The game uses four digits numbers "0000" thru "9999" and looks at the digits. If the number contains a "7" then it is a "Seven"; otherwise it contains four seperate digits (which may have duplicates).

Summary - it's like Craps, if your first roll has a "7" you win, if not remember all the digits you rolled, reroll until either you roll them all again, or you lose when seeing a "7".

(i) Each roll consists of picking a four digit number "0000" thru "9999".

(ii) Each roll that does not contain a "7" will look at the digits chosen rather than the number or total.

(ii) Any roll that does contain a "7" will only consider the "7", either winning on the come out, or losing afterwards.

(iii) On the first "Come Out" roll, if the digit "7" shows then the Player wins.

(iv) If the digit "7" does not show on the come out roll, then the different digits that did come up are remembered.

(v) If a number appeared multiple times (e.g. "1112") you only have to roll the "1" once and "2" once.

(vi) Thus sometimes you will have four numbers to roll, and sometimes fewer.

(vii) The aim is then to keep rolling until all those numbers, in any order, show up before a "7" shows.

(viii) Subject to the above, if you roll more than one number required they all count.

(viii) The player wins if all numbers are re-rolled, and loses if a "7" is rolled beforehand.

The casino is feeling generous, as it wants the House Edge to be similar to Craps, and pays out Evens for all wins, except it pays 10 to 1 if three "7"s or 20 to 1 if four "7"s appear on the Come Out roll.

Mathematically which one would you play.

btw if you did play this, rather than the bonus I think you would prefer to be able to take single "Odds" at favourite rates which would be (N+1) to 1, where N is how many numbers you need to get?

However I also rolled a 7. Is the answer that 1279 is a player win but 1729 is a player loss?

Suppose your comeout roll is 1234.

On your next roll, you roll 5613.

Do you set the 1 and 3 aside, and roll only the 5 and 6, or do you roll all four dice again?

Suppose that, instead of 5613, your second roll was 2928. Which dice do you reroll?

I think we should select at random Tree(3) digits, list them in order, and you win on the come out roll if you can find more than G64 (Graham's number) of subsequences that are the digits of G64. Otherwise, proceed as in your game.Quote:charliepatrick*Very creative idea*

#AprilFools

BTW, I would guess the probability of winning this game is pretty close to 1.

As per rule (ii) (sorry I had two of those!) any roll containing a "7", even if it looks as if the 1 and 2 appeared beforehand, will only consider the "7", either winning on the come out, or losing afterwards. If you had rolled 1249 then you would have won.Quote:oboylereWhat happens when you roll a 1279 after a come out roll of 1112 (as described in v)? Do you win or lose? I rolled the 1 and 2 required to win.

However I also rolled a 7. Is the answer that 1279 is a player win but 1729 is a player loss?

I'd love it to be the other way round, or perhaps a standoff, but then the maths wouldn't work!

Actually it's fairly close to Craps. I even ran a simulation...Quote:teliot...BTW, I would guess the probability of winning this game is pretty close to 1.

Outcome | Count | Pays | Contribution |
---|---|---|---|

Natural | 34 020 468 | 1 | 34 020 468 |

Win 777 | 360 635 | 10 | 3 606 350 |

Win 7777 | 10 050 | 20 | 201 000 |

Win 1 | 37 844 | 1 | 37 844 |

Win 2 | 1 450 257 | 1 | 1 450 257 |

Win 3 | 6 506 410 | 1 | 6 506 410 |

Win 4 | 5 195 541 | 1 | 5 195 541 |

Lose | 52 418 795 | - 1 | -52 418 795 |

-1 400 925 |

your idea is quite brilliant, my comment was about my lame extension of your game to TREE(3) digits.Quote:charliepatrickActually it's fairly close to Craps. I even ran a simulation...

Outcome Count Pays Contribution Natural 34 020 468 1 34 020 468Win 777 360 635 10 3 606 350Win 7777 10 050 20 201 000Win 1 37 844 1 37 844Win 2 1 450 257 1 1 450 257Win 3 6 506 410 1 6 506 410Win 4 5 195 541 1 5 195 541Lose 52 418 795 - 1 -52 418 795 -1 400 925

Quote:charliepatrick(i) Each roll consists of picking a four digit number "0000" thru "9999".

