Broken Dice Game - 333336 & 222555

From the table, the probability of Katherine winning is $\frac{21}{36}$ , which is bigger than Zyan's $\frac{15}{36}$ .

There is 50% probability for Zyan's dice to show 2 and 50% probability to show 5. If Zyan's dice shows 2 then, always Katherine wins as all the numbers on his cube are greater than 2. So, for Zyan to win (or atleast be equal), he should win all the time when 5 shows up on his die, but he can't win all the time because Katherine's die has 6 (>5) on one of the face of his die. So, Katherine wins more than 50% of time.

The probability of Katherine winning = 5/6x1/2 + 1/6x1 = 7/12.

The probability of Zyan winning = 1/2x0 + 1/2x5/6 = 5/12

The question is simply who has the best chance to win. If Katherine's die were all 3s she would win 1/2. Since one face is 6, which would always win the answer is straightforward.

If Katherine has a 3, she wins probability 1/2. If she has a 6, she is guaranteed to win.

Zyan has a 50% chance of losing no matter what Katherine rolls... i.e. if Zyan rolls a 2 (which is 50% probability) he will lose no matter what Katherine... in fact Katherine doesn't even have to roll... so if the game would always start with Zyan rolling first to see if Katherine even has to roll, then 50% of the time Katherine would not even roll her die. So the game is tied without Katherine even rolling... now what if Zyan does roll a 5 making Katherine have to roll her die... well then Katherine has a 1/6 chance of winning on top of the 50% chance of winning determined by how Zyan rolls... Not even doing any grids or thinking too hard... we can see Katherine will win more than lose.

Katharine will win 4/6 Zyan will win 2/6 or 2/3 vs. 1/3

K will win in two cases 1. when Z 's dies show 2 with probability 1/2. 2. when Z's dies show 5 and K's dies show 6 with probability=1/2*1/6 hence winning probability of K will be = 7/12.

Z will win only when K's dies show 3 and Z's dies show 5 with probability (5/6)*(3/6) =5/12

                 hence from above we can say that Katherine  will win more frequently.

P(K)=5/6x3/6+1/6x3/6+1/6x3/6=21/36 P(Z)=3/6x5/6=15/36

probablity that katherine will win is that her dice must show 3 and zyan's will show 2 or her dice show 6 and zyan's dice show up 5. i. e [(5/6) (1/2)+(1/6) (1/2)] =7/12. probablity tha zyan's will win is that his dice must show 5 and katherine's will show 3 i. e [(1/2)*(5/6)] =5/12. So katherine wins.

Pick one person to calculate the winning probability. Let pick Katherine. There are 3 cases that Katherine can win. +Katherine got 3, and Zyan got 2

Chances that Kath got 3 is $\frac {5}{6}$

Chances that Zyan got 2 is $\frac {1}{2}$

Chances that Katherine win with 3 - 2 is $\frac {5}{6} \times \frac {1}{2}$

+Katherine got 6, Zyan got 2

Chances that Katherine got 6 is $\frac {1}{6}$

Chances that Zyan got 2 is $\frac {1}{2}$

Katherine win this case $\frac {1}{6} \times \frac {1}{2}$

+Katherine got 6, Z got 5 : $\frac {1}{6} \times \frac {1}{2}$

Total Katherine's winning probability = Sum of all the case = $\frac {7}{12}$

==> Zyan's winning probability is $\frac {5}{12}$

==> Katherine will win more

If the dice lands on the lowest number for each , katherine has 6 and Zyan has 4, Katherine is more likely to have a larger number

50% of the time Zyan will roll a two and so Katherine will win whenever a two is rolled. When Zyan rolls a 5 there is still a 1 in 6 chance that Katherine will beat Zyan's roll. Therefore Katherine will win at a rate that is slightly greater than 50%

1st Dice is : 3,3,3,3,3,6 2nd Dice is : 2,2,2,5,5,5

meaning the probability of the 1st Dice is: 5/6 -> to come out 3 1/6 ->to come out 6

while the probability of the other: 1/2 -> to come out 2 1/2-> to come out 5

so because we are trying to figure who will win more times we are actually checking who will likely win while both of the dice been thrown at the same time meaning who will likely be the higher number so if the 2nd have a 2 which is 1/2 of the times it doesn't matter what comes out on the 1st , 3 and 6 always higher than 2 while the second 1/2 of the times splits since it comes as 5 on the 2nd: (1/2) (5/6)=5/12 -> 5 of out of 12 games 5 wins 3 but (1/2) (1/6)=1/12->6 wins 5 one would asume but we're asking for the sum meaning Ptotal(1st win)=1/12+1/2=7/12 Ptotal(2nd win)=5/12 7/12>5/12

so the answer Katherine dice wins more she wins 2 games more out of 12 games than Zyan

If Zyan rolls a two (50%) he always loses, If Zyan rolls a 5, he only wins 5 out of 6 times. Essentially 12 possible outcomes, Kathrine wins 7 out of 12

5/6 of the time, Katherine wins half the time. The remaining 1/6, she always wins. Therefore, she wins 7/12 of all outcomes.

Katherine will win if the results are (3,2), (6,2),(6,5) (these results are written in the order (Katherine's result, Zyan's result)). So the probability is 5/6X1/2+1/6X1/2+1/6X1/2=7/12. Zyan will win if the result is only (3,5)(in the same order as previous). So the probability is 1/2X5/6=5/12.... 7/12>5/12... so Katherine will win.

Let's break down both Katherine and Zyan's chances of rolling each side:

Katherine: 5/6 chance of rolling [3] 5/6 chance of rolling [6]

Zyan: 1/2 chance of rolling [2] 1/2 chance of rolling [5]

Assuming each of them rolls 12 times, Katherine will roll 10 of [3] and 2 of [6].

During the 10 rolls of [3] that Katherine has, Zyan will roll 5 of [2] and 5 of [5]. Result: Katherine has 5 wins and 5 losses

During the 2 rolls of [6] that Katherine has, Zyan will roll 1 of [2] and 1 of [5]. Result: Katherine has 2 wins and 0 losses

Summing everything up, Katherine has 7 wins. Zyan has 5 wins. Thus, Katherine will win more frequently than Zyan.