Ace2
Ace2
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March 27th, 2020 at 10:33:11 PM permalink
A coin toss game works as follows:

Coin is flipped once. If itís tails, tails wins and the game ends.

If itís heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise tails wins. Game over

If you could design an unfair coin, what percentage of the time, on average, should it come up heads to make this a fair game?
Itís all about making that GTA
EdCollins
EdCollins
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March 28th, 2020 at 12:18:11 AM permalink
Not sure... but my answer is 70.72% of the time it needs to come up a head to make it a fair game.
ssho88
ssho88
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March 28th, 2020 at 1:54:53 AM permalink

p = heads prob, q = tails prob

If first flip is tails, then you flip again(padding factor), the heads win prob = 0, the tails wins prob = q(p+q)
If first flip is heads, then you flip again, the heads wins prob =p^2, and tails wins prob= pq

To make it a fair game,

0 + p^2 = pq + q(p+q)............Eq1
p+q = 1...................................Eq2

Solve simultaneous equations,

p = 1/(2)^0.5 = 0.707107

Last edited by: ssho88 on Mar 28, 2020
Wizard
Administrator
Wizard
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March 28th, 2020 at 4:16:40 AM permalink
Quote: ssho88


p = heads prob, q = tails prob

If first flip is tails, then you flip again(padding factor), the heads win prob = 0, the tails wins prob = q(p+q)
If first flip is heads, then you flip again, the heads wins prob =p^2, and tails wins prob= pq

To make it a fair game,

0 + p^2 = pq + q(p+q)............Eq1
p+q = 1...................................Eq2

Solve simultaneous equations,

p = 1/(2)^0.5 = 0.707107



I agree
It's not whether you win or lose; it's whether or not you had a good bet.
Ace2
Ace2
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March 28th, 2020 at 10:08:54 AM permalink
Your answers look good, but I misstated the problem. Sorry about that. It should say:

Coin is flipped once. If itís tails, tails wins and the game ends.

If itís heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise itís a push.

If you could design an unfair coin, what percentage of the time, on average, should it come up heads to make this a fair game?

Still an easy ďno beerĒ problem, but the answer is an interesting number, in my opinion anyway
Itís all about making that GTA
ssho88
ssho88
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March 28th, 2020 at 10:27:56 AM permalink
Quote: Ace2

Your answers look good, but I misstated the problem. Sorry about that. It should say:

Coin is flipped once. If itís tails, tails wins and the game ends.

If itís heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise itís a push.

If you could design an unfair coin, what percentage of the time, on average, should it come up heads to make this a fair game?

Still an easy ďno beerĒ problem, but the answer is an interesting number, in my opinion anyway





"If itís heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise itís a push."

Then the equation should be this way, (0 + p^2)*1 = pq*0 + q(p+q)*1

p^2 = q(p+q)

p^2 = 1-p

p^2 + p -1 =0

p = ( 5^0.5 -1 )/2 = 0.618034

EdCollins
EdCollins
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March 28th, 2020 at 10:41:20 AM permalink
With the change to "otherwise it's a push" I think the unfair coin should now come up heads just 61.8% of the time.
Ace2
Ace2
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March 28th, 2020 at 12:11:44 PM permalink
Quote: ssho88



Then the equation should be this way, (0 + p^2)*1 = pq*0 + q(p+q)*1

p^2 = q(p+q)

p^2 = 1-p

p^2 + p -1 =0

p = ( 5^0.5 -1 )/2 = 0.618034

Correct.

0.618 is the golden ratio minus 1
Itís all about making that GTA
charliepatrick
charliepatrick
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March 28th, 2020 at 4:00:46 PM permalink
An easier way to think of it is one side only wins if the same result comes up twice, otherwise the other side wins. Hence q^2 = 1/2. q = 1/SQRT(2) = .707.
Using similar logic p = q^2 = (1-q) which leads to q^2+q-1=0.

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