Two probability questions for nerds

Like this. There are three possibilities for each child. Boy(B), Boy day of birth irrelevant (B), and Girl

The possible outcomes are (BT,BT) (BT,B) (BT,G) (B,BT) (B,B) (B,G) (G,BT) (G,B) (G,G) not all equally likely.

Q2 A lady tells you she has 2 children. You ask her has she a boy born on a Tuesday and she says Yes. What is the probability she has 2 boys?

Of the above possibilities the three in red above three involve two boys, one born on a Tuesday. A BT and a B.

The possibility of a child being a BT 1/14. That is 1/2 for a boy times 1/7 for a Tuesday. So (BT,BT) is 1/14 times 1/14 = 1/196

The possibility of a child being a boy who may or may not have been born on a Tuesday is 3/7 (1/2 less 1/14). So (BT,B) in that order is 1/14 times 3/7 = 3/98 and the same in reverse order.

1/196 + 3/98 + 3/98 = 13/196
Very good but no cigar!
You are right that to begin with the chances are 13/196. But having been told that there is a BT the only other possibilities without 2 boys are BT,G and G, BT which to begin with are 7/196 each, a total of 14/196. So there are 13 chances of BB and 14 chances of either BT,G or G,BT. Thus the probability of 2 boys is 13/(13 + 14) i.e. 13/27 or getting near 50% and is the answer to Q2.
As I say, very good. The judges have decided that you don’t quite win the turkey, but a goose is winging its way to you for Christmas:D

Round of applause please for @cremeegg.
 
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That's a common mistake people make in the coin toss odds. Dependency implies that the first outcome affects the odds of the second event, like pulling a card from a deck affects the next draw as there are fewer cards remaining.

The probabilities change when you have more information so they are dependent. If you don't know the sex of the second child the probability there are two boys is 1/4, i.e. the outcomes are BB, BG, GB and GG, and the BB outcome is one of four.

When you know one child is a boy (i.e. not a girl) it reduces the number of outcomes. The original outcomes are BB, BG, GB and GG. If we are told the other child is a boy we eliminate the GG outcome (i.e. the all girl outcome). So instead of 4 outcomes, we now have 3 outcomes (BB, BG and GB). In two of these outcomes the second child is a girl (i.e. in BG and GB), so the probability the second child is a boy (i.e. the BB outcome), given we know that one child is a boy, is 1 in 3, i.e. 1/3.
 
The probabilities change when you have more information so they are dependent. If you don't know the sex of the second child the probability there are two boys is 1/4, i.e. the outcomes are BB, BG, GB and GG, and the BB outcome is one of four.

When you know one child is a boy (i.e. not a girl) it reduces the number of outcomes. The original outcomes are BB, BG, GB and GG. If we are told the other child is a boy we eliminate the GG outcome (i.e. the all girl outcome). So instead of 4 outcomes, we now have 3 outcomes (BB, BG and GB). In two of these outcomes the second child is a girl (i.e. in BG and GB), so the probability the second child is a boy (i.e. the BB outcome), given we know that one child is a boy, is 1 in 3, i.e. 1/3.
You know, I'm not sure what it says about the quality of my dreams, but I actually woke up in the middle of the night last night realising I didn't just fall for the trap here, I jumped enthusiastically into it!! :rolleyes:
 
You know, I'm not sure what it says about the quality of my dreams, but I actually woke up in the middle of the night last night realising I didn't just fall for the trap here, I jumped enthusiastically into it!! :rolleyes:
It says that your mind is an ordered place and your worries are few.
 
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For further reading, google Leonard Mlodinow and\or Girl Named Florida
 
For further reading, google Leonard Mlodinow and\or Girl Named Florida
Here is that link.
This is a summary of the example.
“Wiki” said:
That's right, the answer is about 49.7%. To summarize:

  • Given that the first child is a girl, the probability is 50% that both children are girls.
  • Given that at least one child is a girl, the probability is 33% that both children are girls.
  • Given that at least one child is a girl named Florida, the probability is 49.7% that both children are girls.


But the third situation is more nuanced still. It is key how you were "given" that information.
1). If you asked “have you a girl named Florida?” and got a Yes then the answer is 49.7%
2). If you asked “have you a girl and if so what is her name?” and got the answer “yes and it’s Florida” then the answer is 33% as you were going to get some answer for the name and so the answer can make no difference to the probability without that info.
If out of the blue you are told “I have a girl and her name is Florida” you ask yourself “why am I being told her name?” If there is something special about being called Florida worth mentioning like being born on Christmas Day then you are closer to (1) above but if you reckon that if she was called Texas or whatever you would have been told that instead then you are closer to (2).
 
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This seems similar to the Monty Hall problem - where additional information alters the odds significantly. In the Monty Hall scenario, the contestant picks a door out of three (the prize is behind the doors). They then open one of the two remaining doors and ask if you want to change your mind. Statistically the contestant should always change their choice because (a) the initial choice was a 33% chance. However, when one option was removed, you are a holding a 33% chance but the other door now has a 50% chance

In this case, the probability of there being two boys (or two girls) is initially 25% - but removing half of the potential outcomes means the odds have changed as there are now only two possible outcomes - BB or BG (assuming the order doesn't matter). The one in three answer only works if BG and GB are two distinct outcomes (and therefore three possible total outcomes) but order doesn't matter
 
This seems similar to the Monty Hall problem - where additional information alters the odds significantly. In the Monty Hall scenario, the contestant picks a door out of three (the prize is behind the doors). They then open one of the two remaining doors and ask if you want to change your mind. Statistically the contestant should always change their choice because (a) the initial choice was a 33% chance. However, when one option was removed, you are a holding a 33% chance but the other door now has a 50% chance

In this case, the probability of there being two boys (or two girls) is initially 25% - but removing half of the potential outcomes means the odds have changed as there are now only two possible outcomes - BB or BG (assuming the order doesn't matter). The one in three answer only works if BG and GB are two distinct outcomes (and therefore three possible total outcomes) but order doesn't matter
A typo. The other door must have a 67% chance to add to a 100%.
Okay take it that order doesn't matter - though you can make order matter if you like and get the same answer of 1/3.
So using order doesn't matter there are 3 possible outcomes to begin with:
2 boys 25%
2 girls 25%
1 boy and 1 girl 50%
The last has to be 50% to add up to 100%.
2 girls are ruled out and so you are left with the other 2 possibilities but in their initial ratios - 25 to 50 i.e, 1/3 chance BB.
 
A typo. The other door must have a 67% chance to add to a 100%.
Okay take it that order doesn't matter - though you can make order matter if you like and get the same answer of 1/3.
So using order doesn't matter there are 3 possible outcomes to begin with:
2 boys 25%
2 girls 25%
1 boy and 1 girl 50%
The last has to be 50% to add up to 100%.
2 girls are ruled out and so you are left with the other 2 possibilities but in their initial ratios - 25 to 50 i.e, 1/3 chance BB.

Yes sorry - 67%

You're right - I think the way I read the question looked like a 100% x 50% probability (i.e. the first is a boy). But BG and GB are valid outcomes as well as BB - so one in three
 
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