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Most realistic strategy game?

Simple dice game. Can you find the best strategy?

  • You play with 3 ordinary 6-sided dice. After rolling the three of them, you have the option of rolling again 0,1,or 2 of them at will. The outcomes of the dice which are rerolled are discarded. The goal of a strategy is to maximize your expected total. In game A, you just add up the three final outcomes. In game B you add the top 2 and disregard the smallest one. What is the best strategy for game A and for game B? What are the average totals in each game with that strategy. This is inspired by http://in.answers.yahoo.com/question/index;_ylt=Ann19lqoMQujIGG4yq4ftZSRHQx.;_ylv=3?qid=20100824095202AAawnal&show=7#profile-info-3EyBkjlGaa

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    First, I need to clarify one point: "The outcomes of the dice which are rerolled are discarded." If this means the rerolled dice are not used at all then the strategy would be to never re-roll. Since you probably did not want a simple problem, I assume this is not what you meant. I am going on the assumption that your meant "The *original* outcomes of the dice which are rerolled are discarded". Since these are ordinary dice they are independent of each other and individual rolls of the same die are also independent. Also, the probability of any outcome of a single die is 1/6. The expected value of a single roll is the sum of the probability of each outcome times the value of that outcome, or 1/6(1+2+3+4+5+6) = 3.5. This means on average over many rolls I can expect an average of 3.5. The best strategy is to always re-roll any number 3 or less if you can because on average you will do better than a three in the future. If 2 or fewer dice on the first roll are 3 or lower re-roll all the 3 or lower dice. If all three dice are 3 or lower re-roll the lowest 2 dice. This strategy is for either game. The expected value for each game is the fun part of the problem. To find this, first let's look at the expected value of a single die. There are six final number events which I will call 1 through 6. There are also several intermediary events which I will define here: F1 = Get a 1 on the first roll and will not re-roll = triple 1's on first roll. P(F1) = 1/216 F2 = Get a 2 on the first roll and will not re-roll = Get a 2 and other dice are 2 or less = 2 and {11,12,21,22}. P(F2) = 1/6 * 4/36 = 4/216 F3 = Get a 3 on the first roll and will not re-roll = Get a 3 and other dice are 3 or less = 3 and {11,12,13,21,22,23,31,32,33}. P(F3) = 1/6 * 9/36 = 9/216 R = Get a 3 or less on the first roll and will re-roll. P(R) = 1/2 - P(F1) - P(F2) - P(F3) = (108 - 1 - 4 - 9)/216 = 94/216 F4 = Get a 4 on the first roll F5 = Get a 5 on the first roll F6 = Get a 6 on the first roll S1 = Get a 1 on the second roll S2 = Get a 2 on the second roll S3 = Get a 3 on the second roll S4 = Get a 4 on the second roll S5 = Get a 5 on the second roll S6 = Get a 6 on the second roll P(F4) = P(F5) = ... = P(S6) = 1/6 Probability of getting a 1 as the final number = P(1) P(1) = P(F1 or (R and S1)) = P(F1) + P(R)*P(S1) = 1/216 + (94/216)*(1/6) = 3/648 + 47/648 = 50/648 P(2) = P(F2 or (R and S2)) = P(F2) + P(R)*P(S2) = 4/216 + (94/216)*(1/6) = 12/648 + 47/648 = 59/648 P(3) = P(F3 or (R and S3)) = P(F3) + P(R)*P(S3) = 9/216 + (94/216)*(1/6) = 27/648 + 47/648 = 74/648 P(4) = P(F4 or (R and S4)) = P(F4) + P(R)*P(S3) = 1/6 + (94/216)*(1/6) = 108/648 + 47/648 = 155/648 P(5) = P(6) = 155/648 NOTE: sum of P(1) to