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Gialmere
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July 6th, 2020 at 12:11:10 PM permalink
Quote: Ace2

Fifteen


Correct!
------------------------------

Q: Why should you cook kale in coconut oil?

A: Makes it easier to slide it right into the trash.
Have you tried 22 tonight? I said 22.
Gialmere
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July 6th, 2020 at 1:11:18 PM permalink
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.]
Have you tried 22 tonight? I said 22.
unJon
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July 6th, 2020 at 1:31:07 PM permalink
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.]



About $1.56 per roll. You could do just as well with less variance but stopping if the first roll is 6 or higher.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere
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July 6th, 2020 at 2:33:15 PM permalink
Quote: unJon



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


We're in the same ballpark, but I'm showing a different return.
Have you tried 22 tonight? I said 22.
ThatDonGuy
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July 6th, 2020 at 3:21:54 PM permalink
Quote: Gialmere

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




Returns:
First roll 12: 1/36 x 36/36 x 12 = 432 / 1296
First roll 11: 2/36 x 36/36 x 11 = 792 / 1296
First roll 10: 3/36 x 36/36 x 10 = 1080 / 1296
First roll 9: 4/36 x 36/36 x 9 = 1296 / 1296
First roll 8: 5/36 x 36/36 x 8 = 1440 / 1296
First roll 7: 6/36 x 36/36 x 7 = 1512 / 1296
First roll 2-6, and:
Second roll 12: 15/36 x 1/36 x 11 = 165 / 1296
Second roll 11: 15/36 x 2/36 x 10 = 300 / 1296
Second roll 10: 15/36 x 3/36 x 9 = 405 / 1296
Second roll 9: 15/36 x 4/36 x 8 = 480 / 1296
Second roll 8: 15/36 x 5/36 x 7 = 525 / 1296
Second roll 7: 15/36 x 6/36 x 6 = 540 / 1296
Second roll 6: 15/36 x 5/36 x 5 = 375 / 1296
Second roll 5: 15/36 x 4/36 x 4 = 240 / 1296
Second roll 4: 15/36 x 3/36 x 3 = 135 / 1296
Second roll 3: 15/36 x 2/36 x 2 = 60 / 1296
Second roll 6: 15/36 x 1/36 x 1 = 15 / 1296

Total Return = 9792 / 1296
Profit = 9792 / 1296 - 6 = 2016 / 1296 = 14 / 9
Note that the extra 1 paid for the second roll is taken into account by reducing the returns for those rolls by 1

charliepatrick
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July 6th, 2020 at 3:42:27 PM permalink
Quote: Gialmere

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

I get the same answer...
Phase 1 you've paid/handed over $6 so stop if you throw 7 thru 12 and take the profit. In Phase 2 (having thrown 2 thru 6) you pay a $1 extra (so with the extra $1 have now handed over a total of $7) and it is a fair bet as your average roll is 7.
Initial rollPermutationsEVContribution
2 - 61500
76+16
85+210
94+312
103+412
112+510
121+66
Total3656

So the return is 56/36.

Another way to look at it, is to see what is a fair price to play the game.
Pay: $7.556
Initial RollCombinationsEVContribution
2
1
-1.556
-1.556
3
2
-1.556
-3.111
4
3
-1.556
-4.667
5
4
-1.556
-6.222
6
5
-1.556
-7.778
7
6
-0.556
-3.333
8
5
0.444
2.222
9
4
1.444
5.778
10
3
2.444
7.333
11
2
3.444
6.889
12
1
4.444
4.444
Total
36
0.000
0.000


charliepatrick
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July 6th, 2020 at 4:10:49 PM permalink
btw If people agree with the above, try the maths for paying £8 and you're allowed to reroll for free. I think there's a new game here!
I get a House Edge of 1/36, so this could be the next internet game revolution!

