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gordonm888
gordonm888 
Joined: Feb 18, 2015
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January 18th, 2021 at 1:41:07 PM permalink
Tuesday Puzzle (one day early)
The casino has a new dice game.

The prize to be won in each round is 11 units. Each roll of a die costs the player 4 units.

The player rolls one die at a time.
- if the sum of the dice equals exactly 4 OR "10 or higher" (>9), the player wins 11 units and the game stops. Reset to another round.
- if the sum of the dice equals 1-3 or 5-9 then player may roll an additional die for another 4 units. The sum of the dice is now recalculated to include all of the dice that player has rolled in that round.

As long as the player keeps rolling dice (as he should) he/she will always win 11 units each round. The unknown in any round is how much he will spend on dice to win the 11 units.

What is the house edge of this game?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
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January 18th, 2021 at 2:59:20 PM permalink
Quote: gordonm888

Tuesday Puzzle (one day early)
The casino has a new dice game.

The prize to be won in each round is 11 units. Each roll of a die costs the player 4 units.

The player rolls one die at a time.
- if the sum of the dice equals exactly 4 OR "10 or higher" (>9), the player wins 11 units and the game stops. Reset to another round.
- if the sum of the dice equals 1-3 or 5-9 then player may roll an additional die for another 4 units. The sum of the dice is now recalculated to include all of the dice that player has rolled in that round.

As long as the player keeps rolling dice (as he should) he/she will always win 11 units each round. The unknown in any round is how much he will spend on dice to win the 11 units.

What is the house edge of this game?



Let E(n) be the expected number of rolls needed to win with a score of n.
Determine the expected number of rolls
E(4) = E(10) = E(11) = E(12) = E(13) = E(14) = E(15) = 0
E(9) = 1
E(8) = 1 + 1/6 E(9)
= 7/6
E(7) = 1 + 1/6 (E(8) + E(9))
= 1 + 1/6 x 13/6
= 49/36
E(6) = 1 + 1/6 (E(7) + E(8) + E(9))
= 1 + 1/6 x 127/36
= 343/216
E(5) = 1 + 1/6 (E(6) + E(7) + E(8) + E(9))
= 1 + 1/6 x 1105/216
= 2401/1296
E(3) = 1 + 1/6 (E(5) + E(6) + E(7) + E(8) + E(9))
= 1 + 1/6 (2401 + 2058 + 1764 + 1512 + 1296 )/1296
= 16,807/7776
E(2) = 1 + 1/6 (E(3) + E(5) + E(6) + E(7) + E(8))
= 1 + 1/6 (16,807 + 14,406 + 12,348 + 10,584 + 9072)/7776
= 109,873 / 46,656
E(1) = 1 + 1/6 (E(2) + E(3) + E(5) + E(6) + E(7))
= 1 + 1/6 (109,873 + 100,842 + 86,436 + 74,088 + 63,504)/46,656
= 714,649 / 279,936
E(0) = 1 + 1/6 (E(1) + E(2) + E(3) + E(5) + E(6))
= 1 + 1/6 (714,649 + 659,238 + 605,052 + 518,616 + 444,528)/279,936
= 4,621,699 / 1,679,616
The expected wager = 4 x the number of rolls = 4,621,699 / 419,904
The expected return is 11
The house edge = (expected wager - expected return) / expected wager
= 2755 / 419,904 = 0.6561%

gordonm888
gordonm888 
Joined: Feb 18, 2015
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January 18th, 2021 at 4:49:43 PM permalink
Quote: ThatDonGuy

Quote: gordonm888

Tuesday Puzzle (one day early)
The casino has a new dice game.

The prize to be won in each round is 11 units. Each roll of a die costs the player 4 units.

The player rolls one die at a time.
- if the sum of the dice equals exactly 4 OR "10 or higher" (>9), the player wins 11 units and the game stops. Reset to another round.
- if the sum of the dice equals 1-3 or 5-9 then player may roll an additional die for another 4 units. The sum of the dice is now recalculated to include all of the dice that player has rolled in that round.

As long as the player keeps rolling dice (as he should) he/she will always win 11 units each round. The unknown in any round is how much he will spend on dice to win the 11 units.

What is the house edge of this game?




