I'm slightly confused I understand that white can win instantly with 6-6 but what if Green rolls a 2-1 or 1-2 wouldn't that give white another chance? For whites odds to be improved on his or her next roll. White needs to roll a double. If however, white rolls a double 1s, white can now win with double fives or sixes. 2/36 If white rolls 2-2, white can win potentially win with sixes, fives, and fours 3/36. With 3-3 Sixes, fives, fours, threes 4/36.With 4-4, twos, three fours, fives, sixes, With 5-5 all doubles give white the win bring white's chances to 6/36 on that turn. White's chances of winning increasing his or her chances is dependent on white rolling a double. 6/36. Something like 16.66 % However, since if white rolls a double. So I'm wondering if I did this right. There a few options. White wins first roll. 2.78 % Green wins on next roll. Since this is dependent on white not rolling a double six, I feel that white's chances of not rolling 6-6 need to multiplied by green's chances of not rolling 1-2 or 2-1. I got 91.08 %. Now, if white fails here, green wins automatically next turn. So the options are white rolled any non-double. White's only chances of winning is still only 6-6. The problem, I believe, is any non-double 30/36 * green rolling 1-2 or 2-1 2/36 * rolling 6-6 1/36. I came up with 0.128 % I'm not entirely sure how to calculate the chances of each scenario. However, I believe is the odds of white rolling that double 1/36 multiplied by green failing 2/36 multiplied by the increased odds by that number 2/36, 3/36, 4/36, 5/36, 6/36. After this I get really confused how to calculate White's percentage from the start, I believe, it's The Chance of winning the first roll OR the chance of winning if white gets no-double, AND green fails, And 1/36, OR white rolls 5-5,AND green fails AND 6/36, white rolls 4-4,AND green fails,AND 5/36, OR white rolls 3-3 and green fails and 4/36, OR 2-2 , green fails, and 3/36, OR 1-1 AND green fails, AND 2/36 1/36 + 30/36 * 2/36 * 1/36 + 1/36 * 2/36 *2/36 + 1/36 + 1/36 * 2/36 *3/36 + 1/36 + 1/36 * 2/36 * 4/36 + 1/36 + 1/36 * 2/36 *5/36 + 1/36 + 1/36 * 2/36 *6/36 = I put this in the calculator and it gave 2.643477101158559e-4 I really like your videos also. I thought I aw the green checkers on the 1 point and the 3 point. Please let me know if I'm looking at it wrong. Also, I wrote this beofre watching the rest of the video so hopefully you didn't revisit this, and I look like an idiot.Okay my horrible math skill probably do that on their own.
Thank you for commenting. You are correct. I should have listed the first example at 1 point- 2 point for green (instead of 1-3), then the percentages would be correct. On the example that is given, White wins 2.78% of the time (Whites 1st roll) and gets a second chance to win an additional 5.4012% of the time (.9722% White does not roll 6-6 * .0556 (Green rolls 1-2, 2-1). Of the 5.4% of the time Green gives White another chance, we would have to take a weighted average (with an excel spread sheet) of the original 35 non-winning rolls. 26 of those 35 remaining rolls could only win with 6-6. The remaining 9 would depend on White using optimal strategy (example; roll 3-3, move 2 pieces in & need 4-6 or better for the remaining to pieces (ie: 3-3, 4-4, 5-5 or 6-6) to win). 5 of the 9 scenarios need 4-6 or better, while the remaining 4 need 5-5+, 4-4+, 6-2+ and 6-1+ respectively. The weighted average for White to win on the second turn is ~10.14%. So, White will win an additional 0.55% (10.14% weighted average win rate * 5.4% second chance roll) of the time and Green wins 100% of the time after that. If we add 0.55% (second chance roll win 10.14% * 5.4%) with the 2.78% (first roll win), we get 3.33%. So, in this example White can win 3.33% of the time and lose the remaining 96.67% of the time to Green (not 2.78% which is stated in the video). My original goal was to start off with an easy example and then tackle a more complicated problem later on (at 4:47). I should have gone with 1-2 to start off. Thanks again for catching my mistake.
In a backgammon game (using online msn free games) .....(means stupid computer randomly will roll the dice for you) .....same computer software gave to my opponent : double 5 (2 dice 5 and 5) , then double 6 , then again double 5 ....can you calculate what are the odds to get that again
If it is complicated usually the best way to do it is to either setup a spreadsheet and set of formulas to handle it that way. . . Otherwise the paper or note card method works good too. . . Have been playing this game on my phone for the last six months. . . Have noticed some moves may be more risky than the standard play. . . But also have a lot of upside that is not easily quantifiable. . . Best way to reach me is through the comment section with any questions. . . Thanks for watching and good luck.
@@garydevries5961 ah well maybe it's not that complicated (for you). If white starts with a 3-1 opening en black with 6-1, what is the probability of winning the game (for either black or white)?
White has 163 spaces left while black has only 160, so black has a slight advantage in numbers. However, white is first to move and can roll a 4 combo with odds of 7-5 or more than half the time, so a slight advantage for white here. Unfortunately, there are too many other random variables to know the exact odds of winning (it still close to 50-50 here). Variables such as playing style where the moves were made as well as other tendencies and situational decisions. In addition to that there is also a wide variance of potential dice rolls (this can be offset sometimes with good strategy). The short answer is that the game is still very close white 163, black 160, with white moving first. Thanks.
