Baffling Math Paradox - Page 7 - Poker Theory

bobf · 06-26-2012, 07:54 AM

Quote:

Originally Posted by Scouse Rob

Isn't this like saying I should fold aces pre-flop rather than call an all-in as when folding P(V>=0)=1

Of course if losing any money meant I would die then I'd throw away the Aces.

But setting a lower death limit on the slip score is surely cheating and changing the rules of the game.

You are basically saying that folding dominates calling because by folding you are 100% sure to not be less than an arbitrary value 0.

In other words by switching anything other than 1 you are taking a gamble, risking some points to (on average) gain more.

So sticking with N>1 is the least risky.

No it's not like saying that at all. It's much stronger than that (if its right). I am saying StayExcept1 beats AlwaysSwitch when trying for $3 and ties AlwaysSwitch when trying for any larger amount. It's like having 2 types of lottery tickets. One has a better chance of winning $3 and and equal chance of winning any of the larger amounts.

I am saying in a single trial

Prob(win at least $1 | StayExcept1) = 1.000
Prob(win at least $1 | AlwaySwitch) = 1.000

Prob(win at least $3 | StayExcept1) = 1.000
Prob(win at least $3 | AlwaySwitch) = 0.750

Prob(win at least $9 | StayExcept1) = 3/8
Prob(win at least $9 | AlwaySwitch) = 3/8

Prob(win at least $27 | StayExcept1) = 3/16
Prob(win at least $27 | AlwaySwitch) = 3/16

Prob(win at least $81 | StayExcept1) = 3/32
Prob(win at least $81 | AlwaySwitch) = 3/32

Prob(win at least 3^T | StayExcept1) = 3/2^(T+1)
Prob(win at least 3^T | AlwaySwitch) = 3/2^(T+1)

I realize this seems to contradict the EV calc. But even if the above is right, I still believe the EV calc is correct but I think the interpretation of it is messed up.

Scouse Rob · 06-26-2012, 09:14 AM

Your probabilities disguise too much information.

We want the most value out of our strategy not just to reach a lower bound.

If the other slip always had 0.5 extra in it then the probabilities of both strategies reaching each of your bounds would be identical whilst switching is clearly a better strategy.

Let E(t) be the expected value gain of switching only for any slip with 3^t or less on it. (Gain over purely sticking.)

Then

E(0) = 1/2
E(1) = 1/2 + 1/4 > E(0)
E(2) = 1/2 + 1/4 + (1/4)(1.5) > E(1)
E(3) = 1/2 + 1/4 + (1/4)(1.5) + (1/4)(1.5^2) > E(2)
....
E(t) = 1/2 + (1/4)(sum[i=1 to t](1.5^(i-1))) > E(t-1)

In fact for any t>=0, E(t) is positive and the larger the value of t the larger the expected gain.

These expected values can be calculated explicitly.

In fact if you repeatedly simulate 2^(t+8) trials you should see the expected gain predicted by the maths above.

Repeated Simulation over groups of 100,000 trials:
E(8) = 12.81
Actual Gains: 12.37, 12.34, 14.08, 13.35, 12.56, 15.33, 11.91, 13.13, ...

stringbettor · 06-26-2012, 01:17 PM

For all cases when n isn't 1

n = 3^T
T = (log n)/(log 3)
T+1 = (log n)/(log 3) +1
T-1 = (log n)/(log 3) - 1

p(T) = 0.5^(T+1) = 0.5^[(log n/log 3) + 1]
p(T+1) = 0.5^[(log n/log 3) + 2]
p(T-1) = 0.5^[(log n/log 3)]

p(T+1)/p(T-1) = 0.5^2 = 0.25

EV(change) = 0.25*(3 n) + 0.75*(1/3 n)
= n
So I agree it doesn't matter unless N =1 when you should switch

bobf · 06-26-2012, 01:18 PM

Quote:

Originally Posted by Scouse Rob

Your probabilities disguise too much information.

We want the most value out of our strategy not just to reach a lower bound.

If the other slip always had 0.5 extra in it then the probabilities of both strategies reaching each of your bounds would be identical whilst switching is clearly a better strategy.

