Quote:OnceDearWhat do you have in mind for ballpark min/max bets?

And will you both have the right to call a halt?

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Based on PM's, I have a high degree of confidence this challenge with LoquaciousMoFW will go off. It's possible we may need to use a proxy, if he can't make it person. Betting has been agreed to at $5 a bet. Of course, anything can be changed by mutual agreement. I never stated my own quitting point, but I am prepared to go for hours and will bring at least $1,000.

Wizard will roll two dice in a cup. If there are no 2s, he will reroll. If there is at least one 2, he will remove a die with a 2 and show it to you. There will be one die still under the cup.

You can bet if that second die is a 2. Wizard will pay you 7:1 if it’s a 2.

Would you make that bet?

Quote:OnceDearwhat would be your motivation for playing this slightly -ev proposition?

To demonstrate that hit and run is a winning strategy.

The relationship between the 2-Dice problem and the Gambler's Fallacy is interesting to ponder.

Gambler's Fallacy: Separate rolls/trials are statistically independent. Given fair systems, the past record does not affect the future. What happens on one dice roll does not affect another dice roll.

Two-Dice Problem: Information about the outcome of an ensemble of distinct events can alter the probability of an individual event's outcome.

And what I am struck by is the similarity between the 2 Dice problem, and the concept of quantum entanglement in physics.

Quantum entanglement: the phenomenon whereby a pair of particles are generated in such a way that the individual quantum states of each are indefinite until measured, and the act of measuring one determines the result of measuring the other, even when at a distance from each other.

Quote:unJonAlan here is a bet for you.

Wizard will roll two dice in a cup. If there are no 2s, he will reroll. If there is at least one 2, he will remove a die with a 2 and show it to you. There will be one die still under the cup.

You can bet if that second die is a 2. Wizard will pay you 7:1 if it’s a 2.

Would you make that bet?

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Unjon I've got a bet for you. Meet me at a craps table with $300. We will both shoot and bet only on our own shooting. After three turns each we'll see who has more money. You can bet pass or DP and as much or as little on yourself as you like.

Oh, the stakes? How about winner takes all?

Your maximum out of pocket is $300 but if one or both of us have winning rolls the winner could walk away with $600 plus the casino's money.

Now that's a worthwhile bet.

Quote:gordonm888Let me try to write something worth reading,

The relationship between the 2-Dice problem and the Gambler's Fallacy is interesting to ponder.

Gambler's Fallacy: Separate rolls/trials are statistically independent. Given fair systems, the past record does not affect the future. What happens on one dice roll does not affect another dice roll.

Two-Dice Problem: Information about the outcome of an ensemble of distinct events can alter the probability of an individual event's outcome.

And what I am struck by is the similarity between the 2 Dice problem, and the concept of quantum entanglement in physics.

Quantum entanglement: the phenomenon whereby a pair of particles are generated in such a way that the individual quantum states of each are indefinite until measured, and the act of measuring one determines the result of measuring the other, even when at a distance from each other.

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So funny, Gordon. I was actually pondering something similar this morning. I was wondering if a mechanic similar to the two dice problem could serve as the basis for a hidden variable that breaches Bell’s Inequality.

Quote:AlanMendelsonQuote:unJonAlan here is a bet for you.

Wizard will roll two dice in a cup. If there are no 2s, he will reroll. If there is at least one 2, he will remove a die with a 2 and show it to you. There will be one die still under the cup.

You can bet if that second die is a 2. Wizard will pay you 7:1 if it’s a 2.

Would you make that bet?

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Unjon I've got a bet for you. Meet me at a craps table with $300. We will both shoot and bet only on our own shooting. After three turns each we'll see who has more money. You can bet pass or DP and as much or as little on yourself as you like.

Oh, the stakes? How about winner takes all?

Your maximum out of pocket is $300 but if one or both of us have winning rolls the winner could walk away with $600 plus the casino's money.

Now that's a worthwhile bet.

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No thanks. I am just trying to engage you to try to get you to see 1/6 vs 1/11 difference. Your bet doesn’t do anything for me. $300 is less than a passline bet with odds.

Quote:UnJonSo funny, Gordon. I was actually pondering something similar this morning. I was wondering if a mechanic similar to the two dice problem could serve as the basis for a hidden variable that breaches Bell’s Inequality.

Hidden variables to undermine Bell's Inequality? Mic drop. Wow! Impressive. Sounds crazy, but it's fun to think about. And you never know . . .

I was wondering whether the probability of an ensemble of seemingly discrete random events that originate at the same point/time could be written as a probabilistic wavefunction in a way that would produce behavior that mirrors the behavior of quantumly entangled particles. Probably not, but . . .

I think your crazy daydream is a bit more fundamental than mine. Shows how a ridiculous thread like this could potentially have value.

Quote:unJonQuote:AlanMendelsonQuote:unJonAlan here is a bet for you.

Wizard will roll two dice in a cup. If there are no 2s, he will reroll. If there is at least one 2, he will remove a die with a 2 and show it to you. There will be one die still under the cup.

You can bet if that second die is a 2. Wizard will pay you 7:1 if it’s a 2.

Would you make that bet?

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Unjon I've got a bet for you. Meet me at a craps table with $300. We will both shoot and bet only on our own shooting. After three turns each we'll see who has more money. You can bet pass or DP and as much or as little on yourself as you like.

Oh, the stakes? How about winner takes all?

Your maximum out of pocket is $300 but if one or both of us have winning rolls the winner could walk away with $600 plus the casino's money.

Now that's a worthwhile bet.

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No thanks. I am just trying to engage you to try to get you to see 1/6 vs 1/11 difference. Your bet doesn’t do anything for me. $300 is less than a passline bet with odds.

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You're a much bigger gambler than I am, and at a -EV game. So you must be a DI to bet so much.

Quote:AlanMendelsonQuote:unJonQuote:AlanMendelsonQuote:unJon

Wizard will roll two dice in a cup. If there are no 2s, he will reroll. If there is at least one 2, he will remove a die with a 2 and show it to you. There will be one die still under the cup.

You can bet if that second die is a 2. Wizard will pay you 7:1 if it’s a 2.

Would you make that bet?

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Unjon I've got a bet for you. Meet me at a craps table with $300. We will both shoot and bet only on our own shooting. After three turns each we'll see who has more money. You can bet pass or DP and as much or as little on yourself as you like.

Oh, the stakes? How about winner takes all?

Your maximum out of pocket is $300 but if one or both of us have winning rolls the winner could walk away with $600 plus the casino's money.

Now that's a worthwhile bet.

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No thanks. I am just trying to engage you to try to get you to see 1/6 vs 1/11 difference. Your bet doesn’t do anything for me. $300 is less than a passline bet with odds.

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You're a much bigger gambler than I am, and at a -EV game. So you must be a DI to bet so much.

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Despite my best efforts, I cannot lay claim to being a DI.