My first instinct is that you want to minimize the number of bets you make, because the probability of winning decreases exponentially the more mint flips you play ( assuming you bet everything you have every prison term ). therefore, the sequence to a successful game is :

- bet 200 (win), bet 400 (win), bet 800 (win) -> win a total of 1600

The probability of this case happening is $ ( \frac { 1 } { 2 } ) ^3=0.125 $

Read more: All About Australian Coins

however, notice that we actually surpass the finish, and if we can bet good the deviation between the finish and what we have we can reach the same target with the same probability, but without risking the crippled ending immediately. If we lose in this case, we can continue to bet and have a find at reaching the target. One such scheme has the stick to sequences of events :

- bet 200 (win), bet 400 (win), bet 200 (win) -> win a total of 1000
- bet 200 (win), bet 400 (win), bet 200 (lose), bet 400 (win) -> win a total of 1000
- bet 200 (win), bet 400 (win), bet 200 (lose), bet 400 (lose), (200 remain, start over…)

I ‘m not quite certain how to calculate probability for this, since font 3 basically resets the game, but if we consider fair the beginning two cases then the probability for reaching the target is at least $ ( \frac { 1 } { 2 } ) ^3+ ( \frac { 1 } { 2 } ) ^4=0.1875 $

sol I believe that this shows the second strategy is better than the first. But this can be extended to the first pair of mint tosses adenine well, so that you do n’t bet your integral toilet on any of the coin flips ( except the last before you run out of money ). My question is how do you analyze this to find what measure to bet at each mint pass, given a target amount \ $ X, for a best scheme that minimizes the number of games played ?

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