But in this game they don't have 0.5 extra written on them.

In this exact game:
- AlwaysSwitch is dominated by StayExcept1
- There is no value that AlwaysSwitch is more likely to win than StayExcept1 on the next trial.
- There is a value ($3) that StayExcept1 is more likely to win that AlwaysSwitch on the next trial.
- After any finite number of games, StayExcept1 is more likely to be ahead of AlwaysSwitch in new worth than vice versa.

Quote:

Let E(t) be the expected value gain of switching only for any slip with 3^t or less on it. (Gain over purely sticking.)

Then

E(0) = 1/2
E(1) = 1/2 + 1/4 > E(0)
E(2) = 1/2 + 1/4 + (1/4)(1.5) > E(1)
E(3) = 1/2 + 1/4 + (1/4)(1.5) + (1/4)(1.5^2) > E(2)
....
E(t) = 1/2 + (1/4)(sum[i=1 to t](1.5^(i-1))) > E(t-1)

In fact for any t>=0, E(t) is positive and the larger the value of t the larger the expected gain.

These expected values can be calculated explicitly.

In fact if you repeatedly simulate 2^(t+8) trials you should see the expected gain predicted by the maths above.

Repeated Simulation over groups of 100,000 trials:
E(8) = 12.81
Actual Gains: 12.37, 12.34, 14.08, 13.35, 12.56, 15.33, 11.91, 13.13, ...

Rob, can you outline how your simulation works?

I trust that the EV calcuatlions are correct. I do not trust the interpretation of those EV calculations. I do not trust the conclusion that says "therefore you should always switch".

One reason I don't trust this conclusion is that I can construct a situation where the EV says "open any envelope with number N written on it" because envelopes with number N written on them are positive EV for all N and yet when you open the envelopes you end up getting poorer and poorer.

bobf · 06-26-2012, 01:29 PM

1 million trials for each GameLength. Who ever is ahead in net worth at the end of the game gets a win.

Code:

