Coin flip problem
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March 27th, 2020 at 10:33:11 PM
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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?
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
March 28th, 2020 at 12:18:11 AM
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Not sure... but my answer is 70.72% of the time it needs to come up a head to make it a fair game.
March 28th, 2020 at 1:54:53 AM
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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
March 28th, 2020 at 4:16:40 AM
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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
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 28th, 2020 at 10:08:54 AM
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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
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
March 28th, 2020 at 10:27:56 AM
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Quote: Ace2Your 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
March 28th, 2020 at 10:41:20 AM
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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.
March 28th, 2020 at 12:11:44 PM
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Correct.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
0.618 is the golden ratio minus 1
It’s all about making that GTA
March 28th, 2020 at 4:00:46 PM
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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.
