I responded very poorly. My apologies.
 
jimmyg said:
We already have two Givens: 1) one child is a boy. 2) the genetic odds of having either a boy or girl are equal, 50/50.

The only unknown is the sex of the second child, and the odds are one in two (50/50) that it will be a boy.

So we are actually only betting whether the second child is a boy or a girl, the additional information is nothing more than intellectual interference, and the relevant odds do not deviate from 50/50.

If the question was asked without knowing that the first child is a boy, then it would become more complicated.

Jim
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More bad news. The odds on having a girl are a little better than 50-50, but please don't tell anyone. :)
 
Geez...this thread went from flipping coins to arguing about the kids..:p

I still remember statistics from college...the first thing when class started, the professor said, "For all intents and purposes, statistics should be renamed: How to Lie With Numbers."

the one thing everyone is forgetting is that statistics (math) leaves out the human factor. call me a cheat or a hustler, but I can tell you over 90% of the time what the coin flip will end on when I do the flipping.

I learned this flipping method at an early age and it is one reason that I never call a coin until it's in the air. a practiced flipper can manipulate the outcome at a really high rate.

I demonstrate this at the poolhall, but leave out the method used. we have too many hustlers trying to get an advantage, any advantage, over his fellow man. :(
 
jay helfert said:
More bad news. The odds on having a girl are a little better than 50-50, but please don't tell anyone. :)
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Is that the same as not only missing the ob when kicking, but also dropping the cue into the pocket?:(

Feels worse, looks terrible, but still only counts as one foul.;)

Gotta go now, my wounds require licking.:)

Jim
 
Patrick Johnson said:
Here's a similar puzzle about probabilities:

On the TV show Let's Make a Deal, contestants were given a choice of three doors to open. Behind one door was the Big Prize, and behind the other two doors was nothing.

After the contestant chose a door, but before it was opened, Monty Hall (the host) always opened one of the remaining two doors - always one with nothing behind it - and gave the contestant the option of sticking with the door originally chosen or switching to the other unopened door.

Since there are two unopened doors and the Big Prize is behind only one of them, it appears that the contestant's odds must be 50/50 and it doesn't matter which door is opened next. But it does.

What are the real odds that the Big Prize is behind the door originally chosen vs. the other one? (Hint: the odds that the Big Prize is behind the originally chosen door haven't changed.)

pj
chgo
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I believe you need to add that if the contestant happens to choose the door with the Big Prize initially, Hall is equally likely to reveal what's behind either of the other two doors.

That's a mighty big hint you gave there!
 
jimmyg said:
The only unknown is the sex of the second child

Jim
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I think you have figured this out already, but your description adds information not given in the original description, viz., it introduces an order to the two children, saying that the first one was a boy. In my original description I just said that one of them is a boy, and did not specify which one. This makes a difference in the odds.
 
PoolBum said:
I believe you need to add that if the contestant happens to choose the door with the Big Prize initially, Hall is equally likely to reveal what's behind either of the other two doors.
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Since he always opens another door but never reveals the prize, that's already a given.

That's a mighty big hint you gave there!
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I'm more interested in illustrating than puzzling.

pj
chgo
 
gwjackal said:
I's take the bong after all this talk is making me go dumdum and want to sit on my bumbum..:rolleyes:

Are we talking quaters, dimes, nickels, or pennies. I prefer to use dice you call high or low or use a pair and play craps for the break....

come on lucky number 7????
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I try to trick my fiance' into playing dice in exchange for 'favours' ......

I offer her 1-6 and I take 7-12. hahaha...... little does she know that she cant roll a 1 on two dice!

(kidding)
 
Patrick Johnson said:
Since he always opens another door but never reveals the prize, that's already a given.

pj
chgo
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Well, no, because he could choose to systematically open the lowest numbered door with nothing behind it each time. For example, in any case where the Prize is behind door number three and the contestant chooses door three initially, Hall could choose to always reveal what's behind door number one and never open door number two. That would mean he's not equally likely to open either remaining door.
 
Patrick Johnson said:
Here's a similar puzzle about probabilities:

On the TV show Let's Make a Deal, contestants were given a choice of three doors to open. Behind one door was the Big Prize, and behind the other two doors was nothing.

After the contestant chose a door, but before it was opened, Monty Hall (the host) always opened one of the remaining two doors - always one with nothing behind it - and gave the contestant the option of sticking with the door originally chosen or switching to the other unopened door.

Since there are two unopened doors and the Big Prize is behind only one of them, it appears that the contestant's odds must be 50/50 and it doesn't matter which door is opened next. But it does.

What are the real odds that the Big Prize is behind the door originally chosen vs. the other one? (Hint: the odds that the Big Prize is behind the originally chosen door haven't changed.)

pj
chgo
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You have seen the movie 21, correct?

It goes over this situation in the movie. first selection is 33% chance of being right, after the door is opened if you take the other door you are looking at a 50% chance of having the prize.
 
