Easy math puzzles

Quote: Ace2

Fifteen

Quote: Gialmere

Too easy eh? Okay...



You're an APr at a church casino night. You end up hanging out at a table run by two sweet old ladies who have designed their own dice game.

You pay $6 to play the game. You roll a standard fair pair of dice. You then have two choices: you can cash out and get paid the dollar amount of the roll, or you can pay $1 to roll the dice again but you must accept the second result.

On average, how much money will you make per game if you use the strategy of stopping if the first roll is 7 or higher?

[I'll leave the ethics of this play up to you.]

Quote: unJon

Show Spoiler

About $1.56 per roll. You could do just as well with less variance but stopping if the first roll is 6 or higher.

Quote: Gialmere

We're in the same ballpark, but I'm showing a different return.

Quote: Gialmere

We're in the same ballpark, but I'm showing a different return.

Quote: Gialmere

This particular puzzle is by mathematician Leslie Green. Here is his solve...

Quote: charliepatrick

I think your answer allows for infinite rerolls for $1, whereas we had assumed you only got one reroll.

I agree with the values

...of 1 20/21 where you change the EV for 2 thru 6 to be EV-1 for the whole game.

Pay: $6.000
Initial Roll	Combinations	EV	Contribution
2	1	0.952 381	0.952 381
3	2	0.952 381	1.904 762
4	3	0.952 381	2.857 143
5	4	0.952 381	3.809 524
6	5	0.952 381	4.761 905
7	6	1.000 000	6.000 000
8	5	2.000 000	10.000 000
9	4	3.000 000	12.000 000
10	3	4.000 000	12.000 000
11	2	5.000 000	10.000 000
12	1	6.000 000	6.000 000
Total	36	1.952 381	70.285 714

Quote: charliepatrick

What lovely ladies they must be offering such fair games. This one reminds me of Crown and Anchor we used to deal at charity events - except it used three dice and paid 1, 2, or 3 to 1 (for those unfamiliar with the game each die has Crown, Anchor, and the four suits).

Quote: Gialmere

(snip) A year later you return to the same church casino night and see the same two sweet old ladies. (snip)

(snip)...What is the house edge for this game?

Quote: charliepatrick

What lovely ladies they must be offering such fair games. This one reminds me of Crown and Anchor we used to deal at charity events - except it used three dice and paid 1, 2, or 3 to 1 (for those unfamiliar with the game each die has Crown, Anchor, and the four suits).

Quote: ksdjdj

The ladies are getting better at this...

the next game they run may have a house edge (?) . Anyway, as stated above by charliepatrick , the game is "fair", so it has a 0% house edge.

Proof:
You are paid 2/1, if one out of two dice match the number you picked
You are paid 5/1, if two out of two dice match the number you picked
You lose for all other rolls

Chance of losing on both dice is: 5/6 x 5/6 = 25/36
Chance of winning on one out of the two is: (5/6 x 1/6) + (1/6 x 5/6) = 10/36
Chance of winning on two out of two is: (1/6)/6 = 1/36

EV = ((2 x 10/36) + (5 x 1/36)) - (25/26) = 25/36 - 25/36 = 0

Note: This should be "mathematically correct", but I probably used too many brackets ("redundant brackets"), ; )

Quote: Ace2

Thirty

Quote: Ace2

Thirty

Quote: unJon

Agree subject to the plane being Euclidean.

Quote: Ace2

Thirty

Quote: Gialmere

In an old western saloon, an argument over a game of cards escalated into all the cowboys in the vicinity drawing their guns and shooting at each other.

When the smoke finally cleared, 90% of the cowboys had been shot in the leg, 85% had been shot in the arm, 80% had been shot in the gut, and 75% had been shot in the head. Surprisingly, only the cowboys who received all 4 types of wounds died in the big gunfight.

What is the smallest possible percentage of cowboys who ended up being laid to rest at Boothill?

Quote: Gialmere

Quote: ThatDonGuy

My Answer

30%

Quote: ThatDonGuy

My Answer

30%

Without loss of generality, assume there are 100 cowboys
10 did not get shot in the leg
15 did not get shot in the arm
20 did not get shot in the gut
25 did not get shot in the stomach
This means at most 10 + 15 + 20 + 25 = 70 did not get shot in all four places, so at least 30 did.

Quote: ChesterDog

Show Spoiler

6 out of 20?
30%

Quote: charliepatrick

Quote: ThatDonGuy

My Answer

30%

I agree using similar logic...

Show Spoiler

I used there were 20 cowboys and saw what happened to them. Ailment A applied to 15, so left 5 unaffected and safe. Ailment B applied to 16, so could apply to the 5 safe ones, meaning 11 were affected by A and B, leaving 4 more safe. Similarly you leave 3 and 2 more safe at the next steps. Total safe = 5+4+3+2 = 14. Total died = 6 (30%)

Quote: Wizard

Answer

I'm getting 30%.

Solution

First, shoot 90% of the cowboys in the leg.

Next, shoot the 10% left standing in the arm. You have 75% more to shoot in the arm, so take them from those already shot in the leg.

