FMM16 - Revision history

Admin at 02:15, 5 December 2019

2019-12-05T02:15:23Z

← Older revision		Revision as of 02:15, 5 December 2019
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	c. What is the probability Paul de Bank will get cleaned out?		c. What is the probability Paul de Bank will get cleaned out?

	d. What is the probability that Petra von Player will lose money on the game?		d. What is the probability that Petra von Player will lose money on the game (assuming Petra von Player pays the amount from part b)?

Admin at 02:11, 5 December 2019

2019-12-05T02:11:46Z

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	\|problem=		\|problem=

	~~Consider the famous Petersburg Paradox:~~		'''BACKGROUND'''

	~~Petra von Player and Paul de Bank agree to play a game based on the toss of a coin. If a head~~ is ~~thrown on~~ the ~~first toss (prob = 1~~/~~2), Paul de Bank pays Petra von Player a dollar, and the game is over~~. ~~If the first toss is tails and the second toss is head (prob = 1~~/~~4), Paul de Bank pays Petra von Player two dollars, and the game is over~~. ~~If the first head appers on the third toss (prob = 1/8), Paul de Bank pays Petra von PLayer four dollars, and so on~~. ~~Each time the winning head does not appear, the payoff doubles~~.		Today's problem is about the famous [https://en.wikipedia.org/wiki/St._Petersburg_paradox St. Petersburg paradox].

	~~Once heads appears~~, the game is over, and the ~~player~~ is ~~paid off~~. ~~The probability that~~ the ~~game is won~~ on the ~~nth~~ toss is 1/2^~~n. Since~~ Petra von Player is the ~~only player and she will always win~~, ~~she must pay to play~~.		In the St. Petersburg paradox, Petra von Player and Paul de Bank agree to play a game based on the toss of a coin. If a head is thrown on the first toss (prob = 1/2), Paul de Bank pays Petra von Player a dollar, and the game is over. If the first toss is tails and the second toss is head (prob = 1/4), Paul de Bank pays Petra von Player two dollars, and the game is over. If the first head appears on the Nth toss (prob = 1/2^N), Paul de Bank pays Petra von Player 2^(N-1) dollars, and so on. Each time the winning head does not appear, the payoff doubles.

	How much should Petra von Player pay ~~Paul de Bank for the privilege~~ of ~~playing the~~ game?		The paradox comes in the form of the question, "How much should Petra von Player be willing to pay to play this game?" The answer is, she should be willing to pay an infinite amount of money to play this game.

			Resolution:

			The resolution to the paradox lies in the fact that the expected winnings of Petra von Player are infinite:

	~~If the game can go on forever, the expected winnings are~~		EW = 1 * 1/2 + 2 * 1/4 + 4 * 1/8 + ... = 1/2 + 1/2 + 1/2 + ... = infinity

	~~1 * 1/2 + 21/4 + 41/8 + ...~~		If Petra is to make money off the game, she should be willing to pay a quantity less than the expected payoff to play the game. But if the expected payoff is infinite, Petra should be willing to pay an infinite amount of money to play the game - hence the paradox. (The resolution comes when we recognize that if Petra von Player can play for 10^100 years, there is a good chance that eventually she will win a payoff that makes it worth the wait, and that Paul de Bank will actually have 10^65 dollars or whatever.)
	~~= 1/2 + 2/4 + 4/8 + ...~~
	~~= 1/2 + 1/2 + 1/2 +~~ ...

	~~which is infinite - meaning if the game goes on forever, the expected winnings are infinite.~~

	If Petra is to make money off the game, she should be willing to pay a quantity less than the expected payoff to play the game. But if the expected payoff is infinite, Petra should be willing to pay an infinite amount of money to play the game - hence the paradox.