Game is as follows, you pay $Y and roll two dice.
You can take the sum $T or pay $X to re-roll.
What is the return for various X and Y?
Pay to reroll: $0
Initial Bet: $8
Initial RollCombinationsEVContribution
2
1
-1
-1
3
2
-1
-2
4
3
-1
-3
5
4
-1
-4
6
5
-1
-5
7
6
-1
-6
8
5
0
0
9
4
1
4
10
3
2
6
11
2
3
6
12
1
4
4
Total
36
House Edge: -2.778%
-1.000

gordonm888
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July 6th, 2020 at 5:06:52 PM permalink
This thread has kinda been dominated by two puzzle solvers. They are certainly two of our top math guys, no doubt. But it is also partly due to schedule -one guy is in Europe at least 4 time zones ahead of everyone else. Anyway, I have very rarely seen a problem on this thread before it has already been solved by Don or Charlie. I am starting to realize why Wizard created his "beer club" and asks former winners to delay for a time.

I'm still enjoying the thread. Don and Charlie are really awesome at this and Thanks to Gialmere! for doing this.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Gialmere
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July 6th, 2020 at 9:43:18 PM permalink
This particular puzzle is by mathematician Leslie Green. Here is his solve for you guys to chew over...

Have you tried 22 tonight? I said 22.
charliepatrick
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July 6th, 2020 at 11:02:43 PM permalink
Quote: Gialmere

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

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

Gialmere
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July 6th, 2020 at 11:22:40 PM permalink
My bad. I was doing them on the fly at work with my cell.
Have you tried 22 tonight? I said 22.
unJon
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July 7th, 2020 at 6:54:07 AM permalink
Quote: charliepatrick

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

...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 RollCombinationsEVContribution
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



Agreed.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere
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July 7th, 2020 at 7:08:20 AM permalink


A year later you return to the same church casino night and see the same two sweet old ladies. They're now running a new dice game of their design.

A player bets $1 on a number from 1 through 6 and gets a single roll of two standard fair dice. If the chosen number comes up on one die, the wager pays 2 to 1. If the number comes up on both dice, the wager is payed 5 to 1. You lose if the number does not appear on either die.

What is the house edge for this game?
Have you tried 22 tonight? I said 22.
charliepatrick
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July 7th, 2020 at 7:49:43 AM permalink
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).
ThatDonGuy
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July 7th, 2020 at 10:16:43 AM permalink
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).


Use regular dice instead of the special ones and you've just described Chuck-A-Luck, which is also an available bet in Sic Bo. For some reason, American Sic Bo pays 3-1 for three of a kind, whereas Australian Sic Bo pays something like 14-1. Anything less than 20-1 is still an advantage for the house.
ksdjdj
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July 7th, 2020 at 6:55:41 PM permalink
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?


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"), ; )
Last edited by: ksdjdj on Jul 7, 2020
Gialmere
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July 8th, 2020 at 9:28:21 AM permalink
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 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"), ; )


Correct!
------------------------------------

Today at the bank, an old lady asked me to help check her balance.

So I pushed her over.
Have you tried 22 tonight? I said 22.
Gialmere
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July 8th, 2020 at 9:52:25 AM permalink


L3 and L4 are two parallel lines in a plane.

If L3 has 3 points equally spaced along its length, and if L4 has 4 points also equally spaced along its length, how many different triangles can be formed by connecting the points on the two parallel lines?

A triangle must be formed using 2 points on one line and 1 point on the other.
Have you tried 22 tonight? I said 22.
Ace2
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July 8th, 2020 at 10:14:08 AM permalink
Thirty
Itís all about making that GTA
unJon
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July 8th, 2020 at 10:45:28 AM permalink
Quote: Ace2

Thirty



Agree subject to the plane being Euclidean.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
IAchance5
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July 8th, 2020 at 11:00:44 AM permalink
Quote: Ace2

Thirty



I agree....not sure why the "points equally spaced on line" matters or just added for fluff....my only question would be if you can't pass through a 3rd point on the same line or not.

Thinking about this with regards to various combinations.....basically picking 2 points from one line and 1 point from the other.....

(4C2 * 3C1) + (4C1 * 3C2) = (6*3) + (4*3) = 18+12 = 30 triangles
ThatDonGuy
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July 8th, 2020 at 11:25:23 AM permalink
Quote: unJon

Agree subject to the plane being Euclidean.