P(6) = 1 as it should be. I skipped some explanations as to why the or probabilities are straight addition and the and are straight multiplication, but they are due to mutually exclusive events in the or case and independent events in the and case. Now finally, the expected value of a single roll = (1*50 + 2*59 + 3*74 + 4*155 + 5*155 + 6*155)/648 = 2715/648 = 4 & 41/216 Game A: 12 & 41/72 Game B looks a little more complicated because the value of a die at the end can now be 0. I do not have the time to work that out right now, but if I get to it later I will edit the answer... EDIT: Game B analysis gianlino, your analysis of the following is correct: “@ Frst Grade Rocks! Ω: When you have 64x, if you replay 4 and x the expectation of sup of 2 dice will be (6*11+5*9+4*7+3*5+2*3+1*1) / 36 = (66+45+28+15+6+1)/36 = 161/36. If you keep 4 and replay only x then your expected max is (2/3)*4 + 1/6*(5+6) = 162/36. Right?” However, this holds for all cases where 4 is the second lowest number, not just the 64x case. You also said “@Dale: in game B when you have 44x with any x<5, if you just replay x, your expected total in the end is 4+4.5 = 8.5 = 306 / 36. But if you replay both 4 and x, your expected total is 310 / 36.” Replaying both 4 and x your expected total is 161/36 + 4 = 305/36, not 310/36. NEW EDIT 8/8/10 17:35 Eastern "@ Dale. For the 310 vs 305 discrepancy, didn't you overlook that if you replay both, then you may discard the first one if it's advantageous. This is why when you have 4 4 x, the chances of getting 55 or better on the reroll makes it advantageous to reroll both. Maybe my wording was ambiguous I apologize." No, you were not ambiguous, I was simply wrong. I hate it when that happens (don't tell my wife, though, because she still thinks I am always right). Not only was I wrong in that, when I looked over my work I realized I had incorrectly calculated the probabilities for P(12) through P(3) even for my slightly less than optimal strategy (but P(2) was right!!). That's what I get for doing the work late at night... ANYWAY: gianlino, you are correct on the 44 first roll strategy. The best strategy on a 44 first roll is to re-roll two dice. I also calculated the 54 first roll expected values for re-roll one and re-roll two dice and both came out to the same value, 9.5. I decided to keep 54 at a single roll because that required fewer changes to my work. Hence, my game B strategy is always re-roll the lowest number, if the second lowest number is <= 3 then re-roll it, and if the highest pair is 44 the re-roll one of the 4's. To clarify how I got my results I will use the following notation: -b = Number 2 or lower -c = Number 3 or lower -d = Number 4 or lower -e = Number 5 or lower -f = Number 6 or lower -F###S## = First roll three numbers descending, Second roll two numbers descending -F###S# = First roll three numbers descending, Second roll single number CORRECTED WORK and CORRECTED STRATEGY P(12) = P(F66fSf or F65eS6 or F64dS6 or F6ccS6f or F5ccS66 or FdddS66) (16/216)*(36/36) + (27/216)*(6/36) + (21/216)*(6/36) + (27/216)*(11/36) + (27/216)*(1/36) + (64/216)*(1/36) = (576 + 162 + 126 + 297 + 27 + 64)/7776 = 1252/7776 P(11) = P(F65eSe or F64dS5 or F6ccS5e or F55eS6 or F54dS6 or F5ccS6e or FdddS65) (27/216)*(30/36) + (21/216)*(6/36) + (27/216)*(9/36) + (13/216)*(6/36) + (21/216)*(6/36) + (27/216)*(10/36) + (64/216)*(2/36) = (810 + 126 + 243 + 78 + 126 + 270 + 128)/7776 = 1781/7776 P(10) = P(F64dSd or F6ccS4d or F55eSe or F54dS5 or F5ccS5e or F4ddS6d or F4ddS55 or FcccS64 or