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Thread: Probability theory

  1. #40


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    Quote Originally Posted by Meistro123 View Post
    should be subtracting 2 and not adding it
    Yup. PEDMAS rules. As a kid, though, I was befuddled. It was sorcery, I tell you.

  2. #41


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    Quote Originally Posted by hypercube View Post
    the math proofs are wrong because there is never any ev of $125.. that is this mistake.. $125 was never a possible outcome so you cant have those two boxes modeled that way. you have $100 and you can either lose 50% or gain 100% which is a net ev of exactly 0 not + $25 as explained:

    it is not better to gain $100 than lose $50 when you already have $100.

    50% loss is exactly equal to 100% gain because if you lose that bet, now you need 100% gain to get back to the original $100 and 400% instead of 100% to get to the $200 from the original $100 we had. The ev to switch is $0,not +$25 and the advantage in edge to switch is also exactly zero. From perspective of a card counter we dont tend to like risking money without an edge so we should never switch. but we all would anyway because its more fun and there zero positive or negative expectancy.
    May I speculate where the idea that the 125 USD EV comes from? Possibly is that we are gaining equity no matter what the outcome is? That is, I gain on average 125 USD per round no matter what? I either make 50 or 200 dollars switching. Each has a .5 probability then the EV(Switch) = E(E_1)*P(E_1) + E(E_2)*P(E_2) = EV/2 = 125 units.

    Where as, the correct EV is the P(2D)*(2D - D)+P(.5D)*(-D+.5D) = 0. Correct?

  3. #42


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    model this as percentage changes everytime you switch not with dollar changes and then you will get why there is no +ev to switching. because the other box always has 1/2x or 2x not a specific dollar amount...

    say you win first switch.. you go from $100 to 200 but then lose second switch you drop back to $100 (half) not 150. it takes 2x win to recover from 1/2 loss. so run this a million times and you have gone nowhere

  4. #43
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    Quote Originally Posted by RCJH View Post
    If this guy thinks he proved that after finding out the amount in one envelope that the decision on which envelope to choose isn't clear he is an idiot. Know what you have to watch out for is the person running the show can choose whether to show you the higher amount or the lower amount. If the amount shown was chosen at random and is equally likely to be the larger amount as the smaller amount choosing the unknown envelope is better.

    From the link in bold:

    1) When you make your choice (C)
    2) Either the envelope containing the lower value (C=lower) or the one that contains the higher (C=higher).
    3) During the game, the value (V) of the content of the chosen envelope is revealed to be a certain value M.

    The paradox above arose because you assumed that
    4) P(C=lower|V=M) = P(C=higher|V=M) = 0.5

    No, the paradox arose because M was a variable. You make it a constant then V is the same constant. That is your number 3 in the bold.
    Let's see why this cannot be the case. (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    No, V is not a variable. If V=M and M is a constant then V is also a constant. You can't treat it like a variable. See your number 3 above. C is a constant as well. C is the envelope you chose. If you know nothing about the contents of either envelope you can view it as a variable but once you know the amount in one envelope it is no longer a variable. It is the envelope that contains the amount M which is a constant. It's value, V, is equal to M which is a constant. So L is your only variable. And you have 3 constants; C,V, and M where V=M.

    The paradox above arose because you assumed that
    P(C=lower|V=M) = P(C=higher|V=M) = 0.5

    No, that was a given. The paradox arose because M was a variable.
    Let's see why this cannot be the case. (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    No, L is the only variable. We Know M is a constant. V=M so V is also a constant. And C is the envelope you chose which contains the amount M, a constant and has the value, V, also a constant.
    P(C=lower|V=M) = P(C=lower)P(L=M)/P(V=M) . . . (2)
    and P(C=higher|V=M) = P(C=higher)P(L=M/2)/P(V=M) . . . (3)

    I skipped to this because with all the constants the preceding was pretty meaningless.
    Given V=M, then P(V=M)=1, this reduces to P(C=lower|V=M) = P(C=lower)P(L=M) and P(C=higher|V=M) = P(C=higher)P(L=M/2). Both terms mean the exact same thing so what is the point both statements are givens.
    From (2), P(C=lower|V=M) is the same as P(C=lower) only if P(L=M) is the same as P(V=M).
    No. The other possibility is the probability that V=M is 1. V=M is 100%, that was how you defined M and V. M is a constant that is the amount in the envelope. V is the value of the contents of the chosen envelope which is the constant M. So the probability that V=M is 1. So P(L=M) does not have to be the same as P(V=M). Your treating constants as variables lead to this false statement. Everything based on this falsehood is wrong.

