Finding out long term betting strategy in the game based on normally distributed random numbers

[Dhamnekar Win,odd]

Suppose we play the game where I draw a N(0, 1) random number A, show it to you, then draw another independent N(0, 1) one B, but before showing it to you ask you to guess whether B will be larger than A or not. If you are right, you win if you are wrong you lose. What is your strategy, and what will be your long term success rate if we keep playing?

Author's Solution:

Since the numbers are zero-mean, the strategy is to guess the sign of the difference B − A as positive if A < 0, and negative otherwise. The long-term success rate will be \(\frac12 P(B-A) > 0| A < 0) + \frac12 P(B-A) \leq 0 | A \geq 0)\)

Due to the symmetry, and the fact that P(A=0)=0 , the above expression simplifies to P(B-A) > 0 |A < 0). Plotting on the (a, b) Cartesian plane the part of the region a < 0 where b − a > 0, we see that this is one and a half quadrants out of the two comprising the full a < 0 region. Therefore, due to the radial symmetry of the Gaussian two-dimensional distribution with zero correlation, we conclude that the success rate is \(\frac34\).

My answer:

The game you’re describing involves guessing whether a second independent draw from a standard normal distribution (N(0,1)) will be larger than the first.

Given that both A and B are drawn from the same distribution, there is no information about B that can be inferred from A. Therefore, the best strategy is to always guess that B will be larger than A, or always guess that B will be smaller than A. Both strategies have the same success rate.

The long-term success rate of this strategy is 50%. This is because, for two independent draws from the same distribution, the probability that the second draw is larger than the first is the same as the probability that the first draw is larger than the second.

In mathematical terms, if A and B are independent and identically distributed (i.i.d.) random variables, then:

P(B > A) = P(A < B) = 0.5

So, no matter how many times we play this game, your long-term success rate will converge to 50%.

 

Whose answer is correct? If Author's solution is correct, would you show me graphically how can the part of the region a < 0 where b-a > 0 is one and half quadrants out of the two quadrants comprising the full a < 0 region? 

Edited January 20 by Dhamnekar Win,odd

[Dhamnekar Win,odd]

My answer is wrong. Author's solution is correct.☺️

[dimreepr]

3 hours ago, Dhamnekar Win,odd said:

My answer is wrong. Author's solution is correct.☺️

What are the chances???

[Dhamnekar Win,odd]

Author has given the success rate in his solution. Please read it.☺️

[dimreepr]

On 1/22/2024 at 12:42 PM, Dhamnekar Win,odd said:

Author has given the success rate in his solution. Please read it.☺️

Not my job, please explain it... 

[Dhamnekar Win,odd]

On 1/23/2024 at 6:43 PM, dimreepr said:

Not my job, please explain it... 

Success rate is 75% because 12P((B−A)>0|A>0)+12P((B−A)≤0|A<0)=0.75 in the long term. If A is +ve, I shall tell B is lower than A and if A is -ve, I shall tell B is larger than A.🤔🤔🤔🤔 ☺️☺️☺️☺️☺️☺️☺️

Edited January 29 by Dhamnekar Win,odd

[dimreepr]

10 minutes ago, Dhamnekar Win,odd said:

Success rate is 75% because 12P((B−A)>0|A>0)+12P((B−A)≤0|A<0)=0.75 in the long term. If A is +ve, I shall tell B is lower than A and if A is -ve, I shall tell B is larger than A. 

Not sure how that would help with my poker strategy.

[Leojames26]

Alright, so here’s the deal. The author’s solution is actually correct, even though it might seem counterintuitive at first. The key is that the strategy isn’t just about guessing randomly or assuming B will be bigger or smaller than A—there’s a bit more to it.

When A is negative (A < 0), B is more likely to be bigger than A, and vice versa when A is positive. This tilts the odds in your favor slightly, but not in a huge way. That’s why the long-term success rate isn’t 50%, but rather 75%.

As for the whole "one and a half quadrants" thing, imagine plotting points on a graph where A is on the x-axis and B is on the y-axis. When A < 0, B is more likely to be bigger, which covers one and a half out of two quadrants where A < 0. This gives you that 3/4 success rate.

So yeah, the author nailed it. You’re right that for two completely independent draws, it would be 50%, but here the strategy adds a bit of an edge. Hope that clears it up!