find the expected number of turns

grobber

I don't know if this is the right place to post this, but I'm sure it will be moved if it's not Smile

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A player wins at each turn an amout uniformly distributed in [ $0,1$ ], and stops when he has won a cumulative total of at least $1$ . What's the expected number of turns that the game lasts?

**Posted:** Sun Dec 12, 2004 1:06 pm

TripleM

I doubt this is correct but..
Let $f(b)$ = the expected number of turns to get a cumulative total at least b. Then

$f(b)=(1-b) + \int^b_0 (f(b-x) + 1) dx$

$= 1+\int^b_0 f(b-x) dx$

$= 1+\int^b_0 f(x) dx$

So
$F'(b) = 1 + F(b) - F(0)$
$F(b) = e^b + C$

So $f(b) = e^b$ and the answer is $f(1) = e$ .

_________________
Stephen

grobber

I got the same answer, but I'll have to read this carefully. What is $f$ exactly? The expected number of turns when you start from $0$ and stop at $b$ , or the expected number of turns when you have $b$ and you want to reach $1$ ?

**Posted:** Sun Dec 12, 2004 1:39 pm

TripleM

The latter. The first formula for f just comes (hopefully) from the fact that you have a (1-b) probability of getting over 1 next turn, otherwise you get a value of x, and the expected value will be 1 (this turn) + f(b-x).

**Posted:** Sun Dec 12, 2004 1:56 pm

_________________
Stephen

grobber

But if you have the probability $1-b$ of getting over $1$ , why do you add this probability to the expected number of turns?

Besides, if you define $f(b)$ as the expected number of turns if you have $b$ and want to reach $1$ , then $f(1)$ should be $0$ , don't you think?

**Posted:** Sun Dec 12, 2004 2:13 pm

TripleM

Gah, edited. I did mean that f(b) is the expected number of turns to get a total of b, not the other one. Then expected total is (1-b) * 1 + other probabilities * turns in those cases.

**Posted:** Sun Dec 12, 2004 2:21 pm

_________________
Stephen

grobber

But with this new definition of $f$ , why are you interested in the probability of getting over $1$ , which is $1-b$ ? Aren't you interested in getting a total of at least $b$ ? What does $1$ have to do with this case? I'm sorry if I start to ask stupid questions, I'm a bit tired right now (it's past my bed time Smile

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**Posted:** Sun Dec 12, 2004 2:31 pm

TripleM

If b = 0.7, say, then you get over b if you get anything from 0.7 to 1, which is a probability of 1-b. Note that I'm "translating everything backwards" after each turn, ie if you first turn is 0.3, then you assume after that you start from 0 and are aiming to get to 0.7.

**Posted:** Sun Dec 12, 2004 2:35 pm

_________________
Stephen

grobber

I got it. Yes, everything seems to be falling into place now Smile

.

**Posted:** Sun Dec 12, 2004 2:36 pm

Kent Merryfield

I have an extension of this, and will start a new thread for it, which will be titled "Expected number of turns (2)."

As for the right forum for this? This seems as good as any, so let's keep it here.

**Posted:** Mon Dec 13, 2004 9:20 am

Kent Merryfield

I'm going to go ahead and move this to "solved." TripleM's approach is correct, and his function, $f(b)$ is indeed $e^b.$ The answer to the original question is $e.$ This approach can be generalized considerably, which is what the "Number of turns (2)" topic is about.

**Posted:** Wed Dec 22, 2004 11:00 pm