A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game”
will eventually empty the pot.
--
Rich
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game”
will eventually empty the pot.
--
Rich
On Tuesday, October 24, 2023 at 5:39:15 PM UTC-4, RichD wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game”
will eventually empty the pot.
--That's because you used percentages. What would happen if you increased/decreased by $1?
Rich
On Wednesday, October 25, 2023 at 1:58:05 PM UTC-7, Tim Norfolk wrote:.
On Tuesday, October 24, 2023 at 5:39:15 PM UTC-4, RichD wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” will eventually empty the pot.
--That's because you used percentages. What would happen if you increased/decreased by $1?
Rich
Hey, dumbfuck....
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game”
will eventually empty the pot.
That's because you used percentages. What would happen if you increased/decreased by $1?
On October 25, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game”
will eventually empty the pot.
That's because you used percentages. What would happen if you increased/decreased by $1?It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
--
Rich
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” >>>> will eventually empty the pot.
That's because you used percentages. What would happen if you increased/decreased by $1?
It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
EV = ??
On October 27, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” >>>> will eventually empty the pot.
That's because you used percentages. What would happen if you increased/decreased by $1?
It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
Not quite. On average, increasing/decreasing by $1 is stationary.Most excellent!
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
You aren't a pro gambler, I take it -
--
Rich
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” >>>>>> will eventually empty the pot.
That's because you used percentages. What would happen if you
increased/decreased by $1?
It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
--> EV = ??
The EV is 0.
On October 28, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” >>>>>> will eventually empty the pot.
That's because you used percentages. What would happen if you
increased/decreased by $1?
It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
You're demented.
--
Rich
On October 28, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%. Obviously, this
is a 50–50 fair game. The pot starts with $100, it must
remain $100, long run.
You play N times, i.e. N tosses. For simplicity, set N = 2
I) Heads, then tails:
Pot = $100 + 10 - 11 = $99
II) Tails, then heads:
Pot = $100 - 10 + 9 = $99
Two equiprobable outcomes both yield a net loss! This “fair game” >>>>>> will eventually empty the pot.
That's because you used percentages. What would happen if you
increased/decreased by $1?
It would remain $100... same as the example I presented,
which used percentages... heh heh
hook, line, sinker -
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
You're demented.
--
Rich
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
On October 31, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1In the Say Red game, in most trials, I can call red at a moment when I'm better than 50% favorite.
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
On October 31, Tim Norfolk wrote:
A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1In the Say Red game, in most trials, I can call red at a moment when I'm better than 50% favorite.
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'm >> better than 50% favorite.A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get.
ahead one bet and quit. That is the bet you dodge and run from.
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'm >>>> better than 50% favorite.A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows
by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get.
ahead one bet and quit. That is the bet you dodge and run from.
That comment doesn't either...
On 11/2/2023 7:13 PM, VegasJerry wrote:.
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get.
ahead one bet and quit. That is the bet you dodge and run from.
That comment doesn't either...
As always, Jerry quacks at Tim runaway..
On Friday, November 3, 2023 at 9:58:14 AM UTC-7, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:.
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'm >>>>>> better than 50% favorite.A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >>>> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...
As always, Jerry quacks at Tim runaway..
Nor does that comment...
Just how many threads are you running from now?
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get.
ahead one bet and quit. That is the bet you dodge and run from.
That comment doesn't either...As always, Jerry quacks at Tim runaway.
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software. www.avg.com
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:As always, Jerry quacks at Tim runaway.
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'm >>>>>> better than 50% favorite.A gambling game, consisting of coin tosses, involves
a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >>>> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't understand, do you?
On 11/3/2023 1:41 PM, VegasJerry wrote:.
On Friday, November 3, 2023 at 9:58:14 AM UTC-7, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:.
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >>>> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...
As always, Jerry quacks at Tim runaway..
Nor does that comment...
Just how many threads are you running from now?
Tim and Jerr ... so happy together.
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote:
On 11/1/2023 5:33 PM, Tim Norfolk wrote:.
On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote:
On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...As always, Jerry quacks at Tim runaway.
[You would not take the bet either.]
.--Only an idiot would take 100:1 bet at even money. You really don't understand, do you?