			'''PROBLEM'''

			Suppose Paul de Bank has a trillion (10^12) dollars.
	The Petersburg Paradox is resolved by recognizing that Petra von Player and Paul de Bank have finite resources, and money is not "infinite" the way that numbers are infinite, so ultimately the game must be limited by de Bank's total amount of money and von Player's capacity to procure (or borrow) money to keep playing.



	~~Now suppose~~ Paul de Bank has a trillion (10^12) dollars.

	a. How many coin tosses would it take Petra von Player to clean out Paul de Bank?		a. How many coin tosses would it take Petra von Player to clean out Paul de Bank?
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	d. What is the probability that Petra von Player will lose money on the game?		d. What is the probability that Petra von Player will lose money on the game?


	\|solution=		\|solution=

Admin at 10:56, 4 December 2019

2019-12-04T10:56:52Z

← Older revision		Revision as of 10:56, 4 December 2019
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	</pre>		</pre>

	But remember that bets start at 1, or <math>2^0</math>, so <math>2^39</math> takes 40 wins to reach.		But remember that bets start at 1, or <math>2^0</math>, so <math>2^{39}</math> takes 40 wins to reach.

	If Petra von Player wins 40 times in a row it would leave de Bank with a measely		If Petra von Player wins 40 times in a row it would leave de Bank with a measely

Admin at 10:56, 4 December 2019

2019-12-04T10:56:29Z

← Older revision		Revision as of 10:56, 4 December 2019
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	'''Part A - number of coin tosses for von Player to clear out de Bank?'''		'''Part A - number of coin tosses for von Player to clear out de Bank?'''

	~~Probability~~ Paul de Bank ~~will get cleared out~~:		We can find the number of times we have to double 1 to reach Paul de Bank's billion dollar fortune:

	<pre>		<pre>
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	</pre>		</pre>

	If Petra von Player wins 39 times in a row it would leave de Bank with a measely		But remember that bets start at 1, or <math>2^0</math>, so <math>2^39</math> takes 40 wins to reach.

			If Petra von Player wins 40 times in a row it would leave de Bank with a measely

	<pre>		<pre>
	>>> 1e12 - np.power(2,39)		>>> 1e12 - np.power(2,40-1)
	450,244,186,112		450,244,186,112
	</pre>		</pre>

	half a trillion dollars left to live on. If Petra von Player wins 40 times in a row, it leaves de Bank underwater		half a trillion dollars left to live on. If Petra von Player wins 41 times in a row, it leaves de Bank underwater

	<pre>		<pre>
	>>> 1e12 - np.power(2,40)		>>> 1e12 - np.power(2,41-1)
	-99,511,627,776		-99,511,627,776
	</pre>		</pre>

Admin at 10:53, 4 December 2019

2019-12-04T10:53:33Z

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	Probability Paul de Bank will get cleared out:		Probability Paul de Bank will get cleared out:

	~~<math>~~
	~~\begin{align}~~
	~~\log_2(10^{12}) & \approx & 39.86 \\~~
	~~2^{39} & = & 549,755,813,888~~
	~~\end{align}~~
	~~</math>~~

	<pre>		<pre>

Admin at 10:53, 4 December 2019

2019-12-04T10:53:12Z

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	<math>		<math>
	10^{12} \sim (10^3)^4 \sim (2^10)^4 \sim 2^{40}		10^{12} \sim (10^3)^4 \sim (2^{10})^4 \sim 2^{40}
	</math>		</math>

Admin at 10:52, 4 December 2019

2019-12-04T10:52:43Z

← Older revision		Revision as of 10:52, 4 December 2019
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	<math>		<math>
	\begin{~~array~~}		\begin{align}
	\dfrac{1}{2}^{40} & \sim & 2^{-40} \sim (2^{10})^{-4} \\		\dfrac{1}{2}^{40} & \sim & 2^{-40} \sim (2^{10})^{-4} \\
	\dfrac{1}{2}^{40} & \sim & (10^3)^{-4} \\		\dfrac{1}{2}^{40} & \sim & (10^3)^{-4} \\
	\dfrac{1}{2}^{40} & \sim & 10^{-12}		\dfrac{1}{2}^{40} & \sim & 10^{-12}
			\end{align}
	</math>		</math>