Agree, if all three vertices have to be among the seven points, as opposed to, say, one side contains a point on each line, but in that case, the answer depends on exactly where the points are located.
Gialmere
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July 8th, 2020 at 11:51:17 AM permalink
Quote: Ace2

Thirty


Correct!

30 it is.

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

Why do math textbooks only ever give you one angle in a triangle?

Just cos.
Have you tried 22 tonight? I said 22.
Gialmere
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July 8th, 2020 at 1:01:53 PM permalink

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?
Have you tried 22 tonight? I said 22.
Ace2
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July 8th, 2020 at 1:25:44 PM permalink
75%

Iím sort of shooting from the hip, though
Itís all about making that GTA
IAchance5
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July 8th, 2020 at 2:04:55 PM permalink
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?




Taking a shot.....my aim could be way off though.....

In order for a cowboy to die, it must have been shot in all places (i.e. have all these probabilities together).....

The probably that a random cowboy died is .9*.85*.8*.75= .46.....but, that might not be the lowest percentage of cowboys dying.....

If throw everything to one side and assume then that of all of the ones that got shot in the head (75%) died, then we can take away the remaining 25% from the random probability......which would leave 46% - 25% = 21%
ThatDonGuy
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July 8th, 2020 at 2:22:03 PM permalink

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.

ChesterDog
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July 8th, 2020 at 2:22:43 PM permalink
Quote: Gialmere




6 out of 20?
30%
charliepatrick
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July 8th, 2020 at 2:47:22 PM permalink
Quote: ThatDonGuy

30%

I agree using similar logic...
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%)
Wizard
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July 8th, 2020 at 3:41:13 PM permalink

I'm getting 30%.



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%

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
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July 8th, 2020 at 4:35:03 PM permalink
Quote: ThatDonGuy


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

6 out of 20?
30%


Quote: charliepatrick

Quote: ThatDonGuy

30%

I agree using similar logic...
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


I'm getting 30%.



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%


Correct!

Today's two easy math problems have been brought to you by the number 30.

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

A cowboy asked me if I could help him round up 18 cattle.

I said, ďSure, thatís 20 cattle.Ē
Have you tried 22 tonight? I said 22.
Wizard
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July 8th, 2020 at 9:02:15 PM permalink
Q: What is the general formula, given any four probabilities of injury?


The general formula seems to be the sum of the four probabilities minus 3. In the case of the OP, .9 + .85 + .8 + .75 - 3 = 3.3 - 3 = 0.3. I suppose if the sum were less than 3, then you could avoid anybody being shot in all four places.

This was arrived at doing a bunch of algebra, but I can't immediately find a common sense answer why this is true.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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July 8th, 2020 at 11:56:05 PM permalink
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.

Let's assume the ailments are A, B, C and D. The logic below can be extended to more ailments.

Assume each cowboy has four slots available for ailments. If you fill all four slots the cowboy is killed, else the cowboy lives.
Step 1 give Ailment A to cowboy 1 to n(A).
AAAAAAAAAAAAAAAAAA--
--------------------
--------------------
--------------------
Step 2 give Ailment B to the next cowboy and wrap.
AAAAAAAAAAAAAAAAAABB
BBBBBBBBBBBBBBB-----
--------------------
--------------------
Steps 3 and 4, continue.
AAAAAAAAAAAAAAAAAABB
BBBBBBBBBBBBBBBCCCCC
CCCCCCCCCCCDDDDDDDDD
DDDDDD--------------
You can see that the number of slots filled correspond to the total of the probabilities, in this case n(A)+n(B)+n(C)+n(D). The number of slots filled on the fourth row is therefore the sum of the probabilities minus 3.

Logically by extension the probabilities for N ailments is the sum of the probabilities - (N-1).