This email has been checked for viruses by AVG antivirus software. www.avg.com
On 11/3/2023 6:48 PM, Tim Norfolk wrote:.
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >>>> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't understand, do you?I agree only an idiot would say that the person proposing the bet in the
Say Red game certainly does have "an advantage" ... wouldn't you say
that, Jim?
[Jerry would not take the bet either.]
On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.
Your comment has nothing to do with what I posted about your game.
It certainly is a 50/50 game ... unless the bet is whether you can get >>>> ahead one bet and quit. That is the bet you dodge and run from..
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't understand, do you?I agree only an idiot would say that the person proposing the bet in the
Say Red game certainly does have "an advantage" ... wouldn't you say
that, Jim?
[Jerry would not take the bet either.]
On Saturday, November 4, 2023 at 1:11:10 PM UTC-4, da pickle wrote:
On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:I agree only an idiot would say that the person proposing the bet in the
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary. >>>>>>>>>>>>> Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.It certainly is a 50/50 game ... unless the bet is whether you can get >>>>>> ahead one bet and quit. That is the bet you dodge and run from.
Your comment has nothing to do with what I posted about your game. >>>>> .
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't understand, do you?
Say Red game certainly does have "an advantage" ... wouldn't you say
that, Jim?
[Jerry would not take the bet either.]
Who is Jim?
On 11/4/2023 5:44 PM, Tim Norfolk wrote:
On Saturday, November 4, 2023 at 1:11:10 PM UTC-4, da pickle wrote:
On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote:I agree only an idiot would say that the person proposing the bet in the >>> Say Red game certainly does have "an advantage" ... wouldn't you say
On 11/2/2023 7:13 PM, VegasJerry wrote:
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>>>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment >>>>>>>>> when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is >>>>>>>>>>>>>> stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or >>>>>>>>>> shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the >>>>>>>>>> probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we >>>>>>>>>> have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln >>>>>>>>>> 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1 >>>>>>>>>> if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n
increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is >>>>>>>>>> still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.It certainly is a 50/50 game ... unless the bet is whether you
Your comment has nothing to do with what I posted about your game. >>>>>> .
can get
ahead one bet and quit. That is the bet you dodge and run from.
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't
understand, do you?
that, Jim?
[Jerry would not take the bet either.]
Who is Jim?
Even Jim knows you are a runner. Just say it, TIM ... you are running
and running and running.
I do have an "advantage" in the Say Red game ... admit it.
On 11/4/2023 5:44 PM, Tim Norfolk wrote:.
On Saturday, November 4, 2023 at 1:11:10 PM UTC-4, da pickle wrote:
On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote: >>>> On 11/2/2023 7:13 PM, VegasJerry wrote:I agree only an idiot would say that the person proposing the bet in the >> Say Red game certainly does have "an advantage" ... wouldn't you say
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1
if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
.And the 'Say Red' game overall is still 50% win, 50% lose.It certainly is a 50/50 game ... unless the bet is whether you can get
Your comment has nothing to do with what I posted about your game. >>>>> .
ahead one bet and quit. That is the bet you dodge and run from.
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software.
www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't understand, do you?
that, Jim?
[Jerry would not take the bet either.]
.Who is Jim?
Even Jim knows you are a runner.....
and running and running.
I do have an "advantage" in the Say Red game ... admit it.
On 11/4/2023 6:25 PM, da pickle wrote:.
On 11/4/2023 5:44 PM, Tim Norfolk wrote:
On Saturday, November 4, 2023 at 1:11:10 PM UTC-4, da pickle wrote:
On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote: >>>>> On 11/2/2023 7:13 PM, VegasJerry wrote:I agree only an idiot would say that the person proposing the bet in the >>> Say Red game certainly does have "an advantage" ... wouldn't you say
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>>>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment >>>>>>>>> when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is >>>>>>>>>>>>>> stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or >>>>>>>>>> shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the >>>>>>>>>> probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we >>>>>>>>>> have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln >>>>>>>>>> 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1 >>>>>>>>>> if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n
increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is >>>>>>>>>> still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
And the 'Say Red' game overall is still 50% win, 50% lose.It certainly is a 50/50 game ... unless the bet is whether you >>>>>>> can get
Your comment has nothing to do with what I posted about your game. >>>>>> .
ahead one bet and quit. That is the bet you dodge and run from. >>>>>> .
