# two dice rolling probability? could use some help

• Jul 10th 2006, 10:18 PM
Dartgen
two dice rolling probability? could use some help
Hello,

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.
• Jul 10th 2006, 11:02 PM
JakeD
Quote:

Originally Posted by Dartgen
Hello,

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.

Hi. I could not follow your calculation with the graph. But I'll calculate the probabilities that A or B wins in two ways.

If i interpret the rules correctly, A wins if an odd product occurs in any of the first three rolls. B wins otherwise. That's the same as saying B wins if an even product occurs in each of the first three rolls. The game never goes beyond 3 rolls.

So B wins if an even product occurs 3 times in a row. The probability is as you say $\displaystyle \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{27}{64} = .42.$ The probability that A wins is $\displaystyle 1 - \frac{27}{64} = \frac{37}{64} = .58.$

To calculate this in another way, use A wins if an odd product occurs in any of the first three rolls. This probability is

$\displaystyle \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{4} + \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{16+12+9}{64} = \frac{37}{64} = .58,$

the same as above.

I hope that helps.
• Jul 11th 2006, 03:59 AM
malaygoel
Quote:

Originally Posted by Dartgen
Hello,

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%

How did you calculate it?

Keep Smiling
Malay
• Jul 11th 2006, 05:27 AM
Quick
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...
$\displaystyle \text{rounds}\times\text{chance of winning}\times\text{points per round}=\text{ending score}$

$\displaystyle 10\times\frac{37}{64}$$\displaystyle \times2=s_A\quad\rightarrow\quad12\approx s_A Now for B \displaystyle \text{rounds}\times\text{chance of winning}\times\text{points per round}=\text{ending score} \displaystyle 10\times\frac{27}{64}$$\displaystyle \times3=s_B\quad\rightarrow\quad13\approx s_B$

So B is predicted to win (by a small margin)
• Jul 11th 2006, 06:34 AM
Soroban
Hello, Dartgen!

JakeD had the best approach.
Let me baby-step through it . . .

Quote:

Player A and Player B take turns rolling two dice until an odd product occurs,
or until even product occur three time in succession.
If a round ends on odd product, player A get 2 points.
If it end in three even products, player B get 3 points.
A game ends after ten rounds.
Who will win the game?

Out of the 36 possible outcomes for a pair of dice,
. . 9 have odd products, 27 have even products.

Hence: .$\displaystyle P(odd) = \frac{9}{36} = \frac{1}{4},\;\;P(even) = \frac{27}{36} = \frac{3}{4}$

So $\displaystyle B$ wins: .$\displaystyle \left(\frac{3}{4}\right)^3 = \frac{27}{64}$ of the time

. . and $\displaystyle A$ wins: .$\displaystyle 1 - \frac{27}{64}\,=\,\frac{37}{64}$ of the time.

$\displaystyle A$ wins $\displaystyle 2$ points with probability $\displaystyle \frac{37}{64}.$
. . . . His expected winnings is: .$\displaystyle (2)\left(\frac{37}{64}\right) = \frac{74}{64}$ points per game.
In 10 games, he can expect to win: .$\displaystyle 10 \times \frac{74}{64} \,= \,11.5625$ points.

$\displaystyle B$ wins $\displaystyle 3$ points with probability $\displaystyle \frac{27}{64}.$
. . . . His expected winnings is: .$\displaystyle (3)\left(\frac{27}{64}\right) = \frac{81}{64}$ points per game.
In 10 games, he can expect to win: .$\displaystyle 10 \times \frac{81}{64} \,=\,12.65625$ points.

Therefore, player $\displaystyle B$ is the expected winner.

• Jul 11th 2006, 08:51 AM
Quick
Just to clear things up, me and Soroban have the same answer, I just approximated mine to the nearest whole number.
• Jul 11th 2006, 11:33 PM
Dartgen
Thank you for your help. I knew that B was going to win but i just couldnt figure out how. I did the calculation and I thought it was too low until i did the tree-like diagram of all possible event i figure out that it was B. again thank you for your help.

Does anyone know what is expected value from Player's A viewpoint. Say that both player put in same amount of money. Since B have slight edge, then that mean i can caluculate the different in probability that you have outline for me.

oh another question, what book did you use to find answer. I wont mind buying a book to help on my probability understanding.
• Jul 12th 2006, 03:41 AM
Quick
Quote:

Originally Posted by Dartgen
Thank you for your help. I knew that B was going to win but i just couldnt figure out how. I did the calculation and I thought it was too low until i did the tree-like diagram of all possible event i figure out that it was B. again thank you for your help.

Does anyone know what is expected value from Player's A viewpoint. Say that both player put in same amount of money. Since B have slight edge, then that mean i can caluculate the different in probability that you have outline for me.

This depends completely on how you're gambling, is the winner of the ten rounds going to get all the money or is the money split up at the end to both players according to how many points they got?
• Jul 14th 2006, 09:28 PM
Dartgen
The expected value
After ten round, whoever got most point win the whole bet. Based on that assumption, do I add the point of 11 divide total (11+12) to get expected value point from A. I know that it isnt 11 or 12 but I am assuming that if A get 11 pt possible doing ten round sthen B is 12pt possible for same
amount of round.
So are the expected valued correct?
• Jul 14th 2006, 10:18 PM
futsalfred
a
Quote:

Originally Posted by Dartgen
Hello,

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.

Coo.
• Jul 15th 2006, 05:14 AM
Quick
The expected points for each player are as follows...

Quote:

Originally Posted by Soroban
$\displaystyle A$ wins $\displaystyle 2$ points with probability $\displaystyle \frac{37}{64}.$
. . . . His expected winnings is: .$\displaystyle (2)\left(\frac{37}{64}\right) = \frac{74}{64}$ points per game.
In 10 games, he can expect to win: .$\displaystyle 10 \times \frac{74}{64} \,= \,11.5625$ points.

$\displaystyle B$ wins $\displaystyle 3$ points with probability $\displaystyle \frac{27}{64}.$
. . . . His expected winnings is: .$\displaystyle (3)\left(\frac{27}{64}\right) = \frac{81}{64}$ points per game.
In 10 games, he can expect to win: .$\displaystyle 10 \times \frac{81}{64} \,=\,12.65625$ points.