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Let's play a simple game to test your probabilistic thinking skill.
You start with $100 to play a betting game.
You are going to flip a coin three times.
For each flip, you bet a fixed percentage of your money in hand.
For a head, you double your bet.
For a tail, you lose your bet.
Here is an example:
Say, if you choose to bet a fixed 10% for each flip.
In your first flip, you have to bet $100 x 10% = $10.
If you win, you get $10 x 2 = $20.
And now have $100 + $20 = $120 in hand.
In your second flip, you have to bet $120 x 10% = $12.
If you win again, you will have $24 + $120= $144.
In the third flip, you have to bet $144 x 10% = $14.4.
But say you lose this time, you end up with $144 - $14.4 = $129.6
Question: statistically, what is the optimal percentage to achieve the maximum gain as long as there is 80% chance that your balance will be greater than half of your initial stake (i.e. $50) in the end?
Obviously, if you bet more, you will win more, but also lose more, and vice versa.
There is a optimal percentage that will give you the best theoretical risk/reward balance.
In order to tackle this question, let us break down the expected scenarios.
In the above game, totally eight scenarios can happen: (H = head; T = tail)
HHH: 1.2 x 1.2 x 1.2 = 1.728 (72.8% gain)
HHT: 1.2 x 1.2 x 0.9 = 1.296 (29.6% gain)
HTT: 1.2 x 0.9 x 0.9 = 0.972 (2.8% loss)
HTH: 1.2 x 0.9 x 1.2 = 1.296 (29.6% gain)
THT: 0.9 x 1.2 x 0.9 = 0.972 (2.8% loss)
THH: 0.9 x 1.2 x 1.2 = 1.296 (29.6% gain)
TTH: 0.9 x 0.9 x 1.2 = 0.972 (2.8% loss)
TTT: 0.9 x 0.9 x 0.9 = 0.729 (27.1% loss)
In other words:
One out of eight times (or 1/8 chance) you will gain 72.8%;
Three out of eight times (or 3/8 chance) you will gain 29.6%;
Three out of eight times (or 3/8 chance) you will lose 2.8%;
One out of eight times (or 1/8 chance) you will lose 27.1%.
On average, if you bet a fixed 10% in the three flips, your expected return is:
(1/8)(1.728) + (3/8)(1.296) + (3/8)(0.972) + (1/8)(0.729) = 1.158
Or, if you play the game repeatedly enough, you will gain about 15.8%.
So, the question is not yet answered: what is the optimal % to bet in this game? If you continue this line of logic, you will find that the answer is 29%, when the expected return is about 50%, and there is a 87.5% chance (7/8) that the final balance is above $50.
Further challenge: how will the result be different if:
A) The number of flips is increased to 100, and
B) We now use a biased coin that there is only 30% chance of getting a head?
You start with $100 to play a betting game.
You are going to flip a coin three times.
For each flip, you bet a fixed percentage of your money in hand.
For a head, you double your bet.
For a tail, you lose your bet.
Here is an example:
Say, if you choose to bet a fixed 10% for each flip.
In your first flip, you have to bet $100 x 10% = $10.
If you win, you get $10 x 2 = $20.
And now have $100 + $20 = $120 in hand.
In your second flip, you have to bet $120 x 10% = $12.
If you win again, you will have $24 + $120= $144.
In the third flip, you have to bet $144 x 10% = $14.4.
But say you lose this time, you end up with $144 - $14.4 = $129.6
Question: statistically, what is the optimal percentage to achieve the maximum gain as long as there is 80% chance that your balance will be greater than half of your initial stake (i.e. $50) in the end?
Obviously, if you bet more, you will win more, but also lose more, and vice versa.
There is a optimal percentage that will give you the best theoretical risk/reward balance.
In order to tackle this question, let us break down the expected scenarios.
In the above game, totally eight scenarios can happen: (H = head; T = tail)
HHH: 1.2 x 1.2 x 1.2 = 1.728 (72.8% gain)
HHT: 1.2 x 1.2 x 0.9 = 1.296 (29.6% gain)
HTT: 1.2 x 0.9 x 0.9 = 0.972 (2.8% loss)
HTH: 1.2 x 0.9 x 1.2 = 1.296 (29.6% gain)
THT: 0.9 x 1.2 x 0.9 = 0.972 (2.8% loss)
THH: 0.9 x 1.2 x 1.2 = 1.296 (29.6% gain)
TTH: 0.9 x 0.9 x 1.2 = 0.972 (2.8% loss)
TTT: 0.9 x 0.9 x 0.9 = 0.729 (27.1% loss)
In other words:
One out of eight times (or 1/8 chance) you will gain 72.8%;
Three out of eight times (or 3/8 chance) you will gain 29.6%;
Three out of eight times (or 3/8 chance) you will lose 2.8%;
One out of eight times (or 1/8 chance) you will lose 27.1%.
On average, if you bet a fixed 10% in the three flips, your expected return is:
(1/8)(1.728) + (3/8)(1.296) + (3/8)(0.972) + (1/8)(0.729) = 1.158
Or, if you play the game repeatedly enough, you will gain about 15.8%.
So, the question is not yet answered: what is the optimal % to bet in this game? If you continue this line of logic, you will find that the answer is 29%, when the expected return is about 50%, and there is a 87.5% chance (7/8) that the final balance is above $50.
Further challenge: how will the result be different if:
A) The number of flips is increased to 100, and
B) We now use a biased coin that there is only 30% chance of getting a head?