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software. >>>>> www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't
understand, do you?
that, Jim?
[Jerry would not take the bet either.]
Who is Jim?
Even Jim knows you are a runner. Just say it, TIM ... you are running
and running and running.
I do have an "advantage" in the Say Red game ... admit it.
I'm still running
On Sunday, November 5, 2023 at 6:02:20 AM UTC-8, da pickle wrote:
On 11/4/2023 6:25 PM, da pickle wrote:.
On 11/4/2023 5:44 PM, Tim Norfolk wrote:
On Saturday, November 4, 2023 at 1:11:10 PM UTC-4, da pickle wrote: >>>>> On 11/3/2023 6:48 PM, Tim Norfolk wrote:
On Friday, November 3, 2023 at 12:58:14 PM UTC-4, da pickle wrote: >>>>>>> On 11/2/2023 7:13 PM, VegasJerry wrote:I agree only an idiot would say that the person proposing the bet in the >>>>> Say Red game certainly does have "an advantage" ... wouldn't you say >>>>> that, Jim?
On Thursday, November 2, 2023 at 4:56:46 PM UTC-7, da pickle wrote: >>>>>>>>> On 11/1/2023 5:33 PM, Tim Norfolk wrote:As always, Jerry quacks at Tim runaway.
.On Tuesday, October 31, 2023 at 4:38:26 PM UTC-4, RichD wrote: >>>>>>>>>>> On October 31, Tim Norfolk wrote:
In the Say Red game, in most trials, I can call red at a moment >>>>>>>>>>> when I'mA gambling game, consisting of coin tosses, involves >>>>>>>>>>>>>>>>>> a fluctuating pot. Each time heads shows, the pot grows >>>>>>>>>>>>>>>>>> by 10%; if tails, the pot shrinks by 10%.
Not quite. On average, increasing/decreasing by $1 is >>>>>>>>>>>>>>>> stationary.
Doing so by percentages decreases to 0.
H,T --> 100 + 10 - 11 = $99
T,H --> 100 - 10 + 9 = $99
H,H --> 100 + 10 + 11 = $121
T,T --> 100 - 10 - 9 = $81
EV = ??
The EV is 0.
You really think the $100 pot will shrink to zero?
Suppose that you consider the problem where the pot grows or >>>>>>>>>>>> shrinks by $1 on heads/tails.
The expected value of the pot after 2n trials is $100, and the >>>>>>>>>>>> probability that the pot is less
than $100 is slightly over 0.5
Now consider your question: After 2n trials, suppose that we >>>>>>>>>>>> have n+k heads and n-k tails.
The value of the pot is 100*\exp ((n+k) \ln 1.1 + (n-k) \ln >>>>>>>>>>>> 0.9) = \exp ((\ln 1.1+\ln 0.9) n + (\ln 1.1 - \ln 0.9) k) < 1 >>>>>>>>>>>> if k < (\ln 1.1+\ln 0.9)n/(\ln 0.9-\ln 1.1)
The probability of that happening increases to 1 as n
increases, using the Central Limit Theorem.
In other words, even though the expected value of the pot is >>>>>>>>>>>> still $100, it is almost certainly going to be less.
better than 50% favorite.
--
Rich
And the 'Say Red' game overall is still 50% win, 50% lose. >>>>>>>>>>It certainly is a 50/50 game ... unless the bet is whether you >>>>>>>>> can get
Your comment has nothing to do with what I posted about your game. >>>>>>>> .
ahead one bet and quit. That is the bet you dodge and run from. >>>>>>>> .
That comment doesn't either...
[You would not take the bet either.]
--
This email has been checked for viruses by AVG antivirus software. >>>>>>> www.avg.com
Only an idiot would take 100:1 bet at even money. You really don't >>>>>> understand, do you?
[Jerry would not take the bet either.]
Who is Jim?
Even Jim knows you are a runner. Just say it, TIM ... you are running
and running and running.
I do have an "advantage" in the Say Red game ... admit it.
You are still running, Tim >Yea, we know....
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