simple question

[BruceZ]

See my solution above.

1st roll: probability = 5/36 = .139
second roll: probability = 31/36 * 5/36 = .120
third roll: probability = 31/36 * 31/36 * 5/36 = .103
...

The probabilities get smaller because you have to get something other than a 6 on all the previous rolls before you get a 6, and you still have to get the 6 on the partcular roll. On the first roll, you just have to roll the 6. The probability of rolling a 6 is the same on every roll, but the probability of getting the 1st 6 is not the same.

Good question, gold star for you (best I could do was a diamond).

[Copernicus]

The work (a little difficult to write out here): The expected number of rolls is

(1*5/36 + 2*(31/36)*5/36 + 3*(31/36)^2*5/36......) = (factoring out the constant 5/36)

5/36* (1+ 2*31/36 + +3*(31/36)^2.....)=5/36*S, where S stands for the big sum.

For explanation, further simplify to S=(1+2r+3r^2....) where r = 31/36.

Then you play a little trick. r*S=(r+2r^2+3r^3...), ie every power of r has a coeffiecient 1 less than it does in S, (except there is no r^0 term)

therefore S-rS=(1+r^2+r^3....). The infinite series on the right is equal to 1/(1-r)=1/(1-31/36)=1/(5/36)=36/5.

ie S-rS=(1-31/36)*S =5/36*S = (see above) 36/5

Solving for S, S=(36/5)^2.

Way back at the beginning we factored out a 5/36 before the Sum, and the expected value was 5/36*S, so our EV is

5/36*(36/5)^2 = 36/5.

[BruceZ]

And you didn't even really use calculus, just algebra (though I consider the infinite series itself to be part of calculus). You essentially found that the sum of 1 + r + 2r^2 + 3r^3 + ... to be 1/(1-r)^2 which is correct, but you got this just from the geometric series 1 + r + r^2 + ... = 1/(1-r) which more people are familiar with, and which you derive the same way. My method for doing the first sum involves taking a derviative of the geometric series to get the other series, hence the calculus.

Still waiting for an explanation of the logical solution from someone.

[Copernicus]

Eureka..now I see the calculus approach...grazi!

[well]

The algebraic approach seemed a little too obvious to look at. But the suggestion of coming up with an answer that uses logic started me to think.

Let's say we throw two dices M times, where M is a really large number. The amount of sixes expected would then be M*5/26. So we rolled until a six came up, and we did that about M*5/36 times (ignoring the last rolls, if the Mth outcome does not equal 6). And now, the average 'length' of a try would be about M/(M*5/36), which is 36/5.

Next Time.

[BruceZ]

Or the way I look at it, which is equivalent to yours, suppose a very large number of people performed this experiment once, rolling until a six comes up. Then you know that out of all of those throws, a six must come up 5/36 of the time. But each person rolls exactly 1 six. So on average that the 1 six represents 5/36 of the total throws. Since 1 is 5/36 of 36/5, it took 36/5 throws on average to get a six.

Here's a variation of this problem that's a little easier to see. This is due to David. Suppose a billion chinese couples have children until they get a boy, and then they stop. How many children will be born on average? Assume each baby has an equal chance of being a boy or a girl. Since each couple will have exactly 1 child, there will be a billion boys born. But since the number of boys equals the number of girls, there will also be a billion girls, so there will be 2 billion children in all.

[well]

It would be a little easier to see if you would have written
'...each couple will have exactly 1 boy' instead of '...one child'...

Next time.

[BruceZ]

Since each couple will have exactly 1 <font color="red">child</font>, there will be a billion boys born.

Should be <font color="blue">boy</font>, right. Let's try this again:

Here's a variation of this problem that's a little easier to see. This is due to David. Suppose a billion chinese couples have children until they each get a boy, and then they stop. How many children will be born on average? Assume each baby has an equal chance of being a boy or a girl. Since each couple will have exactly 1 <font color="blue">boy</font>, there will be a billion boys born. But since the number of boys equals the number of girls, there will also be a billion girls, so there will be 2 billion children in all.

[Copernicus]

"But since the number of boys equals the number of girls"

This is poorly worded as well. It is using the result (that there are a billion boys and a billion girls) to come to the conclusion that there are 2 billion total, which is a tautology.

The real issue of the problem is how does it happen that there ARE equal numbers of boys and girls. At first glance it would seem that there must be more Girls than Boys, since the sequences of children must be

B
GB
GGB
GGGB
GGGGB and so on, and it would seem that Girls overwhelm boys in these sequences. It is only when you get into the math of the infinite sequences that you can come to the conclusion that there are equal numbers of boys and girls and therefore 2 billion total.

[BruceZ]

This is poorly worded as well. It is using the result (that there are a billion boys and a billion girls) to come to the conclusion that there are 2 billion total, which is a tautology. The real issue of the problem is how does it happen that there ARE equal numbers of boys and girls....It is only when you get into the math of the infinite sequences that you can come to the conclusion that there are equal numbers of boys and girls and therefore 2 billion total.

Absolutely not. The fact that there are an equal number of boys and girls on average in a large sample is a result of the fact that each child has an equal probability of being a boy or a girl as stated in the problem. This problem can be solved logically without any knowledge of the result of the series. That's the whole point of the problem. You need to think about this some more.

Here is the original thread by David Sklansky.