59 of 100: Crazy Dice!

If Katherine gets $3$ , the probability of winning is $\dfrac{1}{2}$ (when Zyan gets $2$ ).

If Katherine gets $6$ , the probability of winning is $1$ .

So, the win probability of Katherine is greater than $\dfrac{1}{2}$ , or if they play many times, Katherine will win more frequently.

Simple solution: If Katherine ONLY had threes on her dice, Katherine and Zyan would be equally likely to win. Since Katherine has one six (which always beats Zyan), she clearly has the better chance!

Katherine will win if:

• Zyan rolls a 2: $\frac{1}{2}$
• Zyan rolls a 5 and Katherine rolls a 6: $\frac{1}{2}$ * $\frac{1}{6}$ = $\frac{1}{12}$

So a $\frac{7}{12}$ chance in winning.

On the other hand, Zyan will win if:

• Zyan rolls a 5 and Katherine rolls a 3: $\frac{1}{2}$ * $\frac{5}{6}$ = $\frac{5}{12}$

So a $\frac{5}{12}$ chance in winning.

Katherine wins over time.

Katherine is more likely to win: 21/36

Half the time zyan will throw a 2 and lose. The other half the time he will throw a 5 and usually but not always win. Therefore Katherine will win more often.

If we simplified this problem and said Katherine only had 3s on her dice, Zyan would win half the time with one of his 5s, and lose the other half of the time with one of his 2s: they would both be equally likely to win. But when we give Katherine back her 6, we know she now has the advantage!!!

Katherine has a $\frac{5}{6}$ chance of rolling a $3$ .......[when Zyan gets $2$ ]

now,If Katherine gets $3$ , the probability of winning is = $\frac{1}{2}$

Katherine has a $\frac{1}{6}$ chance of rolling $6$

now, if Katherine gets $6$ , the probability of winning is $1$ .

so, we can easily say that the win probability of Katherine is greater then $\frac{1}{2}$ .that's why if they play many times ,Katherine will win frequently.

There could be another probability----it is seen from the chart that.......Katherine is winning in every chances.so, from here we can say that Katherine will win more frequently.

Most people are looking at the solution from Katherine's perspective, but when looked at Zyan's:

Zyan can only win if he rolls a $5$ (with probability $\frac{1}{2}$ ).

But even still, if Katherine rolls a $6$ with probability $\frac{1}{6}$ Zyan loses.

Then clearly Zyan has less than half probability of winning.

Katherine has a 5/6 chance of rolling a 3. Each of these rolls has a 50% chance of winning (there is a 50% chance Zyan rolls a 2, and a 50% chance Zyan rolls a 5). Katherine has a 1/6 chance of rolling a 6. This roll has a 100% chance of winning. Calculating expected probability, Katherine has a 7/12 chance of winning.

Let's go with the example that they roll the dice 6 times. In this case we can say that Katherine will roll a 6 once. On that turn it is a 100% chance she will win. On all of the other turns she will roll a 3. If Zyan rolls a 5 (which there is a 50% chance of) then he will win that round. But 50% of the time this won't happen (when he rolls a two). Thus if Katherine only had 3's then she would win 50% of the time. But that one six brought her ahead.

Because if you look at katherine die it has larger numbers than zyan die

I find that my solution boils down to the previous one. If there are a large multiple of 36 throws, then with PROBABILITY 1/36 each possibility of Katherine's 6 sides and Zyan's 6 sides will come off. On each of the times that Katherine throws a 3 (and there will be 6x5 = 30 of these) Zyan lose on the times that she throws a 2 (there will be 15 of these) and she will win on the times that she throws a 5 (there will be 15 of these. But on the time that Katherine throws a 6 (and there will be 6 of these) she will win every time. So there is slightly more probability the Katherine will win. Incidentally P(Katherine wins)= (15 + 6) / 36 = 21 / 36 = 7 / 12 = 58.33 %. And the P (Zyan wins) = 15 / 36 = 5 / 12 = 41.66 %. Regards, David

• Let's count the ordered pairs (where each of repeating numbers is distinguished from another, by color for example).
• There are 6 * 6 = 36 total pairs.
• Number of winning pairs for Katherine is (5 * 3) + 6 = 21, from where is not hard to get the answer.