    I can't believe this guy thought he could prove that the odds you chose the envelope with the larger amount or smaller amount can't be equal at .5 and .5 LoL

    The paradox above arose because you assumed that
    P(C=lower|V=M) = P(C=higher|V=M) = 0.5
    Let's see why this cannot be the case.

    Well I guess if you take a given like V=M, and then define V as a variable and M as a constant you can make it look like you did to those that don't try to follow what you did.
    (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    Last edited by Three; 02-18-2018 at 09:26 PM.

  5. #44
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    Quote Originally Posted by RCJH View Post
    the room only cost $25 and hands them 5 $1 bills back. Now each person has $9, with two left over.
    The question is how did $5 become $29.

  6. #45
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    Quote Originally Posted by hypercube View Post
    he math proofs are wrong because there is never any ev of $125.. that is this mistake.. $125 was never a possible outcome so you cant have those two boxes modeled that way.
    EV is your average expectation not the amount in the envelope.Like you flip a far coin and you lose $1 on a heads and win $1 on tails the EV is zero, .5*1 +.5*(-1) = 0. That doesn't mean the coin will never land.
    Quote Originally Posted by hypercube View Post
    it is not better to gain $100 than lose $50 when you already have $100.
    You have $100 and you either end up with $50 or $200. You don't lose $50, you lose $50 of the $100 you have to end up winning a total of $50 instead of a total of $100 or winning $200 instead of $100. The EV for the envelope of (.5*$50 + .5*$200) which equals $125.
    I won't even address the last paragraph. It is just ridiculous.
    Quote Originally Posted by hypercube View Post
    50% loss is exactly equal to 100% gain because if you lose that bet, now you need 100% gain to get back to the original $100 and 400% instead of 100% to get to the $200 from the original $100 we had. The ev to switch is $0,not +$25 and the advantage in edge to switch is also exactly zero. From perspective of a card counter we dont tend to like risking money without an edge so we should never switch. but we all would anyway because its more fun and there zero positive or negative expectancy.

  7. #46


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    Quote Originally Posted by Three View Post
    The guy proves the probability that you picked the envelop with the lower amount is 50%. That never changes with no information on each envelop. It still doesn't change when you know what is in one envelope. You still have a probability of 50% that you chose the larger amount. But the symmetry that creates the paradox gets destroyed once you know the amount in one envelope. If this guy thinks he proved that after finding out the amount in on envelope that the decision on which envelope to choose isn't clear he is an idiot. Know what you have to watch out for is the person running the show can choose whether to show you the higher amount or the lower amount. If the amount shown was chosen at random and is equally likely to be the larger amount as the smaller amount choosing the unknown envelope is better.

    From the link in bold:

    1) When you make your choice (C)
    2) Either the envelope containing the lower value (C=lower) or the one that contains the higher (C=higher).
    3) During the game, the value (V) of the content of the chosen envelope is revealed to be a certain value M.

    The paradox above arose because you assumed that
    4) P(C=lower|V=M) = P(C=higher|V=M) = 0.5

    No, the paradox arose because M was a variable. You make it a constant then V is the same constant. That is your number 3 in the bold.
    Let's see why this cannot be the case. (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    No, V is not a variable. If V=M and M is a constant then V is also a constant. You can't treat it like a variable. See your number 3 above. C is a constant as well. C is the envelope you chose. If you know nothing about the contents of either envelope you can view it as a variable but once you know the amount in one envelope it is no longer a variable. It is the envelope that contains the amount M which is a constant. It's value, V, is equal to M which is a constant. So L is your only variable. And you have 3 constants; C,V, and M where V=M.