Another way of looking at it is (1-n(A) ) are saved by not being infected by A. Similarly for the others. So 4-n(A)-n(B)-n(C)-n(D) are saved. Hence 1-(4-n(A)-n(B)-n(C)-n(D)) die = n(A)-n(B)-n(C)-n(D)-3.
Wizard
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July 9th, 2020 at 7:42:34 AM permalink
Good work Charlie!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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July 9th, 2020 at 9:33:06 AM permalink
I don't know how "easy" this is, but here's one:

There are N cities.
Each pair of cities is connected by a highway that is numbered with a different integer from 1 to 999. For example, if A, B, and C are cities, the A-B highway might be Highway 1, the A-C highway might be Highway 626, and the B-C highway might be Highway 777,
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?

Bonus problem:
Change the condition to:
The highway numbers range from 100 to 999
Given any three cities, of the three highways that connect them to each other, at least one does not have a number between 100 and 399, at least one does not have a number between 400 and 699, and at least one does not have a number between 700 and 999.
Note that, for example, 123, 235, and 678 can be the highways that connect three cities (but 123, 168, and 235 cannot, as all three would be between 100 and 399).
Last edited by: ThatDonGuy on Jul 9, 2020
charliepatrick
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July 9th, 2020 at 12:52:56 PM permalink
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?...

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.
Gialmere
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July 10th, 2020 at 9:13:29 AM permalink


You fall asleep and have a bizarre but very realistic nightmare...

In it, you find yourself standing at the centre of a large, circular field of radius R. The field has a low wire fence around it. Attached to the wire fence (and restricted to running around the perimeter) is a large, rabid, sharp-fanged, hungry dog. You can run at speed v, while the dog can run four times as fast.

You sense that if you can just make it past the demon dog you'll wake up and end the terror. What is your running strategy to escape the field?
Have you tried 22 tonight? I said 22.
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July 10th, 2020 at 1:40:02 PM permalink

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.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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July 10th, 2020 at 4:29:01 PM permalink
Quote: charliepatrick

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.


Correct.


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.

Gialmere
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July 10th, 2020 at 5:38:52 PM permalink
Quote: Wizard


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.


Correct!

Yes. To begin your escape you essentially need to run in an ever widening spiral that keeps the pursuing hell hound on the opposite side of the field.

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

Have you tried 22 tonight? I said 22.
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July 10th, 2020 at 9:02:37 PM permalink
Quote: Gialmere


Correct!



Thanks! To be honest, I saw that one before, a long time ago. Funny how it works without specifying the radius of the circle.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
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July 11th, 2020 at 11:51:30 AM permalink


Once again you go to casino night at the church, and once again the two sweet old ladies are there with a new game they designed. This time it's a card game using a well shuffled standard deck.

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. Cards are not returned to the deck after being drawn. The game ends when you either decide to stop, or when all 52 cards are played.

What is the optimal stopping rule in terms of maximizing expected payoff? Also, what is the expected payoff following this optimal rule?


A complete solve of this problem would be too time consuming and tedious for most people. It also moves well beyond the "easy" of this thread's title. Consider doing only a partial solve or just skipping it altogether.
Have you tried 22 tonight? I said 22.
Ace2
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July 11th, 2020 at 2:45:43 PM permalink

You always take a card if youíre in a negative or neutral position

If youíre in a positive position, let R= red cards remaining in deck and T = total cards remaining in deck.

If combin(2R,R) / combin(T,R) >= 50%, take another card. Otherwise quit

Expected payout 26/27ths of a dollar or about 96 cents

Last edited by: Ace2 on Jul 11, 2020
Itís all about making that GTA
charliepatrick
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July 11th, 2020 at 3:55:44 PM permalink
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?

I remember this being asked by the wizard but I haven't cheated by looking it up. Also I haven't checked my figures with a simulation.
I created a spreadsheet looking at all the cases based on cards left and red cards left. If you stop then you can calculate how much profit or loss you have. If you keep going then at each later stage you could stop or keep going. This is similar to working out how to play BlackJack, you know you'll (usually) stand on 17, decide whether to hit 16, decide whether to hit 15 knowing what you would do if on 16, etc.

