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## Question

You and a friend have loved a girl since tour childhood days to which she has loved you both equally. The three of you have decided to let fate/chance decide who should end up with the girl. All you have available is an pld wooden coin. The coin that you all know will land on heads 50% more of the time.  How can you use this coin in a fair contest to decide who should end up with the girl?

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One way

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One possibility:

- toss the coin 5 fimes, if (head) A scores else B scores
- toss the coin 5 fimes, if (head) B scores else A scores
- if equality, restart over

However, in theory, it can go forever..

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One way

Flip the coin twice. If the outcomes are the same, flip the coin two more times. Eventually you will get HT or TH.
These outcome have equal likelihood. Interpret them as H and T, respectively.

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girls clasps the coin in one hand......pick a hand.

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Short form of Bonanovas solution

Flip the coin until the result is different from previous flip; obey the last flip, per Bonanova's logic

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Short form of Bonanovas solution

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I think that gives the opposite of the first flip.

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Pick the (slightly) lower, or bigger, hand.

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Here's one where you only flip the coin once:

The girl secretly writes one of the friend's names and H or T on a piece of paper. As long as she keeps this secret, the game is fair.

Then the friends discuss who flips. As long as they don't know what is on the paper, the game is fair either way. After deciding, one of the friends flips the coin. If their name matches, the coin has to match. If their name doesn't match, the coin needs to not match for them to win.

I admit that there are other factors besides the coin that decides their fate, but the coin is the deciding factor.

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On 11/18/2015 at 11:17 PM, BMAD said:

You and a friend have loved a girl since tour childhood days to which she has loved you both equally. The three of you have decided to let fate/chance decide who should end up with the girl. All you have available is an pld wooden coin. The coin that you all know will land on heads 50% more of the time.  How can you use this coin in a fair contest to decide who should end up with the girl?

You mentioned the coin lands on heads 50% more of the time. A regular coin lands on heads 50% of the time. 50% more is 100%. If this is what you meant, then I have a solution to the problem.

Spoiler

The girl is blindfolded and holds the coin in her hand. The three of us stand in front of her in a straight line. One straight ahead, one more to the left, and one more to the right. The girl does no know who is where. She spins around to disorientate her direction. Now she throws the coin. Whoever is standing closest to where the coin lands is the winner. The coin obviously will still land on the ground heads up.

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15 hours ago, Buddyboy3000 said:

You mentioned the coin lands on heads 50% more of the time. A regular coin lands on heads 50% of the time. 50% more is 100%.

Not so: 50% more is 'half as much' more, so 50% more than 50% is 75%. That said, I would interpret 'heads 50% more of the time' to be 'heads 50% more often than tails', which would be 60% heads, 40% tails.

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4 hours ago, Logophobic said:

Not so: 50% more is 'half as much' more, so 50% more than 50% is 75%. That said, I would interpret 'heads 50% more of the time' to be 'heads 50% more often than tails', which would be 60% heads, 40% tails.

Thanks Logophobic. I completely forget when I was putting my answer.

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23 hours ago, Logophobic said:

Not so: 50% more is 'half as much' more, so 50% more than 50% is 75%. That said, I would interpret 'heads 50% more of the time' to be 'heads 50% more often than tails', which would be 60% heads, 40% tails.

The puzzle seems to have been solved, although BMAD might be asking for a specific solution, tailored to the given coin bias. If so, the bias should be determined as a prerequisite to the finding that solution. In that vein, when we compare two numbers, "more," (as in five more) generally implies adding, while "times," (as in 5 times) generally implies multiplying. Thus 75 is 50 more than 25. So we would say the probability of heads is 50% more than that of tails when the probability of heads is 75%.

It's clearer, when % is not used for probabilities, to say 0.75 is 0.50 more than 0.25. It's muddier otherwise, since "%", once the question "% of what?" has been answered, generally implies multiplication.

My reservation with CaptainEd's suggestion is that it is biased toward tails. The first result is likely heads; the result that changes it is likely tails.

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Oops! Thanks Bonanova, not only I was backward, but also biased! Durn!

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