Let E(n) be the expected number of rolls needed to win with a score of n.
Determine the expected number of rolls
E(4) = E(10) = E(11) = E(12) = E(13) = E(14) = E(15) = 0
E(9) = 1
E(8) = 1 + 1/6 E(9)
= 7/6
E(7) = 1 + 1/6 (E(8) + E(9))
= 1 + 1/6 x 13/6
= 49/36
E(6) = 1 + 1/6 (E(7) + E(8) + E(9))
= 1 + 1/6 x 127/36
= 343/216
E(5) = 1 + 1/6 (E(6) + E(7) + E(8) + E(9))
= 1 + 1/6 x 1105/216
= 2401/1296
E(3) = 1 + 1/6 (E(5) + E(6) + E(7) + E(8) + E(9))
= 1 + 1/6 (2401 + 2058 + 1764 + 1512 + 1296 )/1296
= 16,807/7776
E(2) = 1 + 1/6 (E(3) + E(5) + E(6) + E(7) + E(8))
= 1 + 1/6 (16,807 + 14,406 + 12,348 + 10,584 + 9072)/7776
= 109,873 / 46,656
E(1) = 1 + 1/6 (E(2) + E(3) + E(5) + E(6) + E(7))
= 1 + 1/6 (109,873 + 100,842 + 86,436 + 74,088 + 63,504)/46,656
= 714,649 / 279,936
E(0) = 1 + 1/6 (E(1) + E(2) + E(3) + E(5) + E(6))
= 1 + 1/6 (714,649 + 659,238 + 605,052 + 518,616 + 444,528)/279,936
= 4,621,699 / 1,679,616
The expected wager = 4 x the number of rolls = 4,621,699 / 419,904
The expected return is 11
The house edge = (expected wager - expected return) / expected wager
= 2755 / 419,904 = 0.6561%


CORRECT


I get that player will invest 11.006561 units to win 11 units, so exact agreement.

Number of dice used
___Probability___
1
0.16666667
2
0.22222222
3
0.36574074
4
0.19135803
5
0.04681070
6
0.00660151
7
0.00057156
8
0.00002798
9
0.00000060


There are 1221 distinct outcomes for player (of widely varying probability) ranging from 4 to 111211116

Let me praise ThatDonGuy for extremely clear answers that are easy to understand and check! I am learning some good math techniques by reading his answers on this thread.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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January 18th, 2021 at 5:21:22 PM permalink
I used EVs at different scores, so looked for the EV at 0. I get a slightly different answer of -2785/419904 = -0.006632. However you now have to divide that by 4 since that is the overall cost for a 4 unit wager. So House Edge = 0.1658%.

Method 1 is to use the EVs of 15 to 10 being 11.
At each stage the EV of a number is the average of the next six numbers less the cost of 4 to make the additional roll.
Thus the EV at 9 is (EV(10)+EV(11)+EV(12)+EV(13)+EV(14)+EV(15))/6-4 = 7.
Repeat and rinse except when at 4 the EV(4) = 11.

Method 2 keeps the numbers
EV(9) = 7 (as above)
EV(8) = (11+11+11+11+11+7)/6-4 = 38/6
EV(7) = 200/36
EV(6) = 1004/216
EV(5) = 4652/1296
EV(4) = 11
EV(3) = 18308/7776
EV(2) = 73724/46656
EV(1) = 220580/279936
EV(0) = -11140/1679616 = -2785/419904 = -0.00663247.
Edit: This answer crossed with your reply above.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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January 18th, 2021 at 5:36:43 PM permalink
Quote: ThatDonGuy

...E(0) = 1 + 1/6 (E(1) + E(2) + E(3) + E(5) + E(6))
= 1 + 1/6 (714,649 + 659,238 + 605,052 + 518,616 + 444,528)/279,936
= 4,621,699 / 1,679,616

Looking at your figures I agree with all of them until the very last stage. I get 714639...444528 adding up to 2942113, but then you need to add 1679616 (for the "1") to get 4621729 rather than 4621699. This would then be the same as my figure.
charliepatrick
charliepatrick
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January 18th, 2021 at 6:41:12 PM permalink
Quote: gordonm888

...There are 1221 distinct outcomes for player (of widely varying probability) ranging from 4 to 111211116...

