Brain-training tasks related to trading in one way or another. Theorist, game theory, etc. - page 12
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this is equivalent to:
p(A)=P
p(B)=1-P=Q
=> P^2+Q^2 >= 2*P*Q
=> (P-Q)^2>=0
hehehe, it's been a long time since I came here - now Reshetov has proved that the square of any number cannot be negative... via a theorist! I'm going down :D
))))))))))))))
I will reveal the essence of Excel. It's simple and obvious.
....
and so on. There is no error here.
You're not taking into account the fact that once the cumulative sum of all previous outcomes becomes negative, the game stops - you can't trade in debt. And your excel approach does exactly that.
Once again, you are arguing with the multiplication table. At the same time you don't even know arithmetic yourself. It's not even funny. 28% is a guaranteed drain.
It depends on the conditions of the problem.
If the chance of winning is 100%, it is necessary to bet 100% of the deposit.
If the chance is close to 100%, it is necessary to bet a significant part of the deposit, etc.
In the conditions of the problem you win 2 coins and lose one. This is a very good trading system.
So 28% of the deposit is good enough.
************************************
Also please note that you can't play for debt here, even if you lose 100 in a row. The sum of the outcomes will never turn negative. Even if you lose 1000 times. Okay?
I will reveal the essence of Excel. It's simple and obvious.
...
100*028=28 we win... 2 coins. 2*28 = 56
the deposit became 156.
156*0.28=43.68 we lost 1 coin -43.68
depo became 112.32
...
There is no error here.
*****************************************
The question is more about using Kelly's formula correctly.
Are we putting the correct values there?
No they are not. Reread your own terms of the problem. Why do we suddenly win 2 coins and lose 1, when you said so earlier:
TVA_11:
...
Let's say we're playing heads/tails.
We lose 2, we win 3. For the sake of simplicity, let's drop the spread.
...You are making mistakes out of the blue. And don't tell us what Exel is all about. You at least need to master arithmetic and learn how to count without error, at least on your own terms.
You do not take into account the fact that as soon as the cumulative sum of all previous outcomes becomes negative, the game stops - you can't trade in debt. And your excel approach does exactly that.
Once again, you are arguing with the multiplication table. At the same time you don't even know arithmetic yourself. It's not even funny. 28% is a guaranteed bust.
this equals:
p(A)=P
p(B)=1-P=Q
=> P^2 + Q^2 >= 2*P*Q
...
Geez, it's been a long time since I came here - now Reshetov has proved that the square of any number cannot be negative... via theorist! I'm going down :D
I'd rather not go in at all, so as not to embarrass myself in algebra lameness:
P^2 + Q^2 <= 1 - 2 * P * Q
The thing is:
P + Q = 1
(P + Q)^2 = P^2 + 2 * P * Q + Q^2 = 1^2 = 1
Consequently, if:
P^2 + 2 * P * Q + Q^2 = 1
then:
P^2 + Q^2 = 1 - 2 * P * Q
I'd rather not come in at all, so as not to embarrass myself in algebra lameness:
P^2 + Q^2 <= 1 - 2 * P * Q
The thing is:
P + Q = 1
(P + Q)^2 = P^2 + 2 * P * Q + Q^2 = 1^2 = 1
Consequently, if:
P^2 + 2 * P * Q + Q^2 = 1
then:
P^2 + Q^2 = 1 - 2 * P * Q
What the hell are you smoking?
for any numbers p and q - not necessarily related, but completely arbitrary - the inequality
(p-q)^2>=0,
and hence (open the brackets and open your eyes at the same time)
p^2+q^2>=p*q+q*p
This is your inequality... lamer yourself.
What the hell are you smoking?
For any numbers p and q - not necessarily related, but completely arbitrary - the inequality
(p-q)^2>=0,
and hence (open the brackets and open your eyes too).
p^2+q^2>=p*q+q*p
This is your inequality... lamer yourself.
I'm sorry. Shit, I thought "=>" meant "follows". Only now I realized it's "greater than or equal to".
This is correct. We have another proof of this inequality, namely the square of any value cannot be negative.
Apologies. Shit, I thought "=>" meant "to follow". Only now I realised it was "more or equal".
That's right. We have another proof of this inequality, namely the square of any value cannot be negative.
28% is not a guaranteed loss, as the loss starts when the Kelly maximum is exceeded by half. I gave a screenshot from Excel on the previous page and it clearly shows that at 28% of the deposit the yield will be about 2 and a bit percent after two coin flips. For this problem, the loss area starts somewhere beyond the 33.4% of deposit bet level.
I ran 10000 simulations for 28% in MATLAB, here is a histogram of the lifetime of this strategy, i.e. before the loss. The vast majority of cases (90%) were lost before the 100th trade. Very few people last longer. I.e. failure is guaranteed.