## 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**

Quote:Ace2If 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.

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?

Bets are made before every new shooter’s first roll. Always then and only thenQuote:ThatDonGuyMy 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.

Quote:ThatDonGuyHere'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.

Quote:rsactuarySince 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.

I found an online calculator and proved that the lines do indeed all intersect at 29.09 years and at 58.18 years.

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.

Quote:rsactuaryone 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.