## Poll

21 votes (46.66%) | |||

14 votes (31.11%) | |||

6 votes (13.33%) | |||

2 votes (4.44%) | |||

12 votes (26.66%) | |||

3 votes (6.66%) | |||

6 votes (13.33%) | |||

5 votes (11.11%) | |||

12 votes (26.66%) | |||

10 votes (22.22%) |

**45 members have voted**

Quote:Ace2Fifteen

Correct!

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

Q: Why should you cook kale in coconut oil?

A: Makes it easier to slide it right into the trash.

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:GialmereToo 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

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.

Quote:GialmereWe'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

I get the same answer...Quote:GialmereWe're in the same ballpark, but I'm showing a different return.

Initial roll | Permutations | EV | Contribution |
---|---|---|---|

2 - 6 | 15 | 0 | 0 |

7 | 6 | +1 | 6 |

8 | 5 | +2 | 10 |

9 | 4 | +3 | 12 |

10 | 3 | +4 | 12 |

11 | 2 | +5 | 10 |

12 | 1 | +6 | 6 |

Total | 36 | 56 |

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 Roll | Combinations | EV | Contribution |
---|---|---|---|

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 |

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 Roll | Combinations | EV | Contribution |
---|---|---|---|

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 |

I'm still enjoying the thread. Don and Charlie are really awesome at this and Thanks to Gialmere! for doing this.

I think your answer allows for infinite rerolls for $1, whereas we had assumed you only got one reroll.Quote:GialmereThis particular puzzle is by mathematician Leslie Green. Here is his solve...

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:charliepatrickI 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: $6.000Initial 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

Agreed.

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?

Quote:charliepatrickWhat 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.

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?

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:charliepatrickWhat 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:ksdjdjthe 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.

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.

Quote:Ace2Thirty

Agree subject to the plane being Euclidean.

Quote:Ace2Thirty

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.

(4C2 * 3C1) + (4C1 * 3C2) = (6*3) + (4*3) = 18+12 = 30 triangles

Quote:unJonAgree 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.

Quote:Ace2Thirty

Correct!

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

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

Just cos.

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?

I’m sort of shooting from the hip, though

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.....

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%

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:Gialmere

30%

I agree using similar logic...Quote:ThatDonGuy30%

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%

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:ChesterDog6 out of 20?

30%

Quote:charliepatrickI agree using similar logic...Quote:ThatDonGuy30%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!

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

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

I said, “Sure, that’s 20 cattle.”

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.

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

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).

A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | - | - |

- | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

- | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

- | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | B | B |

B | B | B | B | B | B | B | B | B | B | B | B | B | B | B | - | - | - | - | - |

- | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

- | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | A | B | B |

B | B | B | B | B | B | B | B | B | B | B | B | B | B | B | C | C | C | C | C |

C | C | C | C | C | C | C | C | C | C | C | D | D | D | D | D | D | D | D | D |

D | D | D | D | D | D | - | - | - | - | - | - | - | - | - | - | - | - | - | - |

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.

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).

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?...

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

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.

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?

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.

Quote:charliepatrickI really liked this until a penny dropped.

It's possible to connect 5 cities as followsNow 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.AB O

AC O BC E

AD E BD E CD O

AE E BE O CE E DE O

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.

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!

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

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.

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?

You always take a card if you’re in a negative or neutral position

If you’re 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

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.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?

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 drawn | Stand 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

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

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?

2) EV = 1 card matches

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.

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.

Quote:charliepatrick(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 = 4.040Parms: 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

Correct!

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

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

It's called, "No".

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?"