## 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**

***: I got 200.00000000002351265624999861789 on the calculator

Note: I used variations of "a2 + b2 = c2" and the "Cosine Rule" to work this out (I forgot how to do it, so had to look up when to use cos and cos-1 on the calculator).

ETA: I figured it out.. the areas aren't proportional. But the sides are so I essentially did x/(100+x) = 100/150. Giving x=200

If the height of the smaller triangle is x (i.e. the distance across the lake), then the height of the larger triangle is 150/100*x = 1.5x.

We're told the difference in heights is 100m;so (1.5x - x)=100m; so x=200m.

Quote:Ace2200 m

Quote:ksdjdjMy answer is 200 m ***

***: I got 200.00000000002351265624999861789 on the calculator

Note: I used variations of "a2 + b2 = c2" and the "Cosine Rule" to work this out (I forgot how to do it, so had to look up when to use cos and cos-1 on the calculator).

Quote:rsactuaryI tried to use similar triangles theorem to compare the bases of the smaller triangle and the larger triangle to the area. I thought they should be in the same proportion? I got 150m, but I'm guessing I've done something wrong. Can anyone enlighten me?

ETA: I figured it out.. the areas aren't proportional. But the sides are so I essentially did x/(100+x) = 100/150. Giving x=200

Quote:charliepatrickSimilar triangles does work (I'm assuming the line across the lake is NS and the others EW).

The base of the big triangle is 150m, and the base of the smaller triangle is 100m.

If the height of the smaller triangle is x (i.e. the distance across the lake), then the height of the larger triangle is 150/100*x = 1.5x.

We're told the difference in heights is 100m;so (1.5x - x)=100m; so x=200m.

Correct!

---------------------------

My dad taught me how to swim by throwing me into a lake.

The swimming part was easy but getting out of the burlap sack took some work.

If anyone cares:Quote:Ace2Two craps players make a side bet on a side bet.

Player A always plays "All Small" and "All Tall". Player B always plays "Make em All".

Starting with a new shooter, A tells B he will win both of his wagers at least twice before B wins his once. In this casino, All Small, All Tall, and Make em All bets can only be made before a new shooter's first roll.

This is an even money bet. They will play until their wager is resolved. What is A's edge (or disadvantage)?

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

However, it’s easier to solve this by integrating the following equation from zero to infinity:

(e^(sx) - sx - 1) * sx/e^x * 2s

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

However, it’s easier to solve this by integrating the following equation from zero to infinity:

(e^(sx) - sx - 1) * sx/e^x * 2s

I’m not quite following. How did you derive the conditional probability of hitting the All given that the Small has just been hit?

I didn’t. There are only 3 relevant outcomes:Quote:unJonI’m not quite following. How did you derive the conditional probability of hitting the All given that the Small has just been hit?

Shooter wins small (only)

Shooter wins tall (only)

Shooter wins all

Quote:Ace2I didn’t. There are only 3 relevant outcomes:

Shooter wins small (only)

Shooter wins tall (only)

Shooter wins all

Oh I see. You actually implicitly have it: 1 - (x - y) / x.

How about this one: what’s the expected number of rolls for this bet between A and B to resolve?