Obviously you can always ensure an even return by keeping going until the end, so you always hit if you are even or losing. The decision is how much up you need to be given how many cards have gone (which also determines how many reds you have found and how many are left). Early in the game you keep going, but later on you settle for a smaller profit. Thus the profit required before you stop gradually goes down as more cards have gone.

Cards drawnStand when profit is...
6-10
+6 or better
11-23
+5 or better
24-34
+4 or better
35-43
+3 or better
44-50
+2 or better
51
+1

Expected profit = 2.624
Last edited by: charliepatrick on Jul 12, 2020
charliepatrick
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July 12th, 2020 at 2:07:38 AM permalink
I've knocked up a simulation and get the same answer.
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
Last edited by: charliepatrick on Jul 12, 2020
ThatDonGuy
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July 12th, 2020 at 7:34:39 AM permalink
Card Craps
There is another thread that shows a California craps table with "card craps."
There are six cards, numbered 1 to 6; they are shuffled (I think they are shuffled when there is a new shooter) and dealt into six boxes, also numbered 1 to 6. When the dice are rolled, each die is "converted" into the value of the card in that numbered box. For example, if the cards are dealt in order (for boxes 1-6) of 3, 1, 4, 2, 5, 6, and the dice are rolled 1-3, the 1 becomes a 3 (as there is a 3 in box #1) and the 3 becomes a 4 (as there is a 4 in box #3), so the roll is a 3-4.

Two questions:

#1 - The cards are shuffled and dealt, and two dice rolled. What is the probability that the numbers on the dice are the same as the numbers on the two cards they represent? Note they do not have to be in order; for example, if the card in position 1 is a 3, and the card in position 3 is a 1, then a roll of 1-3 is a "match" as the result is also a 1 and a 3.

#2 - What is the expected number of cards in a deal that match their assigned numbers?
Ace2
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July 12th, 2020 at 9:18:07 AM permalink
1) 1/12 chance they match.

2) EV = 1 card matches
Itís all about making that GTA
charliepatrick
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July 12th, 2020 at 10:52:33 AM permalink
Agree with the above...
There are 720 ways to organise the numbers 1 thru 6. There are 120 ways to organise the numbers 2 thru 6 (as there are five of them). So the number of ways to have 1xxxxx is 120. Thus the chances of 1 being in front is 1/6.

Similar logic applies to 12xxxx (as there are 24 ways to organise 3 thru 6) and similarly for 21xxxx.

If you roll a double (say 1-1) then Pr(throwing a double) = 1/6, Pr(a match) = 120/720.
If you roll easy (say 1-2) then Pr(throwing an easy) = 5/6. Pr(correct match) = 24/720. Pr(reverse match) = 24/720.

Pr(success) = 1/6*120/720 + 5/6*(24+24)/720 = (120+240)/6/720 = 1/12.

Consider all 720 permutations are lined up on top of each other, each permutation consists of six boxes. In total there are 720*6 boxes.

Let's count how many boxes have the correct number.

Consider column #1: 120 of them have #1 in the correct position (as per above), that's 120 boxes with the correct number.

Similarly for the other columns matching up with 2 thru 6.

So there's a total of 6*120 correct boxes = 720.

That's an average of 1 per permutation.
Gialmere
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July 12th, 2020 at 12:42:52 PM permalink
Quote: charliepatrick

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


Correct!



Interestingly, even if you win the first five cards in a row, you're still better off continuing play.
------------------------------

I've developed a simplified version of the popular card game "Go Fish"...

It's called, "No".
Have you tried 22 tonight? I said 22.
Gialmere
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July 13th, 2020 at 10:53:45 AM permalink


You are a logistical mathematician in the year 1890. Sir Albert Falls invites you to lunch at the famed Explorers Club to ask your advice for his next expedition...

"To reach the lost ruins, in the middle of a barren plateau, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for six days and the farthest we can travel in one day is 25 miles. Also, the city is 125 miles from the starting point."

"What I am trying to figure out is the fewest number of persons, including myself, that I will need in our group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies. How many persons (including myself) will I need to accomplish this mission?"
Have you tried 22 tonight? I said 22.
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