Poll

8 votes (47.05%)
6 votes (35.29%)
3 votes (17.64%)
2 votes (11.76%)
6 votes (35.29%)
2 votes (11.76%)
3 votes (17.64%)
2 votes (11.76%)
8 votes (47.05%)
6 votes (35.29%)

17 members have voted

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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August 10th, 2020 at 10:52:11 AM permalink
Quote: Ace2

If anyone cares:

The overall probability x of winning the Small bet is 0.026354 and the overall probability y of winning the All bet is 0.005258. However, if the only three relevant outcomes are winning Small, Tall or All then the probability s of winning only the Small bet is (x-y)/(2x-y) = 0.44459. If we consider only two outcomes of winning Small or All then the probability m of winning Small (only) is (x-y)/x = 0.80049

Using linear equations where state a is 1 small or tall win, state b is 2 similar wins, state c is 2 distinct wins and state d is 3 wins we can say:

x = 2sa
a = s(b + c)
b = md
c = 2sd
d = m

Solve for x and the probability of player A winning is 0.5347 for a 6.94% advantage


My simulations show that Player A wins about 60% of the time, which is a 20% advantage.

I am working under the following assumption:
(a) Player A always has a Small bet and a Tall bet in play, except when either someone rolls a 7 on a comeout (in which case, both bets lose, and neither can be placed again until the shooter misses a point) or one of the two bets is won (in which case that bet cannot be placed again until the shooter misses a point);
(b) Player B always has the All bet in play (again, unless someone rolls a 7 on a comeout, in which case that bet is lost and the player cannot place it again until the shooter misses a point).

You may be assuming an independence that is not there. For example, if a Small bet is won, then the current All bet must have 2, 3, 4, 5, and 6 made, plus whatever numbers in the current Tall bet have been made.
ThatDonGuy
ThatDonGuy
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August 10th, 2020 at 11:54:53 AM permalink
Here's an easy problem with what may be a trick answer:

In the late 1970s, "biorhythms" were all the rage.
Each person has three "cycles": a 23-day Physical cycle, a 28-day Emotional cycle, and a 33-day Intellectual cycle.
Each cycle is a sine wave with positive and negative values, and a "critical point" at the zero line (so, for example, the Physical cycle goes through a critical point every 11 1/2 days).
All three cycles begin at the critical point, headed in the positive direction.
When all three cycles converge at a critical point, this is Not A Good Thing. How often does this happen?
Ace2
Ace2
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August 10th, 2020 at 11:58:42 AM permalink
Quote: ThatDonGuy

My simulations show that Player A wins about 60% of the time, which is a 20% advantage.

I am working under the following assumption:
(a) Player A always has a Small bet and a Tall bet in play, except when either someone rolls a 7 on a comeout (in which case, both bets lose, and neither can be placed again until the shooter misses a point) or one of the two bets is won (in which case that bet cannot be placed again until the shooter misses a point);
(b) Player B always has the All bet in play (again, unless someone rolls a 7 on a comeout, in which case that bet is lost and the player cannot place it again until the shooter misses a point).

You may be assuming an independence that is not there. For example, if a Small bet is won, then the current All bet must have 2, 3, 4, 5, and 6 made, plus whatever numbers in the current Tall bet have been made.

Bets are made before every new shooter’s first roll. Always then and only then
It’s all about making that GTA
rsactuary
rsactuary
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August 10th, 2020 at 12:17:20 PM permalink
Quote: ThatDonGuy

Here's an easy problem with what may be a trick answer:

In the late 1970s, "biorhythms" were all the rage.
Each person has three "cycles": a 23-day Physical cycle, a 28-day Emotional cycle, and a 33-day Intellectual cycle.
Each cycle is a sine wave with positive and negative values, and a "critical point" at the zero line (so, for example, the Physical cycle goes through a critical point every 11 1/2 days).
All three cycles begin at the critical point, headed in the positive direction.
When all three cycles converge at a critical point, this is Not A Good Thing. How often does this happen?



Since none of the three cycles have a common divisor, it should just be the product of the three, then divided by 2 because the sine waves cross the critical point twice in a cycle? 23x29x33/2 = 10626 days or 29.09 years.

ETA: Looked at Wiki after I posted this and it says I'm wrong, but I'm not convinced.
ThatDonGuy
ThatDonGuy
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August 10th, 2020 at 1:52:39 PM permalink
Quote: rsactuary

Since none of the three cycles have a common divisor, it should just be the product of the three, then divided by 2 because the sine waves cross the critical point twice in a cycle? 23x29x33/2 = 10626 days or 29.09 years.

ETA: Looked at Wiki after I posted this and it says I'm wrong, but I'm not convinced.


No, that is correct, but next time, put answers in a spoiler box.

The trick is that it's not 23/2 x 29/2 x 33/2; you have to think of these in integral numbers of "half-days."

Wikipedia is using the coincidence of the "full cycles," but that doesn't take into account the possibility that the Physical and Intellectual cycles can be ending a full cycle at the same time that an Emotional cycle can be halfway through one.
rsactuary
rsactuary
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August 10th, 2020 at 5:29:30 PM permalink
sorry, I get so excited that I can actually answer a question, and then I forget to spoiler it.

I found an online calculator and proved that the lines do indeed all intersect at 29.09 years and at 58.18 years.
Gialmere
Gialmere
Joined: Nov 26, 2018
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August 11th, 2020 at 8:06:23 AM permalink


On a standard pair of six-sided dice there is one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12. Find another pair of six-sided dice such that:

a) The set of dots on each die is not the standard {1,2,3,4,5,6}.

b) Each face has at least one dot.

c) The number of ways of obtaining each sum is the same as for the standard dice.
Have you tried 22 tonight? I said 22.
rsactuary
rsactuary
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August 11th, 2020 at 8:58:54 AM permalink
deleted.. I really didn't read this well.
Last edited by: rsactuary on Aug 11, 2020
Gialmere
Gialmere
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August 11th, 2020 at 9:02:25 AM permalink
Quote: rsactuary

one die is the standard six faced die with numbers 1 to 6. The other die is 1,1,1,6,6,6


Sorry, no.

Note criteria a.
Have you tried 22 tonight? I said 22.
rsactuary
rsactuary
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August 11th, 2020 at 9:14:24 AM permalink
oooooh

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