FcccS55) (21/216)*(24/36) + (27/216)*(7/36) + (13/216)*(30/36) + (21/216)*(6/36) + (27/216)*(9/36) + (37/216)*(8/36) + (37/216)*(1/36) + (27/216)*(2/36) + (27/216)*(1/36) = (504 + 189 + 390 + 126 + 243 + 296 + 37 + 54 + 27)/7776 = 1866/7776 P(9) = P(F6ccS3c or F54dSd or F5ccS4d or F4ddS5d or F3ccS6c or FcccS54 or FbbbS63) (27/216)*(5/36) + (21/216)*(24/36) + (27/216)*(7/36) + (37/216)*(8/36) + (19/216)*(6/36) + (27/216)*(2/36) + (8/216)*(2/36) = (135 + 504 + 189 + 296 + 114 + 54 + 16)/7776 = 1308/7776 P(8) = P(F6ccS2b or F5ccS3c or F4ddS4d or F3ccS5c or FcccS44 or F2bbS6b or FbbbS53 or F111S62) (27/216)*(3/36) + (27/216)*(5/36) + (37/216)*(7/36) + (19/216)*(6/36) + (27/216)*(1/36) + (7/216)*(4/36) + (8/216)*(2/36) + (1/216)*(2/36)= (81 + 135 + 259 + 114 + 27 + 28 + 16 + 2)/7776 = 662/7776 P(7) = P(F6ccS11 or F5ccS2b or F4ddS3c or F3ccS4c or F2bbS5b or FbbbS43 or F111S61 or F111S52) (27/216)*(1/36) + (27/216)*(3/36) + (37/216)*(5/36) + (19/216)*(6/36) + (7/216)*(4/36) + (8/216)*(2/36) + (1/216)*(2/36) + (1/216)*(2/36) = (27 + 81 + 185 + 114 + 28 + 16 + 2 + 2)/7776 = 455/7776 P(6) = P(F5ccS11 or F4ddS2b or F3ccS3c or F2bbS4b or FbbbS33 or F111S51 or F111S42) (27/216)*(1/36) + (37/216)*(3/36) + (19/216)*(5/36) + (7/216)*(4/36) + (8/2126)*(1/36) + (1/216)*(2/36) + (1/216)*(2/36) = (27 + 111 + 95 + 28 + 8 + 2 + 2)/7776 = 273/7776 P(5) = P(F4ddS11 or F3ccS2b or F2bbS3b or F111S41 or F111S32) (37/216)*(1/36) + (19/216)*(3/36) + (7/216)*(4/36) + (1/216)*(2/36) + (1/216)*(2/36) = (37 + 57 + 28 + 2 + 2)/7776 = 126/7776 P(4) = P(F3ccS11 or F2bbS2b or F111S31 or F111S22) (19/216)*(1/36) + (7/216)*(3/36) + (1/216)*(2/36) + (1/216)*(1/36) = (19 + 21 + 2 + 1)/7776 = 43/7776 P(3) = P(F2bbS11 or F111S21) (7/216)*(1/36) + (1/216)*(2/36) = (7 + 2)/7776 = 9/7776 P(2) = P(F111S11) 1/7776 Notice that the sum of P(2) through P(12) = 1. E(Game B) = (12*1252 + 11*1781 + 10*1866 + 9*1308 + 8*662 + 7*455 + 6*273 + 5*126 + 4*43 + 3*9 + 2*1)/7776 = (15024 + 19591 + 18660 + 11772 + 5296+ 3185+ 1638+ 630+ 172+ 27 + 2)/7776 = 75997/7776 = 9 & 6013/7776 ~= 9.7733 ***************************** "@Dale How come you get less than before as an average? I remember 9.8 something and now 9.77? If anything it should go up, right?" When I looked over my work I realized I had incorrectly calculated the probabilities for P(12) through P(3) even for my slightly less than optimal strategy (but P(2) was right!!). That's what I get for doing the work late at night...

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Unless I misread your question, the game A strategy is pretty simple: Just pick up any die that is less than 4 because you are more likely than not to improve on that particular die thus maximizing your score. If all of the dice are below 4, you just pick up the two lowest. Game B is tougher. And I haven't had time to work out an absolute strategy. But the general strategy should be to pick up any die less than 5 (up to two die, of course). The odds of rolling either a 5 or 6 with 2 die is 5/9. So you are most likely to improve your position. The really interesting question is what to do if you have a 5-5 or a 5-6 combo showing. My intuition is to roll just one die and leave the combo ... but I need to think some more and I have to go now. PS. Farkle is a dice game similar in strategy to your game and the previous dice problems. Loads of fun. http://en.wikipedia.org/wiki/Farkle I couldn't resist not working. Let's examine the odds of improving your lot if you have a 5-5 combo and you pick up two dice (thus getting rid