    The paradox above arose because you assumed that
    P(C=lower|V=M) = P(C=higher|V=M) = 0.5

    No, that was a given. The paradox arose because M was a variable.
    Let's see why this cannot be the case. (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    No, L is the only variable. We Know M is a constant. V=M so V is also a constant. And C is the envelope you chose which contains the amount M, a constant and has the value, V, also a constant.
    P(C=lower|V=M) = P(C=lower)P(L=M)/P(V=M) . . . (2)
    and P(C=higher|V=M) = P(C=higher)P(L=M/2)/P(V=M) . . . (3)

    I skipped to this because with all the constants the preceding was pretty meaningless.
    Given V=M, (P(V=M)=1), this reduces to P(C=lower|V=M) = P(C=lower)P(L=M) and P(C=higher|V=M) = P(C=higher)P(L=M/2). Both terms mean the exact same thing so what is the point both statements are givens.
    From (2), P(C=lower|V=M) is the same as P(C=lower) only if P(L=M) is the same as P(V=M).
    No. The other possibility is the probability that V=M is 1. V=M is 100%, that was how you defined M and V. M is a constant that is the amount in the envelope. V is the value of the contents of the chosen envelope which is the constant M. So the probability that V=M is 1. So P(L=M) does not have to be the same as P(V=M). Your treating constants as variables lead to this false statement. Everything based on this falsehood is wrong.

    I can't believe this guy thought he could prove that the odds you chose the envelope with the larger amount or smaller amount can't be equal at .5 and .5 LoL

    The paradox above arose because you assumed that
    P(C=lower|V=M) = P(C=higher|V=M) = 0.5
    Let's see why this cannot be the case.

    Well I guess if you take a given like V=M, and then define V as a variable and M as a constant you can make it look like you did to those that don't try to follow what you did.
    (In what follows, remember that L, V, C are variables and M is a numerical constant.)
    Of greater interest is the solution to the PGT , calculating both LB and RB variables to insure ride symmetry.

  8. #47
    Senior Member Gramazeka's Avatar
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    Three, I understand you correctly that it is always profitable to change the box?
    Then let's play this game. I will lay out the money 2: 1 on the boxes. At least $ 100 each. You'll at random open boxes, pay me an open amount and change your choice. I'll pay you the amount from the second box. A little nuance. For the right to change the boxes, you'll pay me 1 time (only one!) A dollar. Going?

    p.s.

    EV1 = (1/2) * X + (1/2) * Y = (1/2) * (X + Y);
    EV2 = (1/2) * {X -> Y} + (1/2) * {Y -> X} = (1/2) * (Y + X);
    Curly brackets indicate how the dropped boxes are replaced with alternate ones.
    It can be seen that the expectation does not depend on the chosen strategy!
    Last edited by Gramazeka; 02-18-2018 at 11:14 PM.
    "Don't Cast Your Pearls Before Swine" (Jesus)

  9. #48
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    Quote Originally Posted by Gramazeka View Post
    Three, I understand you correctly that it is always profitable to change the box?
    Then let's play this game. I will lay out the money 2: 1 on the boxes. At least $ 100 each. You'll at random open boxes, pay me an open amount and change your choice. I'll pay you the amount from the second box. A little nuance. For the right to change the boxes, you'll pay me 1 time (only one!) A dollar. Going?

    p.s.

    EV1 = (1/2) * X + (1/2) * Y = (1/2) * (X + Y);
    EV2 = (1/2) * {X -> Y} + (1/2) * {Y -> X} = (1/2) * (Y + X);
    Curly brackets indicate how the dropped boxes are replaced with alternate ones.
    It can be seen that the expectation does not depend on the chosen strategy!
    I am a little unfamiliar with your symbols but if I understand you right the EV of both boxes is the same. Paying $1 to change boxes would be losing $1 in EV each time.

  10. #49


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    Quote Originally Posted by Three View Post
    I am a little unfamiliar with your symbols but if I understand you right the EV of both boxes is the same. Paying $1 to change boxes would be losing $1 in EV each time.
    This thread is what happens when we dont argue about counts.

    Can someone post a trip report so it can be suggested that they might have done better using hiopt2

  11. #50


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    But if I decide to switch 10 times and 5 times I get $50 and 5 times I get $200 then I have $1250.
    Why do you always start with $100? That is basically ignoring the downside of switching, which is that you can get a lower result and then will have the potential for a much lower result.

  12. #51
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    Quote Originally Posted by Meistro123 View Post
    Why do you always start with $100?
    Because that was the condition the OP set.

  13. #52
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    Quote Originally Posted by Three View Post
    Because that was the condition the OP set.
    Ok, no problem... ))
    "Don't Cast Your Pearls Before Swine" (Jesus)

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