GameLength = 1
  Prob(AlwaysSwitch wins) = 0.25014
  Prob(StayExcept1 wins)  = 0.49995
GameLength = 2
  Prob(AlwaysSwitch wins) = 0.35453
  Prob(StayExcept1 wins)  = 0.54158
GameLength = 3
  Prob(AlwaysSwitch wins) = 0.40725
  Prob(StayExcept1 wins)  = 0.54599
GameLength = 4
  Prob(AlwaysSwitch wins) = 0.42678
  Prob(StayExcept1 wins)  = 0.54098
GameLength = 5
  Prob(AlwaysSwitch wins) = 0.43995
  Prob(StayExcept1 wins)  = 0.53689
GameLength = 6
  Prob(AlwaysSwitch wins) = 0.44713
  Prob(StayExcept1 wins)  = 0.53387
GameLength = 7
  Prob(AlwaysSwitch wins) = 0.45272
  Prob(StayExcept1 wins)  = 0.53137
GameLength = 8
  Prob(AlwaysSwitch wins) = 0.45698
  Prob(StayExcept1 wins)  = 0.52948
GameLength = 9
  Prob(AlwaysSwitch wins) = 0.46070
  Prob(StayExcept1 wins)  = 0.52717
GameLength = 10
  Prob(AlwaysSwitch wins) = 0.46310
  Prob(StayExcept1 wins)  = 0.52647
GameLength = 11
  Prob(AlwaysSwitch wins) = 0.46613
  Prob(StayExcept1 wins)  = 0.52452
GameLength = 12
  Prob(AlwaysSwitch wins) = 0.46746
  Prob(StayExcept1 wins)  = 0.52416
GameLength = 13
  Prob(AlwaysSwitch wins) = 0.46907
  Prob(StayExcept1 wins)  = 0.52357
GameLength = 14
  Prob(AlwaysSwitch wins) = 0.47089
  Prob(StayExcept1 wins)  = 0.52242
GameLength = 15
  Prob(AlwaysSwitch wins) = 0.47142
  Prob(StayExcept1 wins)  = 0.52266
GameLength = 16
  Prob(AlwaysSwitch wins) = 0.47294
  Prob(StayExcept1 wins)  = 0.52174
GameLength = 17
  Prob(AlwaysSwitch wins) = 0.47478
  Prob(StayExcept1 wins)  = 0.52010
GameLength = 18
  Prob(AlwaysSwitch wins) = 0.47425
  Prob(StayExcept1 wins)  = 0.52110
GameLength = 19
  Prob(AlwaysSwitch wins) = 0.47516
  Prob(StayExcept1 wins)  = 0.52056
GameLength = 20
  Prob(AlwaysSwitch wins) = 0.47588
  Prob(StayExcept1 wins)  = 0.52023
GameLength = 21
  Prob(AlwaysSwitch wins) = 0.47734
  Prob(StayExcept1 wins)  = 0.51905
GameLength = 22
  Prob(AlwaysSwitch wins) = 0.47860
  Prob(StayExcept1 wins)  = 0.51814
GameLength = 23
  Prob(AlwaysSwitch wins) = 0.47828
  Prob(StayExcept1 wins)  = 0.51845
GameLength = 24
  Prob(AlwaysSwitch wins) = 0.47895
  Prob(StayExcept1 wins)  = 0.51797
GameLength = 25
  Prob(AlwaysSwitch wins) = 0.48000
  Prob(StayExcept1 wins)  = 0.51720
GameLength = 26
  Prob(AlwaysSwitch wins) = 0.48017
  Prob(StayExcept1 wins)  = 0.51718
GameLength = 27
  Prob(AlwaysSwitch wins) = 0.48033
  Prob(StayExcept1 wins)  = 0.51711
GameLength = 28
  Prob(AlwaysSwitch wins) = 0.48153
  Prob(StayExcept1 wins)  = 0.51609
GameLength = 29
  Prob(AlwaysSwitch wins) = 0.48212
  Prob(StayExcept1 wins)  = 0.51562
GameLength = 30
  Prob(AlwaysSwitch wins) = 0.48270
  Prob(StayExcept1 wins)  = 0.51511
GameLength = 31
  Prob(AlwaysSwitch wins) = 0.48319
  Prob(StayExcept1 wins)  = 0.51483
GameLength = 32
  Prob(AlwaysSwitch wins) = 0.48286
  Prob(StayExcept1 wins)  = 0.51519
GameLength = 33
  Prob(AlwaysSwitch wins) = 0.48293
  Prob(StayExcept1 wins)  = 0.51524
GameLength = 34
  Prob(AlwaysSwitch wins) = 0.48253
  Prob(StayExcept1 wins)  = 0.51570
GameLength = 35
  Prob(AlwaysSwitch wins) = 0.48392
  Prob(StayExcept1 wins)  = 0.51444
GameLength = 36
  Prob(AlwaysSwitch wins) = 0.48352
  Prob(StayExcept1 wins)  = 0.51489
GameLength = 37
  Prob(AlwaysSwitch wins) = 0.48368
  Prob(StayExcept1 wins)  = 0.51478
GameLength = 38
  Prob(AlwaysSwitch wins) = 0.48424
  Prob(StayExcept1 wins)  = 0.51425
GameLength = 39
  Prob(AlwaysSwitch wins) = 0.48525
  Prob(StayExcept1 wins)  = 0.51331
GameLength = 40
  Prob(AlwaysSwitch wins) = 0.48471
  Prob(StayExcept1 wins)  = 0.51388

DarkMagus · 06-26-2012, 01:51 PM

Quote:

Originally Posted by bobf

I'm not using EV's. I deliberately avoided EV's because the EV of the game diverges. I am using probabilites

I am saying that
- if you need $1 or you die, any strategy works 100% of the time.
- if you need $3 or you die, StayExcept1 beats AlwaysSwitch by 100% to 75%.
- if you need any V>$3 or you die, StayExcept1 = AlwaysSwitch, probabilitywise

No matter what amount I need, StayExcept1 is at least as good as AlwaysSwitch.

StayExcept1,3 does not dominate StayExcept1. If I need exactly $3 StayExcept1 always works. StayExcept1,3 fails 25% of the time.