I agree with your probabilities but...
Contribution is $4*no of rolls*probability.
No of rollsTimesProbabilityContribution
1
1
.166 666 667
0.666 666 667
2
8
.222 222 222
1.777 777 778
3
79
.365 740 741
4.388 888 889
4
248
.191 358 025
3.061 728 395
5
364
.046 810 700
0.936 213 992
6
308
.006 601 509
0.158 436 214
7
160
.000 571 559
0.016 003 658
8
47
.000 027 983
0.000 895 443
9
6
.000 000 595
0.000 021 433
Totals
1.000 000 000
11.006 632 468
Gialmere
Gialmere
Joined: Nov 26, 2018
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January 23rd, 2021 at 11:26:12 AM permalink


Fill in the boxes of the grid below with the other numbers from 1 to 16 by making only the moves of a knight in chess.

The numbers follow each other in numerical order.

Not all of the boxes in the grid will be used. The answer is unique.

Have you tried 22 tonight? I said 22.
chevy
chevy
Joined: Apr 15, 2011
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Thanks for this post from:
Gialmere
January 23rd, 2021 at 11:52:26 AM permalink



--, 14, 9, 4, --
2, 7, 12, 15, 10
13, 16, 3, 8, 5
--, 1, 6, 11, --


other than it was the first walk through I tried, I have not not tried all the other solutions to see at what stage they break down.

ThatDonGuy
ThatDonGuy 
Joined: Jun 22, 2011
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Thanks for this post from:
Gialmere
January 23rd, 2021 at 12:16:02 PM permalink
Quote: chevy




--, 14, 9, 4, --
2, 7, 12, 15, 10
13, 16, 3, 8, 5
--, 1, 6, 11, --


other than it was the first walk through I tried, I have not not tried all the other solutions to see at what stage they break down.



Label the squares:
A, 14, 9, 4, B
C, D, E, F, G
H, 16, J, K, M
N, 1, 6, 11, P
(note I, L, and O were not used as they may be confused with numbers)

"Touching" in this case means that the two squares are a knight's move apart
The only empty square touching both 14 and 16 is F, so F = 15
The only empty square touching both 9 and 11 is G, so G = 10
C cannot be 8, as there would be no empty square touching 6 and 8 in which to put 7;
the only other empty square touching 9 is K, so K = 8
The only still-empty sqaure touching both 6 and 8 is D, so D = 7
The only still-empty square touching both 4 and 6 is M, so M = 5
E cannot be 2, as there would be no empty square touching 2 and 4 in which to put 3;
the only other empty square touching 1 is C, so C is 2
The only still-empty square touching both 2 and 4 is J, so J = 3
This leaves 12 and 13; E touches 11, so E = 12, and H touches 12 and 14, so H = 13

The solution is unique

Gialmere
Gialmere
Joined: Nov 26, 2018
  • Threads: 41
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January 23rd, 2021 at 12:31:30 PM permalink
Quote: chevy




--, 14, 9, 4, --
2, 7, 12, 15, 10
13, 16, 3, 8, 5
--, 1, 6, 11, --


other than it was the first walk through I tried, I have not not tried all the other solutions to see at what stage they break down.


Quote: ThatDonGuy


Label the squares:
A, 14, 9, 4, B
C, D, E, F, G
H, 16, J, K, M
N, 1, 6, 11, P
(note I, L, and O were not used as they may be confused with numbers)

"Touching" in this case means that the two squares are a knight's move apart
The only empty square touching both 14 and 16 is F, so F = 15
The only empty square touching both 9 and 11 is G, so G = 10
C cannot be 8, as there would be no empty square touching 6 and 8 in which to put 7;
the only other empty square touching 9 is K, so K = 8
The only still-empty sqaure touching both 6 and 8 is D, so D = 7
The only still-empty square touching both 4 and 6 is M, so M = 5
E cannot be 2, as there would be no empty square touching 2 and 4 in which to put 3;
the only other empty square touching 1 is C, so C is 2
The only still-empty square touching both 2 and 4 is J, so J = 3
This leaves 12 and 13; E touches 11, so E = 12, and H touches 12 and 14, so H = 13

The solution is unique


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

Have you tried 22 tonight? I said 22.

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