I got this problem from one of my friend. The problem is:

A player starts of with 10 points. If he reaches 1000 points he wins.

Every time he plays, the chance of him winning is 40% Maximum everytime, he can only double the amount that he has put in.

That means that for example if he puts in 10 points and wins he can only reach 20 points and if he puts in that 20 points again and wins he reaches 40 points. However, if he loses those 20 points then he reaches back to 0 again and he can start of again with 10 points.

Thus the game is that he can always start of with 10 points if he has reached 0.

The question is how many times would he have to play so that on average he can reach 1000 points the fastest way possible. Thus, if he reaches 100 and then puts in only 80 points, and reaches 160 and uses the remaining 20 points originally to increase his overall amount to 180 and if that helps him in reaching 1000 points faster than if he put all those initial 100 points than so be it.

I am not able to find any easy way to do this.I tried to solve using markov chains, but it was going very complex. Every time he plays, the chance of him winning is 40% Maximum everytime, he can only double the amount that he has put in.

That means that for example if he puts in 10 points and wins he can only reach 20 points and if he puts in that 20 points again and wins he reaches 40 points. However, if he loses those 20 points then he reaches back to 0 again and he can start of again with 10 points.

Thus the game is that he can always start of with 10 points if he has reached 0.

The question is how many times would he have to play so that on average he can reach 1000 points the fastest way possible. Thus, if he reaches 100 and then puts in only 80 points, and reaches 160 and uses the remaining 20 points originally to increase his overall amount to 180 and if that helps him in reaching 1000 points faster than if he put all those initial 100 points than so be it.

I think this problem will have two stage.

- Strategy of putting money ( I guess 10 ,20,40,80,... is the best strategy)
- Finding the distribution of number of trials

Regards

Richie