You've modified the game though. If you only care about getting at least V dollars, then the payouts aren't 1, 3, 9, 27, etc., they are a step function 0, 0, 0, 1, 1, 1, ... This new game is going to have a different strategy and a finite EV.

bobf · 06-26-2012, 02:01 PM

Quote:

Originally Posted by DarkMagus

You've modified the game though. If you only care about getting at least V dollars, then the payouts aren't 1, 3, 9, 27, etc., they are a step function 0, 0, 0, 1, 1, 1, ... This new game is going to have a different strategy and a finite EV.

I'm not saying you only care about getting V dollars. I am saying there exists no dollar amount preference at all for which AlwaysSwitch beats StayExcept1. But there is a dollar amount for which StayExcept1 beats AlwaysSwitch.

Choosing between AlwaysSwitch and StayExcept1 is exactly like being given a choice between two lottery tickets who payout structures are identical except that StayExcept1 will pay you at least $3 always but AlwaysSwitch will sometimes pay you only $1.

All of this ^^^ assumes I didn't blunder in some calculation. So if someone can confirm the following that would be nice....

Prob(win at least $1 | StayExcept1) = 1.000
Prob(win at least $1 | AlwaySwitch) = 1.000

Prob(win at least $3 | StayExcept1) = 1.000
Prob(win at least $3 | AlwaySwitch) = 0.750

....

Prob(win at least 3^T | StayExcept1) = 3/2^(T+1)
Prob(win at least 3^T | AlwaySwitch) = 3/2^(T+1)

for T >1 on a single game trial.

DarkMagus · 06-26-2012, 02:16 PM

Quote:

Originally Posted by bobf

I'm not saying you only care about getting V dollars.

Yes, this is exactly what you're saying. Your logic assumes that getting $3 100% of the time is better than getting $9 99% of the time and $1 1% of the time.

bobf · 06-26-2012, 02:22 PM

Quote:

Originally Posted by DarkMagus

Yes, this is exactly what you're saying. Your logic assumes that getting $3 100% of the time is better than getting $9 99% of the time and $1 1% of the time.

No

In the next trial StayExcept1 is just as likely to win $9 as AlwaysSwitch.
In the next trial StayExcept1 is just as likely to win $27 as AlwaysSwitch.
In the next trial StayExcept1 is just as likely to win $81 as AlwaysSwitch.
In the next trial StayExcept1 is just as likely to win $243 as AlwaysSwitch.
...
forever

But
In the next trial StayExcept1 is more likely to win $3 than AlwaysSwitch.

R Gibert · 06-26-2012, 02:23 PM

Quote:

Originally Posted by DarkMagus

You've modified the game though. If you only care about getting at least V dollars, then the payouts aren't 1, 3, 9, 27, etc., they are a step function 0, 0, 0, 1, 1, 1, ... This new game is going to have a different strategy and a finite EV.

It's the same game, but he is computing a different metric. AlwaysStayExcept1 does better 75% of the time, but this is very misleading. It performs better only when what is at stake is minimal. The rest of the time it is break-even. It's what happens the rest of the time that matters most.

I think that considering AlwaysStayExcept1 is needlessly confusing. If you compare AlwaysStay with AlwaysSwitch, it is easier to see that these perform equivalently. They each do better than the other 50% of the time. The improvement in AlwaysStayExcept1 is so miniscule in dollar terms that it can be considered virtually negligible.

bobf · 06-26-2012, 02:39 PM

Repeat 1 million times
1. Flip coins until H comes. Let T = Number of tails.
2. Write down A = 3^T and B=3^(T+1)
3. Fiip an additional coin to decide who gets what slip.
4. Give AlwaysSwitch A on heads, B on tails
5. Give StayExcept1 B on heads, A on tails
6. Apply strategies
7. Accumulate net worths

StayExcept1 will win this game with Prob close to 1.

