Give me some good paradoxes

[vhawk01]

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E.g.: Player always moves all-in preflop with AA. In such a situation 22 cannot do better than AA.

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The player need not even go all-in preflop, now I may be rusty due to playing Omaha, but I am pretty sure if AA just bets more than 1/6 of his stack (assuming no menaingful pot here) 22 is a net loser.

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that doesnt work because if you're playing 600nl and you open for $100, you only win the blinds; thus 22 is still more profitable.

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Huh? The AA is always profitable in this situation. The 22 is never profitable in this situation. How does your example contradict that? How is 22 profitable if AA opens for 100 in a 600 NL game?

[Dan.]

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E.g.: Player always moves all-in preflop with AA. In such a situation 22 cannot do better than AA.

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The player need not even go all-in preflop, now I may be rusty due to playing Omaha, but I am pretty sure if AA just bets more than 1/6 of his stack (assuming no menaingful pot here) 22 is a net loser.

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that doesnt work because if you're playing 600nl and you open for $100, you only win the blinds; thus 22 is still more profitable.

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Huh? The AA is always profitable in this situation. The 22 is never profitable in this situation. How does your example contradict that? How is 22 profitable if AA opens for 100 in a 600 NL game?

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What he was trying to say originally is that AA will only win the pot that developed preflop, since 22 won't continue without a set. However, for the times the 22 does make a set, he'll stack the player with AA. Thus, the 22 is more profitable over the long run.

But if the AA player just constantly opens for $100, I don't ever see the 22 player calling, so 22 will always be losing blinds slowly making it less profitable than AA.

[Copernicus]

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Zeno's paradox.

Take a set distance, then halve it, repeat until your hair's glossy and shiny and has that TV commercial bounce.

You never get there, do you? But when you try it in a room, you're going to bang your head on the wall every time.

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not a paradox at all.

[madnak]

What if you can make him fold AA on the river every time? I bet my 72 would hold up then.

[eviljeff]

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I saw this one in a book called Keys to Infinity.

the probability that an integer with n-digits contains a 3 is 1 - (9/10)^n, therefore as n tends to infinity the probability that an integer contains a 3 is 1, yet there will be infinitely many integers containing no 3.

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dude. you just like. blew my mind.

[ShakeZula06]

related,

Is the universe infinite or finite?
1)If finite, where does in end? Is there a wall? What's on the other side other side of the wall?
2) If Infinite, well, how can something go on forever?

[ShakeZula06]

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Zeno's paradox.

Take a set distance, then halve it, repeat until your hair's glossy and shiny and has that TV commercial bounce.

You never get there, do you? But when you try it in a room, you're going to bang your head on the wall every time.

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not a paradox at all.

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yes it is

[holmansf]

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related,

Is the universe infinite or finite?
1)If finite, where does in end? Is there a wall? What's on the other side other side of the wall?
2) If Infinite, well, how can something go on forever?

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What if it's finite, but doesn't have an end. Ya know, like a circle.

[LadyWrestler]

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How to get a job without experience, and how to get experience without a job.

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Volunteer. [img]/images/graemlins/tongue.gif[/img]

[PairTheBoard]

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Here is the paradox:
Sleeping Beauty is to be given on Sunday a drug that causes her to sleep. A fair coin will be tossed. If it lands tails, she will be awoken on monday, put back to sleep, and awoken again on tuesday, and then again put back to sleep to await prince charming's kiss. If the coin lands heads, she will be awoken only on tuesday and then again put to sleep to await charming. The sleeping drug has the secondary consequence of short term memory loss: Sleeping Beauty will not remember on tuesday any monday awakening that might have occured. At each awakening she is to be offered a gamble on the outcome of the coin toss, at 1:1 odds, she taking the affirmative on the proposition "the coin landed heads". On sunday afternoon, Sleeping Beauty clearly ought regard the chance of heads as .5, and so ought to regard each potential wager as fair. Upon being awoken, should she change her estimation of the probability the coin lands heads?
Options:
(1) No. The prior on heads was .5, no new information has been learned, so the updated probability ought to be .5.

(2) Yes. If she takes the probabilty of heads to be .5, she ought regard the wagers as fair; but if she accepts such wagers, her expectation is negative (since she will be offered the wager twice when the coin is tails, but only once when heads). Hence, she must revise her estimate of the probability the coin landed heads.
Solution below....





















Various philosophers and computer scientists have in all seriousness argued for 2, and thought they could learn something about probability from the case. They can't of course (the solution is related both to Parrondo's paradox and to value betting), but it is not obvious to non-gamblers. This is just one of the many situations in which fair odds come appart from the probability of winning. In value betting, this happens when we are a) in a must call situation, b) do not fear a raise, and c) judge that our opponent will more often call with a loosing hand than bet with a loosing hand. Optimal action no longer depends on the chance that we win a wager if contracted, depending instead on the chances (freqeuncies ) that we contract a wager under winning (or loosing) conditions. These two chances-- the chance of winning given a contract and the chance of contracting under winning conditions-- are only nicely distinguished and often equal, but not always.

Bruce

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I think this is involves the same principle as the Two Envelope Problem. You have a 50% chance of choosing the envelope with the smaller amount and would be willing to bet a fixed amount at even money that you will do so. After choosing the envelope, opening it and looking at the amount you glean no information as to whether it is the larger or smaller amount so you would still be willing to bet the fixed amount at even money that you are looking at the smaller amount. But you would NOT be willing to bet the amount in the envelope at even money that it's the smaller amount. The reason being that the amount of your bet has been dictated by the already determined outcome. You are forced to wager twice as much when you are losing. Choosing to switch envelopes amounts to betting half the envelope amount that it it is the smaller of the two. It's a bad bet at even money and 0 EV at the 2-1 odds being given.

Sleeping beauty is in the same situation.

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If she takes the probabilty of heads to be .5, she ought regard the wagers as fair

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Although the "probability" as we commonly use the term is .5, the outcome has really been determined already and dictates that she wager twice as much when the outcome was tails. So, although she correctly takes the "probabilty" of heads to be .5 she shouldn't bet on it.

I think a good example of this principle is the following proposition. I will deal a card from a standard deck and let you bet that it's a black card. I'll even give you 3-2 odds to induce your action. I deal the card face down. As you prepare to place your bet I stop you. At this point we're both in agreement that the "probabilty" the card is black is 50%. I then shine a black light on the back of the card and a dollar amount appears. I tell you that if you want to bet you must bet that dollar amount. I'm hoping you will be a sleeping beauty and say to yourself, "If I take the probabilty of black to be .5, I ought regard the wager as good at 3-2 odds".

PairTheBoard