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I got one approach for this,not sure if it is optimum.But with this algorithm in worst case also player will have more money at the end of the game what was with him at the begining.
- shaw.ravi July 08, 20121- Determine LossCount=n and WinCount=n
2- Determine the money to be bet.
a- If LossCount =0 then bet the whole money
b - Else devide the remaining money by (LossCount + 1)
3- Play the card.
a- If it is loss then LossCount-- and RemaningMoney = RemainingMoney - BetMoney.
b - Else WinCount-- and RemainingMoney=RemainingMoney+BetMoney
4- Go to Step 2 and repeat this till nth time.
Logic for choosig the divider as (LossCount +1) - We need to survive in the game till we start wining.Suppose we have X money and n win and n Loss chnaces.In worst case for the first nth time we lose.So for the next (n+1)th time we need minimum money to play.