So, now we're at:

Leg only 15% (90% - 75%)
Arm only 10%
Both 75%
Neither 0%

Total leg: 90%
Total arm: 85%

Next, let's go onto the gut injuries (80%). Shoot the 25% with just one injury in the gut. We have 80%-25% = 55% more people to shoot. We will take that 55% from the pool of people with both injuries. So now we're at:

Leg & gut 15%
Arm & gut 10%
Leg & arm 20% (75% - 55%)
All three 55%
One injury 0%
Zero injuries 0%

Finally, consider the 75% with head injuries. First, shoot the 45% with exactly two injuries. We have 30% more to go, so get them from the 55% with all three injuries. That leaves:

Head, Leg & gut 15%
Head, Arm & gut 10%
Head, Leg & arm 20%
Leg, arm & gut: 25% (55% - 30%)
All four 30%
Zero injuries 0%
One injury 0%
Two injuries 0%

Total leg: 15% + 20% + 25% + 30% = 90%
Total arm: 10% + 20% + 25% + 30% = 85%
Total gut: 15% + 10% + 25% + 30% = 80%
Total head: 15% + 10% + 20% + 30% = 75%

Quote: Wizard

Q: What is the general formula, given any four probabilities of injury? ... I can't immediately find a common sense answer why this is true.

Quote: ThatDonGuy

...here's one:...There are N cities.
Each pair of cities is connected by a highway that is numbered with a different integer...Given any three cities, the three highways that connect them to each other include at least one even-numbered highway and at least one odd-numbered highway.
What is the largest possible N?...

AB O
AC O BC E
AD E BD E CD O
AE E BE O CE E DE O

Quote: charliepatrick

nice puzzle as you have to create your own logical answer, not obvious.

I really liked this until a penny dropped.

It's possible to connect 5 cities as follows

AB O
AC O BC E
AD E BD E CD O
AE E BE O CE E DE O

Now consider 6 cities then city A and it's connection with cities B,C,D,E and F. Three of these must be the same, either Odd or Even. Suppose AB AC and AF were odd.
Then consider ABC then BC must be even.
Then consider ABF then BF must be even.
Then consider ACF then CF must be even.
However now consider BCF - BC BF CF are all even.
So it is impossible for A to have three routes all being odd.

Thus you can't connect six cities and five is the limit.

Bonus question solution

If three "types" of highways are allowed, the maximum number is 16.
To make it easier to explain, color the highways numbered 100-399 red, the ones numbered 400-699 blue, and the ones numbered 700-899 green.

Here is a coloring of 16 cities:


As for 17, every city has highways touching 16 others; if there were no more than 5 of any color, then there could be no more than 15 highways from that city, so at least one color has at least six highways.
Let AX, BX, CX, DX, EX, and FX all be red.
If any of AB, AC, AD, ..., DE, DF, EF are red, then those two cities and X are connected by three red highways.
If none of them are red, then this becomes the "six cities and two types" problem, which has already been shown to be impossible.

Quote: Wizard

Wiz answer

Imagine a line from the dog to the center of the circle. Endeavor to be as far away from the dog as possible on this line. You will be able to do this until you reach 1/4 of the radius of the circle.

Once at this point, with the dog as far away as possible, make a dash towards the closest point on the fence.

Let's say your speed is 1. Then

Time it takes you = 0.75*R/1 = 0.75R
Time it takes the dog = pi*R/4 = 0.7853982R (which is greater than 0.75R)

Whatever the radius of the circle, it will take you less time to reach the fence.

Joke Meme

Quote: Gialmere

Correct!

Quote: Gialmere

...You draw cards from the deck one by one. A red card pays you a dollar. A black one loses you a dollar. You can stop any time you want...What is the optimal stopping rule in terms of maximizing expected payoff? Also, what is the expected payoff following this optimal rule?

Quote: charliepatrick

simulation results for 1m hands

Parms: ndx:1 sh:1000000 Time:10:3:22:801
Count: 6 Exp: 6 Hands: 11340 Win: 68040
Count: 8 Exp: 6 Hands: 17186 Win: 103116
Count: 10 Exp: 6 Hands: 19426 Win: 116556
Count: 11 Exp: 5 Hands: 44719 Win: 223595
Count: 13 Exp: 5 Hands: 32893 Win: 164465
Count: 15 Exp: 5 Hands: 28431 Win: 142155
Count: 17 Exp: 5 Hands: 25640 Win: 128200
Count: 19 Exp: 5 Hands: 23035 Win: 115175
Count: 21 Exp: 5 Hands: 21042 Win: 105210
Count: 23 Exp: 5 Hands: 19572 Win: 97860
Count: 24 Exp: 4 Hands: 41611 Win: 166444
Count: 26 Exp: 4 Hands: 28239 Win: 112956
Count: 28 Exp: 4 Hands: 23647 Win: 94588
Count: 30 Exp: 4 Hands: 20975 Win: 83900
Count: 32 Exp: 4 Hands: 19475 Win: 77900
Count: 34 Exp: 4 Hands: 17874 Win: 71496
Count: 35 Exp: 3 Hands: 39522 Win: 118566
Count: 37 Exp: 3 Hands: 27081 Win: 81243
Count: 39 Exp: 3 Hands: 23471 Win: 70413
Count: 41 Exp: 3 Hands: 20940 Win: 62820
Count: 43 Exp: 3 Hands: 19272 Win: 57816
Count: 44 Exp: 2 Hands: 46247 Win: 92494
Count: 46 Exp: 2 Hands: 32951 Win: 65902
Count: 48 Exp: 2 Hands: 29451 Win: 58902
Count: 50 Exp: 2 Hands: 26028 Win: 52056
Count: 51 Exp: 1 Hands: 92964 Win: 92964
Count: 52 Exp: 0 Hands: 246968 Win: 0
Overall Result: Exp: 2.624832 Hands: 1000000 Win: 2624832
Parms: ndx:1 sh:1000000 Time:10:3:26:719 

(The second simulation also got 2.625) Also interesting when I ran a simulation where I remember where the best possible place to stop would have been, the EV = 4.040