Coin Flip Odds

av84fun said:
That's called a Martingale or sometimes Chinese Doubles system and it is seriously flawed.

You bet $5 and you lose.

Your next bet is $10. If you lose:
Your next bet is $20. If you lose:
Your next bet is $40. If you lose:
Your next bet is $80. If you lose:
Your next bet is $160. If you lose:
Your next bet is $320. If you lose:
Your next bet is $640

In this system, you become progressively dumber. For example, on the last bet, you are risking $640 to win $5

Such streaks are not at all uncommon.

Be careful out there.
(-:

Jim
Click to expand...

You are not risking $640 to win $5. At that point you are already stuck $635, and you are risking $640 to climb out of a $635 hole and come out a $5 winner. It is still an even money bet. But you are right that the system is seriously flawed. You can lose your entire bankroll using a system (as you pointed out, such losing streaks are not uncommon) that will never win as big.

The theory is unbeatable, but you have to have an unlimited amount of money with no bet limit to correctly implement it. Neither of those givens are ever in place.

Kelly
 
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
Click to expand...

The unknown is not the sex of the second child. It's still the sex of both children. The question is not "our first child is a boy, what's the sex of the second child" that's 50/50 b/g. The question is ONE of our children is a boy.

As previously stated, with two children, the options are:

BB
BG
GB
GG

They're giving you information to elimination just the fourth option from that set of possibilities. Each of the other three sets is equally likely. Two of the three have the other child as a girl.

BB
BG
GB

So if you know at least one of the kids is a boy, only 1/3 of the time will the other kid be a boy.
 
jay helfert said:
STFU! :)
Well, there's your answer. I wasn't planning on telling this to everybody but Jack did! When you see a coin spinning on a table top. it's about 11-10 or better it will land Tails up.
Click to expand...

How sure are you of this? I've defiinitely heard this before, but I've never seen conclusive proof. I've also seen it used to argue that HEADS is actually slightly more likely, as since that side is heavier it will rotate through it's down position quicker, and through it's up position slower (kind of like those big boats that flip around at amusement parks).
 
So, I'll pay someone on this forum $100 today, but they have to pay me $.01 the next day, and pay me for one month doubling my payment each day.

Any takers?
 
The original set of facts and question is:Quote:
Originally Posted by PoolBum
Speaking of probabilities, here is an interesting one. Suppose you know that your neighbors have two children, and that at least one of them is a boy. Suppose further that for any given birth the odds are exactly 50/50 of a boy versus a girl.

Given what you know, what are the odds that both of your neighbor's children are boys?

Rocket354 said:
The unknown is not the sex of the second child. It's still the sex of both children. The question is not "our first child is a boy, what's the sex of the second child" that's 50/50 b/g. The question is ONE of our children is a boy.

As previously stated, with two children, the options are:

BB
BG
GB
GG

Here lies your error. Choices 2 and 3 are identical. GB & BG are, in fact, the same, the order of birth is not an issue in this discussion and does not change the possibilities. There are only two real possibilities here, GG (#4) is automatically eliminated by the given that one is a boy, and since BG
(#2) and GB (#3) are the same, we can eliminate either one (let's eliminate #3). That leaves only two choices, BB (1) and BG (2). The order of birth is just another irrelevant interference factor, no different than reversing BB to create another BB, they are the same.

They're giving you information to elimination just the fourth option from that set of possibilities. Each of the other three sets is equally likely. Two of the three have the other child as a girl.

BB
BG
GB

So if you know at least one of the kids is a boy, only 1/3 of the time will the other kid be a boy.
Click to expand...

Sorry, 50/50.
Jim
 
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jimmyg said:
The original set of facts and question is:Quote:
Originally Posted by PoolBum
Speaking of probabilities, here is an interesting one. Suppose you know that your neighbors have two children, and that at least one of them is a boy. Suppose further that for any given birth the odds are exactly 50/50 of a boy versus a girl.

Given what you know, what are the odds that both of your neighbor's children are boys?



Sorry, 50/50.
Jim
Click to expand...

Sorry, 1/3.
 
To elaborate, options 2 and 3 are NOT identicaly. Of all the families that have two children, they can go

BB
BG
GB
GG

As a result, 25% of families with two kids have two boys. 25% of families with two kids have two girls. And 50% of families with two kids have a boy and a girl. That's the key.

If you know one of the kids is a boy that eliminates 25% of the possibilities. So there's 75% remaining. 50/75 is BG (in some order), 25/75 is BB. In other words, 1/3 is BB, 2/3 is BG (in some order).

This is the last I will say on the subject. If you want to convince yourself, look the problem up online or go talk to any math professor nearby.
 
The Martingale system is a strategy employed in craps too. If you can hold on and just keep doubling your bet till you win once, all your losses will be cleared and you will be up the amount of the original bet. Sounds good, but the only problem is casino tables have a maximum bet and it doesn't take too long to get there.

Do the math with any structure of betting and you'll see how it works.

Julius Ceaser was known to bet this way and lost a lot of money and property betting with the Martingale.
 
Kelly_Guy said:
You are not risking $640 to win $5. At that point you are already stuck $635, and you are risking $640 to climb out of a $635 hole and come out a $5 winner. It is still an even money bet. But you are right that the system is seriously flawed. You can lose your entire bankroll using a system (as you pointed out, such losing streaks are not uncommon) that will never win as big.

The theory is unbeatable, but you have to have an unlimited amount of money with no bet limit to correctly implement it. Neither of those givens are ever in place.

Kelly
Click to expand...

