[bushindo]

Mr. Smith is your friend. The last time you saw him was at his wedding, where he confided that he wants to have 4 sons. His plan is to have as many children as needed until he gets 4 sons. Upon the birth of his 4th son, he would stop having kids.

You lost contact with Mr. Smith, and many years later you happened to meet him again. He tells you that he got the family he wanted. He then offers you a simple chance to make money. If you can correctly guess his total number of children, he would give you 100 dollars. You were about to guess when one of these two scenario happen

1) Two girls walk in and said hi. Mr. Smith introduced them to you as his first and second children. In order words, they are his two eldest children. What is your best guess of the total number of children now?

2) In this scenario, two girls walk in and said hi. Mr. Smith introduced them to you as his children. What is your best guess of the total number of children now?

[Guest]

i am somehow not convinced with the solvability of this problem. The author could have thought up of any number and then would come up with a way to justify it.

This is a pure gamble and no logic puzzle. In either case he has atleast 6 children. and atmax is anybody's guess.

[Guest]

i am somehow not convinced with the solvability of this problem. The author could have thought up of any number and then would come up with a way to justify it.

This is a pure gamble and no logic puzzle. In either case he has atleast 6 children. and atmax is anybody's guess.

He is asking what your best geuss would be for telling his friend how many kids he has. yeah in both scenarios he obviously has at least 6 kids, but would you put money on it that he has exactly 6?

[Guest]

i get

equal between 9 and 10 .2734 for the first part havent look at the second

[Guest]

same probabilities but

7,8

[Guest]

This is a tricky problem. I haven't got the second question sorted yet, but I believe that the answer to part one is 8 or 9, both having equal probability (see spoiler for calculations). However, that doesn't take into account the fact that people who begin with two children of the same sex are more likely to have another one of the same sex. (Something to do with womb environment and strngth of sperm).

Assuming that the probability of boys and girls are equal, then there are the following scenarios:

1) Next 4 children are all boys. Probability = 0.5 x 0.5 x 0.5 x 0.5 = 0.0625.

2) 3 boys and a girl, then a boy. This is 0.0625 x 0.5 x 4 = 0.125

3) 3 boys and 2 girls, then a boy. Using combinations, we get an answer of 0.15625.

4) 3 boys and 3 girls, then a boy. This gives the same answer, 0.15625

5) 3 boys and 4 girls, then a boy. The probability of this is 0.13671875, which shows that we have passed the peak, so no more calculations are necessary.

Scenarios 3 and 4 are the most likely, that is 8 or 9 children.

[bonanova]

Six with no clues.

[1] Eight - starting with 2 girls has same cases as if starting with no children, thus add 2.

[2] Seven - no-clue case has a few cases with 0 or 1 girls.

[Guest]

I think I've got the answer for the second question. It would be a bit tedious to write down all the fractions involved, so I'll just show a summary on the spoiler below.

The probability of 6 children, given that the last one must be a boy, I calculate to be 10/32 = 0.3125.

7 children (3 girls and 4 boys) works out at 20/64, the same answer.

Thereafter, the probabilities start to reduce.

4 boys and 4 girls gives 35/128 = 0.2734375,.....

So, again there are two choices with equal possibility. This time the best gamble is 6 or 7 children.

However, I wouldn't take the bet on. Statisticians never gamble. We know there's always a chance of being wrong

[Guest]

Six with no clues.

[1] Eight - starting with 2 girls has same cases as if starting with no children, thus add 2.

[2] Seven - no-clue case has a few cases with 0 or 1 girls.

Are the chances of having 4 boys the same as having 2 girls and 2 Boy? If so then the answer to #1 should be:

6. In an earlier post about this it was determined to say four to the man after stopping at 2 boys and 2 girls. So how you put it you would add 2 for 6.

[bonanova]

Are the chances of having 4 boys the same as having 2 girls and 2 Boy? If so then the answer to #1 should be:

6. In an earlier post about this it was determined to say four to the man after stopping at 2 boys and 2 girls. So how you put it you would add 2 for 6.

In this case the first two girls don't happen by chance. They are given.

If it were known that the first one thousand children were girls, you'd have [just] the same cases for the next six children as you have for the first six children in the general case.

[Guest]

wow completeley missed the stop at 4 boys i was just looking at likely number if he had only 4 boys

part 1. 8or9 at .15625

part 2. 6 or 7 at same

lack of sleep strikes again