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	To make back $20, we would need to win a power of 2 quantity more than 16 - which is 32		To make back $20, we would need to win a power of 2 quantity more than 16 - which is 32

			<pre>
	>>> np.log2(20)		>>> np.log2(20)
	4.32		4.32
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	>>> np.power(2,5)		>>> np.power(2,5)
	32		32
			</pre>

	To make $32 we have to win 6 times (remember, the first bet is $1, so we are starting at <math>2^0</math> and it takes 6 wins to receive a payoff of at least <math>2^5</math>), which has the probability:		To make $32 we have to win 6 times (remember, the first bet is $1, so we are starting at <math>2^0</math> and it takes 6 wins to receive a payoff of at least <math>2^5</math>), which has the probability:

Admin at 10:51, 4 December 2019

2019-12-04T10:51:51Z

← Older revision		Revision as of 10:51, 4 December 2019
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	<math>		<math>
	10^{12} ~ (10^3)^4 ~ (2^10)^4 \sim 2^{40}		10^{12} \sim (10^3)^4 \sim (2^10)^4 \sim 2^{40}
	</math>		</math>

Admin at 10:51, 4 December 2019

2019-12-04T10:51:21Z

← Older revision		Revision as of 10:51, 4 December 2019
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	Probability Paul de Bank will get cleared out:		Probability Paul de Bank will get cleared out:

			<math>
			\begin{align}
			\log_2(10^{12}) & \approx & 39.86 \\
			2^{39} & = & 549,755,813,888
			\end{align}
			</math>

			<pre>
	>>> np.log2(1e12)		>>> np.log2(1e12)
	39.86		39.86
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	>>> np.power(2,39)		>>> np.power(2,39)
	549,755,813,888		549,755,813,888
			</pre>

	If Petra von Player wins 39 times in a row it would leave de Bank with a measely		If Petra von Player wins 39 times in a row it would leave de Bank with a measely

			<pre>
	>>> 1e12 - np.power(2,39)		>>> 1e12 - np.power(2,39)
	450,244,186,112		450,244,186,112
			</pre>

	half a trillion dollars to live on. If Petra von Player wins 40 times in a row, it leaves de Bank underwater		half a trillion dollars left to live on. If Petra von Player wins 40 times in a row, it leaves de Bank underwater

			<pre>
	>>> 1e12 - np.power(2,40)		>>> 1e12 - np.power(2,40)
	-99,511,627,776		-99,511,627,776
			</pre>

	and de Bank ends up with 99 billion dollars of debt.		and de Bank ends up with 99 billion dollars of debt.


	~~Alternative way to think about it:~~

	2^10 ~ 10^3		An alternative way to think about it is to use the back of the envelope approximation:

			<math>
			2^{10} \sim 10^3
			</math>

	so it follows that		so it follows that

	10^12 ~ (10^3)^4 ~ (2^10)^4 ~ 2^40		<math>
			10^{12} ~ (10^3)^4 ~ (2^10)^4 \sim 2^{40}
			</math>

	~~so approx~~. 40		Our back of the envelope estimate is about 40 times, which matches the more precise calculations above.


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	'''Part b - amount to play?'''		'''Part b - amount to play?'''

	A maximum of 40 tosses leads to an expected payoff of		Given our maximum of 40 tosses, and the 50% probaability, a maximum of 40 tosses leads to an '''expected payoff''' of

			<pre>
	E = 1/2 + 1/2 + ... + 1/2		E = 1/2 + 1/2 + ... + 1/2
	E = 40*1/2		E = 40*1/2
	E = $20		E = 20
			</pre>

			So we should expect to win $20 from the game.


