Suppose you play a game which has equal probabilities of winning and loosing, example the game of tossing a coin. Every time before tossing the coin, set a particular amount as a bet. If it’s heads then you win and you get double the money. But if it’s tails, then you loose the bet amount.
So, if you are required to play this game, what strategy must be applied so that you are in net profit? Before revealing the solution let’s take a deeper look at the problem.
If I stake 1 coins (let say) every time, if I win at n th step I would bear a loss of n-2 coins. So, what I should think is that if I only have one chance to recover all previous loses, every time I loose, I should increase my bet at the next chance. But the question is by what factor? This brings us to the solution with Martingale Strategy.
Martingale Strategy
Double the stake (bet) after every loss. Example: place bet like a, 2a, 4a, 8a… where a is minimum 1 coin. If you win at k th toss, you get a net profit of a. But why double? If I triple my bet after every loss, soon I would exhaust all my money. That’s the drawback of Martingale Strategy. Since the bet if increased exponentially, doubling the bet everytime is the best option.
Proof:
Let say first time bet amount is a and you bet according to the Martingale Strategy. Suppose win takes place at k th step. Then loss occured till (k-1) th step
= a+2a+4a+….+2^(k-2)a
= a[2^(k-1) – 1]
Since at k th step the stake is a[2^(k-1)], the profit is a[2^(k-1)]. So, net profit is a. So, even if this strategy has limitation that all your money is exhausted if you don’t win after few steps but it’s good because it always lands you with profit.
That’s all you got! Thanks for reading. For more consider supporting Mathhexagon. Share and follow me on social platforms.
Mathhexagon 