of one of the 5s): 6-6, 1/36, improve by 2 6-x, 10/36,improve by 1 5-5, or 5-x, 9/36 even (assumes x ≠ 6) 4-4 or 4-x, 7/36, regress by 1 (assumes x ≠ 5, 6) Anything else, regress by ≥2 Weighted values 2*1/36 1*10/36 0* 9/36 -1 * 7/36 -2 *5/36 -3 * 3/36 -4 *1/36 Sum = -0.5 So by rolling at 5-5 your average score will decrease by 0.5. If your initial roll was 6-5, it would be even worse. Just roll one die (or none if you rolled a 6-6) ********* Let me summarize game B. If your top two numbers are all 5s and 6s, reroll 1 die and then take the best two dice. Exception if the top pair is 6-6 you cannot improve your score by rerolling. As shown above in the 5-5 example, rerolling more than one die will result in a net decrease in your score. Rerolling just one die will result in a 1/6 chance of improving your score if you have a 5-5 or a 5-6 combo, which means a net improvement of 0.1667 If your top two dice are not all 5s and 6s, keep your best die and then reroll two dice. Your score will on average improve. You have a 5/9th chance that one of the dice you roll will be a 5 or a 6, which means that when combined with the die you left on the table, your lot will improve. ********** I'm sorry. I never even read your line: "What are the average totals in each game with that strategy." Just stopped where it said to figure the strategies: Odds First roll 12: 6,6, x P = 16/216 11: 6,5, x P = 27/216 (x ≠ 6) 10: 5,5, x P = 13/216 (x ≠ 5, 6) (For the below, x≠ 5, 6 or any number higher than the first number) 6,x,x P = 48/216 5,x,x P = 48/216 4,x,x P= 36/216 3,x,x P = 19/216 2,x,x P = 8/216 1,1,1 P = 1/216 Odds after second roll P12 = 16/216 + 27/216 * 1/6 + 48/216 * 11/36 + 112/216 * 1/36 = 1378/6^5 = 17.72% P11 = 27/216 *5/6 + 13/216 * 1/6 + 48/216 *(11-1)/36 +48/216*9/36 + 64/216* 2/36 =1928/6^5 = 24.79% P10 = 13/216 *5/6 + 48/216* 7/36 + 48/216 * 9/36 + 36/216*(11-3)/36 + 28/216* 3/36 = 1638/6^5 = 21.06 % P9 = 48/216* 5/36 + 48/216 * 7/36 + 36/216*9/36 + 19/216 * (11-5)/36 + 9/216 * 4/36 = 1037/6^5 = 13.36% P8 = 48/216* 3/36 + 48/216 * 5/36 + 36/216*7/36 + 19/216 * (9-1)/36 + 8/216 *(11- 7)/36 + 1/216 * 5/36 = 825/6^5 = 10.61% P7 = 48/216* 1/36 + 48/216 * 3/36 + 36/216*5/36 + 19/216 * 7/36 + (8)/216 *(9-3)/36 + 1/216 * (11-9)/36 = 555/6^5 = 7.14% P6 = 48/216 * 1/36 + 36/216*3/36 + 19/216 * 5/36 + 8/216 *7/36 + 1/216 * (9-5)/36 = 311/6^5 = 3.99% P5 = 36/216*1/36 + 19/216 * 3/36 + 8/216 *5/36 + 1/216 * (7-1)/36 = 139/6^5 = 1.79% P4 = 19/216 * 1/36 + 8/216 *3/36 + 1/216 * 5/36 = 48/6^5 = 0.62% P3= 8/216 *1/36 + 1/216 * 3/36 = 11/6^5 = 0.14% P2 = 1/6^5 = 0.01% (error corrected) Average Total: 9.8659979... ***** I rolled over last night in bed and asked myself if 6-4 worked. I was able to easily convince myself that everything else was not a problem. But I was still unsettled. Wake up this morning and I see you beat me to the punch. (I should have known given that 6-4 > average expected return) The expected return for keeping a 6-4 combo is 378/36 = 10.5 whereas dumping the 4 results in a return of 377/36 = 10.4722... The probability of having a 6-4 combo high is 21/216 This drives the average expected return to: 9.86869856 ... The following probabilities need to be corrected: P12 = (1378-5*21)/6^5 = 1273/ 6^5 = 16.37% P11 = (1928- 3*21)/6^5 = 1865/6^5 = 23.98% P10 = (1638 + 17*21) / 6^5 =1995/6^5 = 25.66% P9 = (1037 - 5*21)/6^5 = 932/ 6^5 = 11.99% P8 = (825 - 3*21)/6^5 = 762/6^5 = 9.80% P7 = (555 - 21)/6^5 = 534/6^5 = 6.87%

Frst Grade Rocks! Ω

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