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	'''Part c - What is probability Paul de Bank will get cleaned out?'''		'''Part c - What is probability Paul de Bank will get cleaned out?'''

	1/~~2^n =~~ 1/2^40 ~~= 9.09e-13~~		The probability of getting a heads once is <math>\dfrac{1}{2}</math>; twice is <math>\dfrac{1}{4}</math>; and so on. The probability of 40 heads in a row is:

	~~Alternatively,~~		<math>
			\dfrac{1}{2}^n = \dfrac{1}{2}^{40} = \dfrac{1}{2^{40}} = 9.09 \times 10^{-13}
			</math>

	1/2^40 ~ 2^-40 ~ (2^10)^-4		Alternatively, using our back of the envelope estimate of
	1/2^40 ~ (10^3)^-4
	1/2^40 ~ 10^-12		<math>
			2^{10} \sim 10^3
			</math>

			we get a back of the envelope estimate of

			<math>
			\begin{array}
			\dfrac{1}{2}^{40} & \sim & 2^{-40} \sim (2^{10})^{-4} \\
			\dfrac{1}{2}^{40} & \sim & (10^3)^{-4} \\
			\dfrac{1}{2}^{40} & \sim & 10^{-12}
			</math>

	if Paul de Bank paints a single one dollar bill red, and then put all of his dollars into one giant pile, it's the probability of pulling out that one single dollar bill		if Paul de Bank paints a single one dollar bill red, and then put all of his dollars into one giant pile, it's the probability of pulling out that one single dollar bill
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	'''Part d - What is the probability that Petra von Player will lose money on the game?'''		'''Part d - What is the probability that Petra von Player will lose money on the game?'''

	~~> 97%~~		To make back $20, we would need to win a power of 2 quantity more than 16 - which is 32

	to make back $20, we would need to win a power of 2 quantity more than 16 - which is 32

	>>> np.log2(20)		>>> np.log2(20)
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	32		32

	To make $32 we have to win 6 times, which has the probability		To make $32 we have to win 6 times (remember, the first bet is $1, so we are starting at <math>2^0</math> and it takes 6 wins to receive a payoff of at least <math>2^5</math>), which has the probability:

	1/2^6 = 0.0156		<math>
			\dfrac{1}{2^6} \sim 0.0156
			</math>

	so probability of Petra von Player losing money on the game is		so probability of Petra von Player losing money on the game is

			<pre>
	>>> 1-(1/np.power(2,6))		>>> 1-(1/np.power(2,6))
	0.9843		0.9843
			</pre>

	}}		}}

Admin at 10:38, 4 December 2019

2019-12-04T10:38:12Z

← Older revision		Revision as of 10:38, 4 December 2019
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	\|solution=		\|solution=

	~~---~~		'''Part A - number of coin tosses for von Player to clear out de Bank?'''
	Part a - number of coin tosses for von Player to clear out de Bank?

	Probability Paul de Bank will get cleared out:		Probability Paul de Bank will get cleared out:
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	so approx. 40		so approx. 40

	~~---~~
	Part b - amount to play?
			'''Part b - amount to play?'''

	A maximum of 40 tosses leads to an expected payoff of		A maximum of 40 tosses leads to an expected payoff of
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	~~---~~
	Part c - What is probability Paul de Bank will get cleaned out?		'''Part c - What is probability Paul de Bank will get cleaned out?'''

	1/2^n = 1/2^40 = 9.09e-13		1/2^n = 1/2^40 = 9.09e-13
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	if Paul de Bank paints a single one dollar bill red, and then put all of his dollars into one giant pile, it's the probability of pulling out that one single dollar bill		if Paul de Bank paints a single one dollar bill red, and then put all of his dollars into one giant pile, it's the probability of pulling out that one single dollar bill

	~~---~~
	Part d - What is the probability that Petra von Player will lose money on the game?
			'''Part d - What is the probability that Petra von Player will lose money on the game?'''

	> 97%		> 97%