Interesting question, looks like this is the opposite of “double down” strategy in gambling. It's a “half down” strategy, seems still a losing one though.
Let's say the player wins x times among the 30 tosses, and loses 30-x times. Then the money at the end of 30 tosses will be 10000*1.5^x*0.5^(30-x). We want 10000*1.5^x*0.5^(30-x) > 10000, solve it gives us
x > 30*ln2/ln3 = 18.9, so x =19,20,21, … 30.
This is a binomial process, thus the total probability is
[C(30,19)+C(30,20) +C(30,21) +…+C(30,30)]*(1/2)^30 = 107636432/1073741824
Which is about 10%, so the poor guy has only 10% chance to come out ahead after 30 losses.
For 100 losses,x>63, same method applies. My guess is that the guy's chance will get even dimmer.
Let's say the player wins x times among the 30 tosses, and loses 30-x times. Then the money at the end of 30 tosses will be 10000*1.5^x*0.5^(30-x). We want 10000*1.5^x*0.5^(30-x) > 10000, solve it gives us
x > 30*ln2/ln3 = 18.9, so x =19,20,21, … 30.
This is a binomial process, thus the total probability is
[C(30,19)+C(30,20) +C(30,21) +…+C(30,30)]*(1/2)^30 = 107636432/1073741824
Which is about 10%, so the poor guy has only 10% chance to come out ahead after 30 losses.
For 100 losses,x>63, same method applies. My guess is that the guy's chance will get even dimmer.
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