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What would you pay...? |
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| Nov20-11, 06:42 PM | #18 |
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What would you pay...?What is amusing is that the same theory that correctly says all casino gambling and all lottery games (unless the winning amount rolls unclaimed a number times) are foolish, claims that you should be willing to bid any amount to play this game once. |
| Nov20-11, 09:41 PM | #19 |
Recognitions:
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| Nov22-11, 11:39 AM | #20 |
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The game host should pay me at least 51 cents to play his game, of course! (Since 1+1+1+...=-1/2) lol ;) kidding...
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| Nov22-11, 11:58 AM | #21 |
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I was tempted to post this answer myself some time ago. |
| Nov25-11, 06:00 PM | #22 |
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This is discussed here:
http://www.physics.harvard.edu/acade.../problems.html If anyone is interested. It's toward the bottom, Week 6. |
| Nov25-11, 07:02 PM | #23 |
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I think the answer is that you should play the game if you have to pay 4 dollars or less. (If you pay 2 dollars, you are guaranteed to always at least break even. If you pay $4, you have a 50-50 shot at at least getting your money back or making more. If you pay anything over $4, you have less than a 50% chance to win, so you shouldn't play.
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| Nov27-11, 08:21 PM | #24 |
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Since the question is what I would pay, I would pay zero, so that any winnings would be pure profit.
Frankly, I think the question (and the game rules) are both poorly worded. |
| Nov28-11, 08:26 AM | #25 |
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I would pay $1.99. Guaranteed to win, baby!
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| Nov28-11, 08:34 AM | #26 |
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| Nov28-11, 08:48 AM | #27 |
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In my opinion, this is a boring game... I would elect to low-ball. |
| Nov28-11, 09:06 AM | #28 |
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One way to look at games of chance, and, in general, all forms of gambles in your life, is to view all the gambles in your life as part of an overall game. Then, even though expectation is an average of an infinite number of trials, and strictly doesn't apply to individual decisions, by taking advantage of winning expectation gambles as they present themselves, over your life you may come out ahead.
This leads to an important corollary: If the chance of winning is too small, there cannot be enough comparable opportunities in your lifetime. Thus, there is a threshold probability below which expectation is irrelevant. This can be quantified if you can guess the number of similar opportunities you might have in your life. For example, for a lottery, you can compute the most likely outcome of the number of plays you can reasonably make in your life, of tickets that have developed a winning expectation (as happens sometimes when a lottery round has no winner, and the prize rolls over). You find that for this finite game, your lifetime winnings are almost certain to be less than your cost. Even if you include other types of opportunities, you conclude that such a lottery play is still not rational. This can be combined with observations about reality check (as DH first pointed out, the stated game is unimplementable); and also that you really need to think about your personal utility function of dollars, not just raw dollars. In my view, then, choices like this are fully quantifiable in principle, the practical difficulty being lack of definition of things like 'lifetime similar opportunities' and 'personal utility function for money'. This is not a game like poker, where guessing human behavior is paramount; nor like the running of lottery (though guessing mass behavior here is pretty trivial). For playing in a lottery or this game, there is no human behavior component involved. [Edit, putting all this together for me, I would be willing to bid up to $10 to play this game]. |
| Nov28-11, 09:35 AM | #29 |
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| Nov28-11, 02:47 PM | #30 |
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- performing activities that one need to do accomplish things in one's daily life, knowing those activities carry a risk of failure, and - taking a risk purely for the thrill of the possible win. Strictly using the term gambling, I would apply it to the latter but not the former. |
| Nov28-11, 02:54 PM | #31 |
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| Dec26-11, 08:25 PM | #32 |
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I have an objection to the reasoning that the expected gain is infinite.
Rather, the expectation value [itex]E(X)[/itex] is infinite. However, one must ask "how does [itex]E(X)[/itex] acquire its usual meaning of the expected gain?". The answer is: "due to some limit theorem". These limit theorems (e.g. law of large numbers) requires that E(X) is finite. Consequently, the math doesn't seem to tell us anything about the expected gain. |
| Jan1-12, 10:55 AM | #33 |
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I'd bet 4$. You have a 50/50 chance of losing 2$ on the first toss. If you don't lose you have a 50/50 chance of winning 4$ (8-4=4) on the next. You also will keep doubling up for every consecutive tails you throw after the first one. If you throw 3 tails in a row you have 16$ and a 50/50 chance to double it on the next toss, as well as after every additional tails thrown.
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