# Optimal number of goes

1. Aug 21, 2007

### jessica123

A game has two outcomes win or lose, the probability of winning is 1/3 and
the probability of losing is 2/3, if i win i will receive 3 units but if i
lose i will lose 1 unit, I've calculated that the average rate of return on
this game is 33.3%. Over time it becomes more likely that i will receive my 33%
return. My question is if i play the game 1000 times which go would give me the best bang for my buck in other words, what is the optimal number

Ps. Any help would gratefully be appreciated,my thoughts are that a graph with probability of gething 33% return on the y-axis(scale 0 to 1) and No of goes on the x-axis would produce a sigmoidal like fuction with the optimal number being at the point of inflection but how do i find that point!

2. Aug 21, 2007

### CRGreathouse

In general, playing more times will give you more money. After 30 plays your expected earnings are 10; after 3000 they are 1000. The problem becomes more interesting if you have a certain amount of starting money and can't go negative. For example, if you start with 1, you have a 2/3 chance of going broke after your first try.

What do you mean by this?

3. Aug 22, 2007

### jessica123

The Sigmoid like function i referred to can be seen
here(http://en.wikipedia.org/wiki/Sigmoid_function )

If 1 on the y-axis represents certain probability of us
getting our 33.3% return.
Then as the function tends toward 1
it becomes more likely we shall receive our 33.3%

I apologize the sentence should have read
Over Number of goes it becomes more likely that i

4. Aug 22, 2007

### matt grime

No, that still doesn't really make sense. I think you're attempting to invoke some rule that says in the long run things tend to the expected value. Assuming you can lose arbitrarily many times, that is true. Some of us might go broke in the interim.

Your question is still ill-posed. You assert you're going to play 1,000 times and then ask how many times to play. That makes no sense.

At each stage you expect to gain, so play as many times as the sucker who's running the game will let you.

Last edited: Aug 22, 2007
5. Aug 22, 2007

### CRGreathouse

Why do you care about what average return you get? You just want to maximize total winnings, don't you? That is, you'd prefer to make 100 units in 300 plays (average return: about 33%) than 10 units in 10 plays (average return: 100%), right?

6. Aug 24, 2007

### jessica123

The question came in three parts and was verbally stated so
please excuse my poor interpretation, i will try to provide more clarity

Part1 was to work out average rate of return, given the stated stakes and probabilities, which i worked out as 33.3%.

The second part was to draw a graph with probability of getting the 33.3% return on the y-axis and number of times played on the x-axis
This graph tend towards one the more you play and looks similar to
the wikipedia sigmoid function that is linked in my 2nd post.

Part3 was to find the point on the graph that gives the
player the best value for money i was informed that this
is the point of inflection.

I cannot do part 3 and am unsure if i need the information from
part 1 to find the point that is why i stated the question as such in my first post thank you very much for replying, I truly appreciate it.

7. Aug 24, 2007

### CRGreathouse

I still don't know what you mean. Is this the chance of getting at least 1/3 return, or of exactly 1/3?

Tries - At least - Exact
0 - 1 - 1
1 - 1/3 - 0
2 - 5/9 - 0
3 - 7/27 - 0
4 - 11/27 - 0
5 - 17/81 - 0
6 - 233/729 - 0
7 - 379/2187 - 0
8 - 1697/6561 - 0
9 - 6883/19683 - 448/2187
10 - 4195/19683 - 0
11 - 17065/59049 - 0
12 - 31483/177147 - 0
. . .

8. Aug 24, 2007

### EnumaElish

How do you know it has an inflection point? A convergence result would look like an "r" shape, rather than an "s" shape, IMO.

Last edited: Aug 24, 2007