This is one of those half full/half empty issues...but make that last bet, you take $640 out of your pocket and post it. That puts you AT RISK for ANOTHER $640...the outcome of which will be that you will be up $5.00 or down $1,275.00

As we agree that the bet...on its own...is still 50/50 but the system has to be viewed in its overall context. We agree that the unlimited sum of money never exits and if it did, the opponent would never permit the system to be worked which is WHY casino tables have betting limits.

They are NOT going to let Bill Gates play a Martingale on them.

So, the system does become progressivley dumber as the betting limit is reached.

Regards,
Jim
 
... BG (#2) and GB (#3) are the same...
The order of birth is just another irrelevant interference factor, no different than reversing BB to create another BB, they are the same.
Click to expand...

Try it this way:

Older Boy/Younger Boy
Older Boy/Younger Girl
Older Girl/Younger Boy

OB-YB and YB-OB are the same. OB-YG and OG-YB are not.

pj
chgo
 
BWTadpole said:
So, I'll pay someone on this forum $100 today, but they have to pay me $.01 the next day, and pay me for one month doubling my payment each day.

Any takers?
Click to expand...

Old one but good one.

About $20 million if I remember correctly.

(-:
 
PoolBum said:
The odds of nine tails in a row are no less and no more than any other series of nine outcomes. For example, the odds of THTHTHTHT are exactly the same as TTTTTTTTT.
Click to expand...

These strings of two different digits are the same as binary numbers. Instead of "H" and "T", make heads = "0" and tails = "1". The number of possible outcomes for 7 flips is the value of the 7-digit number 1111111 in base 2 (= 2^7):

1111111 = 128 (2^7)
1111110 = 127
1111101 = 126
1111100 = 125
1111001 = 124
... and so forth, down to ...
0001000 = 8 (2^3)
0000111 = 7
0000110 = 6
0000101 = 5
0000100 = 4 (2^2)
0000011 = 3
0000010 = 2 (2^1)
0000001 = 1 (2^0)

pj
chgo
 
Rocket354 said:
To elaborate, options 2 and 3 are NOT identicaly. Of all the families that have two children, they can go

BB
BG
GB
GG

As a result, 25% of families with two kids have two boys. 25% of families with two kids have two girls. And 50% of families with two kids have a boy and a girl. That's the key.

How you reach the above conclusion, aside from convenience, escapes me. For purposes of this discussion, I can only conclude 33% have two boys, 33% have two girls, and 33% have a boy and a girl. If there are statistics proving otherwise I am not aware of them, and based on the problem as presented, they hold no weight in this discussion. Reversing the order of birth does not add another group, nor does it change the possibilities.

If you know one of the kids is a boy that eliminates 25% of the possibilities. So there's 75% remaining. 50/75 is BG (in some order), 25/75 is BB. In other words, 1/3 is BB, 2/3 is BG (in some order).

This is the last I will say on the subject. If you want to convince yourself, look the problem up online or go talk to any math professor nearby.
Click to expand...

Including or creating new factors into a discussion to substantiate your position doesn't make it correct.

Have a good day.
Jim
 
jimmyg said:
Including or creating new factors into a discussion to substantiate your position doesn't make it correct.
Click to expand...
But Rocket and Patrick and anyone else that's been arguing for 1/3 are right.

50% of two-child families have a boy and a girl (in any order), while 25% have two boys. So 75% of all two-child families have at least one boy. 25/75 equals 1/3.

Jim
 
Not my fault if you lack the basic requirements to understand fundamental math. If you don't see that 50% of families with two kids have a boy and a girl then I can only conclude one of two things:

1) you are a troll

2) you need to go back and repeat 6th grade

I can absolutely conclude one thing, however:

before you decide that pulling numbers out of your butt constitutes math, speak to someone you trust to know these things. That does not include that buddy of yours you've known since 4th grade that managed a 550 on the math portion of his SAT 20 years ago, but rather maybe a math teacher/professor neighbor, or that guy with a math degree in some other department at your job that you've never really liked because he always thinks he's right about stuff.

Really, this is too funny. Using rhetorical arguments in a math discussion is like bringing a knife to a gun fight. Heck, a pea-shooter to a nuclear war.
 
jimmyg said:
The original set of facts and question is:Quote:
Originally Posted by PoolBum
Speaking of probabilities, here is an interesting one. Suppose you know that your neighbors have two children, and that at least one of them is a boy. Suppose further that for any given birth the odds are exactly 50/50 of a boy versus a girl.

Given what you know, what are the odds that both of your neighbor's children are boys?



Sorry, 50/50.
Jim
Click to expand...

When using probabilities in logic theory, you have to consider ALL possible combinations to determine the answer. In this case its BB,BG,GB and GG. Since it cannot be GG, it has to be one out of the other three possibilities...which is 1/3. Cant explain it much more simple than that.

Southpaw
 
Rocket354 said:
Not my fault if you lack the basic requirements to understand fundamental math. If you don't see that 50% of families with two kids have a boy and a girl then I can only conclude one of two things:

1) you are a troll

2) you need to go back and repeat 6th grade

I can absolutely conclude one thing, however:

before you decide that pulling numbers out of your butt constitutes math, speak to someone you trust to know these things. That does not include that buddy of yours you've known since 4th grade that managed a 550 on the math portion of his SAT 20 years ago, but rather maybe a math teacher/professor neighbor, or that guy with a math degree in some other department at your job that you've never really liked because he always thinks he's right about stuff.

Really, this is too funny. Using rhetorical arguments in a math discussion is like bringing a knife to a gun fight. Heck, a pea-shooter to a nuclear war.
Click to expand...

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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The Monty Hall Problem

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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