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Topic: Perplexing Probability Proposal (Read 1443 times) |
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Sir Col
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Perplexing Probability Proposal
« on: Jan 10th, 2005, 12:32am » |
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Lucky Larry and Dim Dora are playing a game involving spinning coins: the person who spins the most heads is the winner.
Larry suggests to Dora, "To prove that I am lucky by name, and lucky by nature, I propose that you spin one extra coin. Obviously you have the advantage, but let's see how often you win."
Dora happily agrees, but is amazed that it doesn't seem to matter how many coins Larry spins, if she spins one extra coin she still only seems to win 50% of the time.
How does Larry achieve this feat of "luck"?
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towr
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Re: Perplexing Probability Proposal
« Reply #1 on: Jan 10th, 2005, 1:21am » |
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If Larry doesn't win all of the time, he isn't spinning correctly. Spinning is much more sensitive to hguidance than flipping.
But I suppose you're assuming each coin behaves fairly.
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Grimbal
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Re: Perplexing Probability Proposal
« Reply #2 on: Jan 10th, 2005, 1:40am » |
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The extra coin has no face.
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andyram
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Re: Perplexing Probability Proposal
« Reply #3 on: Jan 10th, 2005, 9:09am » |
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Consider the probability of Larry getting H heads out of N throws to be P. The same applies to dora's chances of getting H heads in N throws. So, for Dora to win she needs a head on the extra throw she gets, the probability of which is 1/2. This is y she wins only 50% of the time. the rest can also be draws. Hope this is correct.
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ThudnBlunder
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Re: Perplexing Probability Proposal
« Reply #4 on: Jan 10th, 2005, 9:23am » |
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Quote:
Yes, "Hope this" is correct.
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| « Last Edit: Jan 10th, 2005, 10:18pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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WombatDeath
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Re: Perplexing Probability Proposal
« Reply #5 on: Jan 10th, 2005, 10:12am » |
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I wouldn't call Larry 'lucky' - Dora might 'only' win half the time, but Larry doesn't do even that well. In the case where Larry gets one spin we have eight possible outcomes:
L D1 D2
H H H
H H T
H T H
H T T
T H H
T H T
T T H
T T T
...so Dora wins half the time and Larry only an eighth. In the general case I agree with andyram's reasoning.
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ThudnBlunder
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Re: Perplexing Probability Proposal
« Reply #6 on: Jan 10th, 2005, 10:56am » |
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on Jan 10th, 2005, 10:12am, WombatDeath wrote:| In the case where Larry gets one spin... |
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But the question says:
on Jan 10th, 2005, 12:32am, Sir Col wrote:
...it doesn't seem to matter how many coins Larry spins, if she spins one extra coin she still only seems to win 50% of the time. |
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Admittedly, the question is not well-posed, perhaps deliberately.
For example, it could be that Dora is initially given one less spin than Larry, so that giving her an extra spin gives them an equal number of total spins.
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| « Last Edit: Jan 10th, 2005, 8:44pm by ThudnBlunder » |
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Sir Col
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Re: Perplexing Probability Proposal
« Reply #7 on: Jan 10th, 2005, 12:12pm » |
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So that there is no confusion...
If Larry spins n coins, Dora will spin n+1 coins.
Dora will win 50% of the time.
Of course, WombatDeath, you picked-up on a crucial insight: it does not mean that Larry necessarily wins the other half. However, Dora is dim by name, dim by nature: "I can't believe Larry's luck. Surely I should win more than only 50% of the time?"
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WombatDeath
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Re: Perplexing Probability Proposal
« Reply #8 on: Jan 10th, 2005, 12:21pm » |
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My theory is that Dora can expect to win 50% of the time if she is given one more spin than Larry. As the number of spins increases I think that Larry's expected victories get closer to 50% (and the number of expected draws decreases).
So my answer to the original question is "Yes, that's supposed to happen". Unfortunately, my maths isn't up to proving my claim. If I have time after dinner I might 'prove' it by running a simulation 
Edit: In light of Sir Col's statement I'm going to award myself a point for "half-baked conjecture which happened to coincide with the truth" 
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| « Last Edit: Jan 10th, 2005, 12:24pm by WombatDeath » |
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rmsgrey
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Re: Perplexing Probability Proposal
« Reply #9 on: Jan 10th, 2005, 7:28pm » |
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A proof:
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Before adding the extra coin, there are three classes of result: Larry Up, Dora Up, and Even.
There are equally many (equiprobable) outcomes in each of Larry Up and Dora Up due to the symmetry of the situation, and, in general, a different number of (equally probable) outcomes in Even.
If the extra coin comes up T, then the outcome is unchanged, but if the extra coin comes up H, then all Even results become Dora's win, and an unspecified number of Larry Ups results become draws.
So the situations where Dora wins the new game are: Dora Up, T; Dora Up, H; and Even, H. The situations where Dora doesn't win are: Larry Up, T; Larry Up, H; and Even, T.
The two Dora Up cases are as probable as the two Larry Up cases, and the two Even cases are equally probable (assuming the extra coin gives H half the time), so Dora wins half the time.
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Sir Col
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Re: Perplexing Probability Proposal
« Reply #10 on: Jan 12th, 2005, 11:13am » |
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Good reasoning. A slightly more formal way to express your idea...
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For the first n coins there are three possible outcomes: D>L, D=L, or D<L. As P(D>L) = P(D<L) = a, P(D=L) = 1-2a.
If Dora is ahead after n coins, she will win regardless of final coin.
The only other way for Dora to win is to be tied after n coins then spin a head.
Therefore P(D wins) = P(D>L)+P(D=L)*0.5 = a+(1-2a)*0.5 = 0.5
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Isn't it wierd that the probability of Dora winning is independent of n. It does, however, lead on to two more challenging problems...
(E1) Find the probability that Larry wins.
(E2) Find P(D wins) if she spins k more coins than Larry. What about P(L wins)?
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