(ii) Each roll that does not contain a "7" will look at the digits chosen rather than the number or total.

(ii) Any roll that does contain a "7" will only consider the "7", either winning on the come out, or losing afterwards.

(iii) On the first "Come Out" roll, if the digit "7" shows then the Player wins.

(iv) If the digit "7" does not show on the come out roll, then the different digits that did come up are remembered.

(v) If a number appeared multiple times (e.g. "1112") you only have to roll the "1" once and "2" once.

(vi) Thus sometimes you will have four numbers to roll, and sometimes fewer.

(vii) The aim is then to keep rolling until all those numbers, in any order, show up before a "7" shows.

(viii) Subject to the above, if you roll more than one number required they all count.

(viii) The player wins if all numbers are re-rolled, and loses if a "7" is rolled beforehand.

I think I understand the rules now, but could you just confirm this? I inferred from (v) that you didn't roll all four dice each time.

Anyway, here's how I understand it:

If the comeout roll has any 7s in it, you win.

Otherwise, make a note of the distinct digits in the comeout roll.

Roll all four dice, as many times as necessary, until a roll has:

(a) one or more 7s, which is a loss;

(b) no 7s, but each of the digits in the comeout roll at least once, which is a win.

For example, if the comeout roll is 1234, you have to roll a 1234 (in any order - e.g. 4123 and 3124 also count) before any number with a 7 to win.

If the comeout roll is 1699, you have to roll any number that includes (at least) a 1, and 6, and a 9 before any number with a 7 to win.

I have a feeling that last part is wrong - otherwise, there is a 30.24% chance that your comeout roll has four different digits, none of which are 7s, in which case, each subsequent roll has a 24/10,000 chance of winning and a 3439/10,000 chance of losing.

If, say, the comeout roll is 1234, and my next two rolls are 1303 and 6524, do I win since I have rolled each needed digit at least once in the two rolls combined?

The original idea is you pick a four-digit number, so all digits appear at the same time. Obviously there are 10-sided dice, so you could use those but roll four at the same time. (If you're lucky enough to have one of Stephen's dice then you have 00 0 and 1 - 18! - I haven't done the maths. If you use these where any 0/00 would be the loser; although the dice do have 7 and 11! Perhaps there's an idea there.)

e.g. 1234 (you now need to roll 1, 2, 3 and 4 before any 7 shows)

Next roll 1256 - you now only need 3 and 4

Next roll 2839 - you now only need the 4

(a) Next roll 3945 - you win

or (b) Next roll 3947 - you lose (as the 7 appeared as one of the digits in the four digit number)

Quote:charliepatrickActually it's fairly close to Craps. I even ran a simulation...

Outcome Count Pays Contribution Natural 34 020 468 1 34 020 468Win 777 360 635 10 3 606 350Win 7777 10 050 20 201 000Win 1 37 844 1 37 844Win 2 1 450 257 1 1 450 257Win 3 6 506 410 1 6 506 410Win 4 5 195 541 1 5 195 541Lose 52 418 795 - 1 -52 418 795 -1 400 925

Here are some exact numbers:

Win 1 = 9/10,000 x 2,465/5,904

Win 2 = 504/10,000 x 2,150,405/7,477,416

Win 3 = 3024/10,000 x 34,314,535/159,518,208

Win 4 = 3024/10,000 x 64,236,744,953/373,870,800,000

Win Probability = 49,410,914,885,427/103,853,000,000,000 (47.5777%)

House Edge = 734,506,164,573/51,926,500,000,000 (1.4145112%)

In 100 million plays (these are rounded):

Natural: 34,020,000

3 7s: 360,000

4 7s: 10,000

Win 1: 37,576

Win 2: 1,449,437

Win 3: 6,505,035

Win 4: 5,195,696

Lose: 52,422,256