Code:

Flips   Slips  AS Sees  SX1 Sees  AS Wins  SX1 Wins   Delta
Hh      (1,3)      1        3        3        3          0
Ht      (3,1)      3        1        1        3         +2
THh    (3,9)       3        9        9        9          0
THt    (9,3)       9        3        3        3          0
TTHh  (9,27)       9       27       27       27          0
TTHt   (27,9)     27        9        9        9          0
.....
all the rest of the deltas are zero

Scouse Rob · 06-26-2012, 03:49 PM

Quote:

Originally Posted by bobf

Rob, can you outline how your simulation works?

I trust that the EV calcuatlions are correct. I do not trust the interpretation of those EV calculations. I do not trust the conclusion that says "therefore you should always switch".

One reason I don't trust this conclusion is that I can construct a situation where the EV says "open any envelope with number N written on it" because envelopes with number N written on them are positive EV for all N and yet when you open the envelopes you end up getting poorer and poorer.

I simply create 100,000 trials where the number of consecutive tails is constructed.
(Limit maximum tails to 21, which is more than enough as any number of tails greater than t in E(t) contributes zero as we stick by definition.)

Then place the two values randomly for each trial.

Then to calculate E(t) add the gains (positive or negative) for each switch when the number on the envelope is less than or equal to 3^t.

Of course when t gets too high the variance of the small amount of the higher numbers takes over making the simulated gain range from hugely positive to hugely negative and anything in between.

But if t is low enough so there are enough expected occurrences of 3^t then the average simulated gain is close to E(t).

I am VERY interested by your result that over repeated small numbers of trials that always switch perform so badly.

My intuition tells me this may be because the rare highest numbers we see over all the of trials simulated are weighted 2-1 to lose big rather than win big and this skews the results due to the fact they are geometrically bigger than anything else.

Increasing the number of trials or reruns of trials to get a better average for this high number of tails just creates a small number of the next amount of tails and the problem re-manifests itself.

(Complete speculation. And only regarding the result above. The other reason is that always switching gains you nothing as switching and sticking are equivalent over all envelopes, but always changing 1 is a gain. Thus the paradox.)

I never stated that always switch should be better. (Or maybe I did but not in that last reply.

)
Only that E(t) is better than E(0) for any integer t>0.
(And this can be simulated given enough trials.)

I am convinced by my simulation that this is true.
I could PM the excel file to you if you wish but it is easy enough to construct, and probably even easier for you to do in code.

Just modify your program from always switch with 1 to always switch with 1 or 3 and see the difference over the million trials.

And so on to about 3^12.

You'll probably need more trials for higher numbers of t but by this stage you'll be convinced.

DarkMagus · 06-26-2012, 04:03 PM

Quote:

Originally Posted by bobf

Repeat 1 million times
1. Flip coins until H comes. Let T = Number of tails.
2. Write down A = 3^T and B=3^(T+1)
3. Fiip an additional coin to decide who gets what slip.
4. Give AlwaysSwitch A on heads, B on tails
5. Give StayExcept1 B on heads, A on tails
6. Apply strategies
7. Accumulate net worths

StayExcept1 will win this game with Prob close to 1.

Code:

Flips   Slips  AS Sees  SX1 Sees  AS Wins  SX1 Wins   Delta
Hh      (1,3)      1        3        3        3          0
Ht      (3,1)      3        1        1        3         +2
THh    (3,9)       3        9        9        9          0
THt    (9,3)       9        3        3        3          0
TTHh  (9,27)       9       27       27       27          0
TTHt   (27,9)     27        9        9        9          0
.....
all the rest of the deltas are zero

Ok but the probability of winning does not detemine what makes the best strategy. The EV does. I could sit at a NLHE table and go allin every hand and I'd win very often, does that make it the best possible strategy?

Both strategies have an undefined EV so saying one is "better" than the other just doesn't make any mathematical sense.

bobf · 06-26-2012, 04:33 PM

Quote:

Originally Posted by Scouse Rob

I simply create 100,000 trials where the number of consecutive tails is constructed.
(Limit maximum tails to 21, which is more than enough as any number of tails greater than t in E(t) contributes zero as we stick by definition.)

Then place the two values randomly for each trial.

Then to calculate E(t) add the gains (positive or negative) for each switch when the number on the envelope is less than or equal to 3^t.

