Ace2 Joined: Oct 2, 2017
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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 it’s tails, tails wins and the game ends.

If it’s 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?
It’s all about making that GTA
EdCollins Joined: Oct 21, 2011
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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 Joined: Oct 16, 2011
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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 Joined: Oct 14, 2009
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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 Joined: Oct 2, 2017
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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 it’s tails, tails wins and the game ends.

If it’s heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise it’s 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 “no beer” problem, but the answer is an interesting number, in my opinion anyway
It’s all about making that GTA
ssho88 Joined: Oct 16, 2011
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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 it’s tails, tails wins and the game ends.

If it’s heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise it’s 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 “no beer” problem, but the answer is an interesting number, in my opinion anyway

"If it’s heads, coin is flipped one more time. If it comes up heads again, heads wins. Otherwise it’s 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 Joined: Oct 21, 2011
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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 Joined: Oct 2, 2017
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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
It’s all about making that GTA
charliepatrick Joined: Jun 17, 2011
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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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