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bushindo
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Mr. and Mrs. Smith are friends whom you haven't seem for a long time. The last time you saw them was at their wedding ceremony, where they confided that they wanted to have 2 daughters and 2 sons, and they would have as many kids as needed until they have a minimum of 2 daughters and 2 sons. (For instance, suppose they have 2 daughters in a row, then would then give birth as many times as necessary until they get 2 sons, at which point they will immediately stop having kids. Likewise, suppose their first three kids are son, daughter, and son. They would then try again until their number of daughters reach 2. ) Assume that the probability of getting a son is .5.

You haven't seen the couple for years since the wedding. One day, you happened to meet Mr. Smith, who told you that his wife and he succeeded in getting the family they wanted. He then offered you a simple game. If you can correctly guess his total number of children, he would give you 100 dollars. You can only guess once. What is the best guess?

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Well if the probability of having a son is .5 (which it isn't) then the probability of having a daughter is also .5 (which it also isn't). If there is a 1:1 chance of having a son or a daughter, so the most probable, barring triplets or quadruplets of the same sex, is that they were able to get their ideal family with only four children. Males are more probable than 50% though...

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Ok, the best bet will actually be 4, and here is why.

Probability of having 2 sons and 2 daughters out of any 4 children = .375

Probability of having 2 sons and 2 duaghters out of any 5 children = .625

however, the possibilities for 2 sons and 2 daughters in 4 children also exist in the possibility for 2 sons and 2 daughters in 5 children, so...

Probability of NOT having 2 sons and 2 daughters in 4 children, but then acheiving this in 5 = .625-.375 =.25

If we take this out to 6 total children, it becomes more likely that with 6 kids, there will be 2 sons and 2 daughters, but the odds of not getting them in 4 or 5 and then getting them with # 6 will be even less than .25

I'd write out the statistics that are used for this, but the truth is I don't remember the formulas, so I just did it longhand, basically wrote out every possibility outcome for 4 and 5 total children.

The best bet you could make on this one is 4 total kids, and the odds of you winning are 3 in 8

Correct me if I'm wrong

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Correct me if I'm wrong

Ok, the best bet will actually be 4, and here is why.

Probability of having 2 sons and 2 daughters out of any 4 children = .375

Probability of having 2 sons and 2 duaghters out of any 5 children = .625

however, the possibilities for 2 sons and 2 daughters in 4 children also exist in the possibility for 2 sons and 2 daughters in 5 children, so...

Probability of NOT having 2 sons and 2 daughters in 4 children, but then acheiving this in 5 = .625-.375 =.25

If we take this out to 6 total children, it becomes more likely that with 6 kids, there will be 2 sons and 2 daughters, but the odds of not getting them in 4 or 5 and then getting them with # 6 will be even less than .25

I'd write out the statistics that are used for this, but the truth is I don't remember the formulas, so I just did it longhand, basically wrote out every possibility outcome for 4 and 5 total children.

The best bet you could make on this one is 4 total kids, and the odds of you winning are 3 in 8

OBoyOBoyle got it. Well done.

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One approach is to consider all possible ways of having six kids. It is very rare for a family not to satisfy these condition in 6 kids or less. There are 64 different permutations, and those who are familiar with logic tables or probability should be able to construct such a table quickly. Once you have such a table, go down row by row and circle the 'stopping point', or the point at which the family satisfies the condition. For instance, suppose the row is this- S D D D S D- you would circle the 5 kid, since that is the point where the family satisfies the condition and stops.

Once done, look at each column and count how many stopping points you circled in that column. From there, we should have a reasonable estimate of what is the best guess.

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Well if the probability of having a son is .5 (which it isn't) then the probability of having a daughter is also .5 (which it also isn't). If there is a 1:1 chance of having a son or a daughter, so the most probable, barring triplets or quadruplets of the same sex, is that they were able to get their ideal family with only four children. Males are more probable than 50% though...

actually that is the likely hood to have a son and daughter. By doing a four square box to multiply xx by xy you get two xx and two xy. If it wasnt 50/50 it would throught the population off and the human race would become extince eventually

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actually that is the likely hood to have a son and daughter. By doing a four square box to multiply xx by xy you get two xx and two xy. If it wasnt 50/50 it would throught the population off and the human race would become extince eventually

Well... I don't know if we'd become extinct quite so easily. Perhaps, we could devise some sort of contest to kill off the excess males and restore the balance. A war, perhaps?

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the population would only die out if the all people had less than or equal to an average of 2 offspring as long as that nots true ur fine the ugly people just wont have mates

really the number has to be the 1/the lower chance but greater then two is ball park

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