Of course when t gets too high the variance of the small amount of the higher numbers takes over making the simulated gain range from hugely positive to hugely negative and anything in between.

But if t is low enough so there are enough expected occurrences of 3^t then the average simulated gain is close to E(t).

I am VERY interested by your result that over repeated small numbers of trials that always switch perform so badly.

My intuition tells me this may be because the rare highest numbers we see over all the of trials simulated are weighted 2-1 to lose big rather than win big and this skews the results due to the fact they are geometrically bigger than anything else.

Increasing the number of trials or reruns of trials to get a better average for this high number of tails just creates a small number of the next amount of tails and the problem re-manifests itself.

(Complete speculation. And only regarding the result above. The other reason is that always switching gains you nothing as switching and sticking are equivalent over all envelopes, but always changing 1 is a gain. Thus the paradox.)

I never stated that always switch should be better. (Or maybe I did but not in that last reply.

)
Only that E(t) is better than E(0) for any integer t>0.
(And this can be simulated given enough trials.)

I am convinced by my simulation that this is true.
I could PM the excel file to you if you wish but it is easy enough to construct, and probably even easier for you to do in code.

Just modify your program from always switch with 1 to always switch with 1 or 3 and see the difference over the million trials.

And so on to about 3^12.

You'll probably need more trials for higher numbers of t but by this stage you'll be convinced.

Ok thanks. I agree that switching for any value N produces a positive EV relative to staying. But I believe that it is wrong to conclude anything about how we should play the game from that fact.

Let me try to illustrate. Let's generate some number pairs we might encounter when trying to calculate these EV's.

(1,3) (1,3) (1,3) (1,3) (1,3)(1,3)(1,3)(1,3)
(3,9) (3,9) (3,9) (3,9)
(9,27)(9,27)
(27,81)

an EV calculator for switching would conclude
EV(switch | see 1) = 2
EV(switch | see 3) = (-16 + 24) / 12 = 2/3
EV(switch | see 9) = (-24 + 36) / 6 = 2
EV(switch | see 27) = (-36 + 54) / 3 = 6
EV(switch | see 81) = -54 OOPS

that last "oops" will alwyas be there while you are playijng AlwaysSwitch for a finite length of time. Play longer and EV(switch | 81) will converge correctly but a new "oops" will pop up for higher and higher N's. No matter how long you play, you never realize any advantage from switching even though more and more N's are converging to their proper EV.

Quote:

My intuition tells me this may be because the rare highest numbers we see over all the of trials simulated are weighted 2-1 to lose big rather than win big and this skews the results due to the fact they are geometrically bigger than anything else.

If I understand you, I believe this is the case too. But I believe that it's the EV calculator that is scewing the results by not mimicing reality of actually playing. In the actual game, no matter how long we play, there is always a largest number on any slip of paper so far. Any switch from that paper is guaranteed to lose a huge sum of money because there is no larger number to switch to yet. In my example above, a person playing AlwaysSwitch loses $54 when he sees 81 and switches. The EV calculator just says, that's ok, I know this will eventually be corrected by switching to some 243's. And that's true. That will happen. But by the time that happens, a new biggest number will appear, like 243, with a new guaranteed loss.

DarkMagus · 06-26-2012, 04:38 PM

Quote:

Originally Posted by bobf

Ok thanks. I agree that switching for any value N produces a positive EV relative to staying. But I believe that it is wrong to conclude anything about how we should play the game from that fact.

Let me try to illustrate. Let's generate some number pairs we might encounter when trying to calculate these EV's.

(1,3) (1,3) (1,3) (1,3) (1,3)(1,3)(1,3)(1,3)
(3,9) (3,9) (3,9) (3,9)
(9,27)(9,27)
(27,81)

an EV calculator for switching would conclude
EV(switch | see 1) = 2
EV(switch | see 3) = (-16 + 24) / 12 = 2/3
EV(switch | see 9) = (-24 + 36) / 6 = 2
EV(switch | see 27) = (-36 + 54) / 3 = 6
EV(switch | see 81) = -54 OOPS

Here you are taking the results and then calculating the EV based on already knowing the results. This makes no sense. You really need to rethink your logic on this.