two dice rolling probability? could use some help
How is everyone doing? I am new to this forum so if I make some error, please let me know.
So here is problem. Player A and Player B, they each take turn rolling two dice until an odd product occurs, or until even product occur three time in succession. if round ends on odd product, player A get 2 point, if end in three even product, player b get 3 points. A game ends after ten round. Who will win the game.
At first, i was thinking Player A will win but i know somehow it is player B have higher percentage of wining.
Here what I know. to get odd product, I solve as (3/6)*(3/6) = 9/36 =>1/4 ~ to get odd, the only way is both dice is odd. I consider 1*1 an odd number. I even drew a table of 6 by 6. I know that this is correct.
to get even product on first time, it is 1-9/36 = 27/36 = 3/4
Here where I get confuses: In order to get even product three time in row, I thought you have to muliple three time of 27/36 (aka 3/4) = 19683/46656 => 42%
however, when i did the table count number of total possiblility and it end up with total possible even:23361 and favorable event: 17496. Percentage is 73%
Here how I did the graph. First roll aka First table, you have 6 by 6 which give you 36 possible event. out of that 27 event you can continue on and rest of 9 event that you cant. then on 2nd roll,out of last 27 event, you have a 6 by 6 table for each event. so 27*36 = total possible event = 972 event. out of each table, you get 9 event that you cant continue on. so 9*36 = 324. therefore only 648 event can continue to next roll ( I done by 972-324). The final roll, for each event you get 6 by 6 table . so total possible event including non continue event is 648 *36 = 23328.
The 23328 is not correct event until you add in non-continue event from 1st roll and 2nd roll ( 9+324 respectively) so real total possible event is 23661. The favorable event out of that total possible event is 17496 which I did by 648 *27. where 648 is total event but 27 of that is favorable.
so final probability of get B three time in row is 73%
Now I am pretty confident this is correct answer but I would like to express this answer in a math term not in graph. like for example get odd product, i demonstrate that is 3/6 for each dice to get odd number then to get odd product which is (3/6)*(3/6)
sorry for long post. after that I was suppose to find expected value from player A viewpoint. But I am more interest in solving Player B probability in math expression rather than by drawing a 6 by 6 box.
thank you for your help.
The other part of the question
we need to find who will be the winner... To find the number of points A is probably going to win, you multiply the number of rounds by the chance A will win, and then you multiply by how many points A will get per round...
Now for B
So B is predicted to win (by a small margin)