One thing I know about the Martindale system is you can go to Vegas and find a roulette table that has a dollar minimum and bet $1 on black. If you lose bet two if you lose that bet 4 then 8 then 16 then 32 dollars. Whenever you win a bet you start back at a dollar again. So if you bet black every time you win $1 every time it hits black and you lose $63 if it hits red 6 times in a row.

Here is the thing: It takes quite a while to hit red 6 times in a row and in the meantime you are getting free drinks and $1 everytime it hits black. And when it finally does hit red 6 times in a row you have already won a certain amount of money so you aren't actually losing $63

If you don't trust yourself just walk to the table with $63

This system is great for the guy that wants to have some fun gambling and get a bunch of free drinks before hitting the club (dance or strip) where the drinks are $8

When I did it in vegas they had to start giving me different colored chips because they ran out of the color I was using.
 
Since he always opens another door but never reveals the prize, that's already a given.
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Well, no, because he could choose to systematically open the lowest numbered door with nothing behind it each time.
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If he did that and you knew it then your odds would go up to 100%. What are the odds that he does it?

pj
chgo
 
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Jaden said:
I have even been mentioned in Card Player for winning a qualifier to the WSOP where I did quite well. I have won several smaller tournaments as well.

Jaden....

p.s. anytime you want to play me some, I'll be happy to....
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Check your sarcasm detector as I think it's a little off.
 
Luxury said:
One thing I know about the Martindale system is you can go to Vegas and find a roulette table that has a dollar minimum and bet $1 on black. If you lose bet two if you lose that bet 4 then 8 then 16 then 32 dollars. Whenever you win a bet you start back at a dollar again. So if you bet black every time you win $1 every time it hits black and you lose $63 if it hits red 6 times in a row.

Here is the thing: It takes quite a while to hit red 6 times in a row and in the meantime you are getting free drinks and $1 everytime it hits black. And when it finally does hit red 6 times in a row you have already won a certain amount of money so you aren't actually losing $63

If you don't trust yourself just walk to the table with $63

This system is great for the guy that wants to have some fun gambling and get a bunch of free drinks before hitting the club (dance or strip) where the drinks are $8

When I did it in vegas they had to start giving me different colored chips because they ran out of the color I was using.
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Different color chips? You should've just pulled out $6300 and done it with $100 bills.

On the game in casino where they spin the wheel and you can bet on the one dollar spot, 2 dollar spot, 5 dollar spot, etc. - - it is the same principal with the same % as the roulette (red, black, or green) if you bet the dollar spot each time. I have done it and you can average about $19 per hour. It was all going well til you had to lay the $240 to win a $1. That cured me into thinking this system was a good idea.
 
What cured me was the next morning I decided to start with $5 and I bet on black every time and it hit black 8 times in a row. I was $40 dollars richer but I then realized that it COULD happen.

I am going in July and I'm going to do my $63 limit Martindale system to get free drinks before hitting the strip club.
 
Patrick Johnson said:
Here's a similar puzzle about probabilities:

On the TV show Let's Make a Deal, contestants were given a choice of three doors to open. Behind one door was the Big Prize, and behind the other two doors was nothing.

After the contestant chose a door, but before it was opened, Monty Hall (the host) always opened one of the remaining two doors - always one with nothing behind it - and gave the contestant the option of sticking with the door originally chosen or switching to the other unopened door.

Since there are two unopened doors and the Big Prize is behind only one of them, it appears that the contestant's odds must be 50/50 and it doesn't matter which door is opened next. But it does.

What are the real odds that the Big Prize is behind the door originally chosen vs. the other one? (Hint: the odds that the Big Prize is behind the originally chosen door haven't changed.)

pj
chgo
Click to expand...
2/3'rds if the contestant switches. There's a 2/3'rds chance it's behind one of the two unselected doors. Monty eliminates one of them, leaving the 2/3'rds chance it's behind the remaining unselected door.

Jim
 
Jal said:
2/3'rds if the contestant switches. There's a 2/3'rds chance it's behind one of the two unselected doors. Monty eliminates one of them, leaving the 2/3'rds chance it's behind the remaining unselected door.

Jim
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Not really 2/3rds. Each door has its own chance of winning.

Say you select door #1, and he opens door #3 and its out of the picture.

When you selected door #1 it was a 33% chance that it had the prize. (1 in 3) If he then offers you a switch, you are upping you chance at winning by 17% (50/50 shot) by switching.

Gotta play the numbers.
 
D_Lewis said:
Not really 2/3rds. Each door has its own chance of winning.

Say you select door #1, and he opens door #3 and its out of the picture.

When you selected door #1 it was a 33% chance that it had the prize. (1 in 3) If he then offers you a switch, you are upping you chance at winning by 17% (50/50 shot) by switching.

Gotta play the numbers.
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JAL is correct, as usual. This problem has been around a long time and there are many, many Websites that deal with it. I think the Wikipedia explanation is pretty good and you can see that here: http://en.wikipedia.org/wiki/Monty_Hall_problem

And here is a link that will help you understand why your answer, while certainly appearing sensible, is wrong: http://www.youtube.com/watch?v=mhlc7peGlGg
 
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