Zyan loses automatically half of the time (when they roll a 2), and the rest of the time Katherine sometimes wins, hence Zyan wins less than half the time, so Katherine should win more.

Katherine always wins when her die shows a 6. P(6 and 2) + P(6 and 5) = (1/6)(1/2) + (1/6)(1/2) = 1/6 Katherine will also win if her dies shows a 3 and Zyan's die shows a 2. P(3 and 2) = (5/6)(1/2) = 5/12 Thus, Katherine's probability to win is 1/6 + 5/12 = 7/12.

Or, the only way that Zyan could win is for his die to show a 5 and Katherine's die to show a 3. P(3 and 5) = (5/6)(1/2) = 5/12.

Thus, Katherine will win more frequently than Zyan.

I used expected value and found 10.9 for Katherine and 10.1 for Zyan

I used the Partition theorem. Let the event of Katherine winning be denoted by $K$ and the event of Zyan winning by $Z$ . We'll give the roll that Katherine and Zyan make the symbols $K_N$ and $Z_N$ respectively, where $N$ is the dice roll. Then Katherine's total winning probability is given by:

$P(K) = P(K|K_3)P(K_3) + P(K|K_6)P(K_6)$

$=\frac{1}{2}\cdot\frac{5}{6} + 1\cdot\frac{1}{6}$

$=\frac{7}{12}$

However, Zyan's winning probability is given by:

$P(Z) = P(Z|K_2)P(Z_2) + P(Z|Z_5)P(Z_5)$

$=0\cdot\frac{1}{2} + \frac{5}{6}\cdot\frac{1}{2}$

$=\frac{5}{12}$

Thus, $P(K) > P(Z)$ so Katherine has a higher probability of victory.

We can also note that the total probability must equal one, since someone has to win in this case, so from calculating $P(K) = \frac{7}{12} > \frac{1}{2}$ it can be deduced that she has a greater winning probability than Zyan.

P(Katherine wins) = P(She rolls 3|Zyan rolls a 2)P(Zyan rolls a 2) +

P(She rolls a 6) = (5/6)(3/6) + 1/6 = 5/12 + 2/12 = 7/12.

P(Zyne wins) = P(Zyne rolls a 5|Katherine rolls a 3)P(Katherine rolls a 3)

(3/6)(5/6) = 5/12.

Therefore Katherine wins.

$K_{win} = -\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{2} = \frac{7}{12}$

Half of the time, Zyan is guaranteed to lose, since a two will never beat a three or a six. The other half of the time, Zyan has a relatively high likelihood of winning. However, since Katherine has a six, which will win every time, and will also win every time Zyan rolls a two, Katherine has a higher likelihood of winning.

Suppose that Katherine has a bit similar die with faces (3, 3, 3, 3, 3, 3) and Zyan has the die with faces (2, 2, 2, 5, 5, 5). In this case, it's clear that each one of them has a 1/2 chance of winning. Now consider the case mentioned in the problem where we increase the number on one of the faces of Katherine's die which clearly result in an advantage for Katherine.

Katherine wins if :

• Katherine throws 3 and zian throws 2 With probability (5/6) * (3/6) = (5/6) * (1/2) = 5/12

Or

• Katherine throws 6 and zian throws 2 or 5 It is easy to see that if Katherine throws 6, she will win irrespective of zian’s throw. The probability of the above is 1/6 = 2/12

The total probability of Katherine winning is 5/12 + 2/12 = 7/12 which is greater than 1/2.

Hence, it is more likely that Katherine will win

Zyan has $1/2$ a chance of rolling a 2 which would see him lose regardless of the value on Katherine's dice.

Zyan has $1/2$ a chance of rolling a 5, but this would only see him win $5/6$ of the time this happens.

Therefore Katherine has more than 50-50 odds of winning each game.

Katherine has a 21/36 likelihood to win over ZYan.

Half of the rolls of Zyan's die lose regardless of Katherine's roll. The other half still lose to Katherine's 6, so Katherine will win more than Zyan.

As usual, I wrote simple script.

http://ideone.com/wFCig9

Here we just simulate the game between two players using standard library function randint.

If Zyan gets 2 he loses. The probability for Zyan getting 2 is 1/2. So there's no chance that Zyan will win more frequently. Two answers are left. If there's one situation where Zyan gets 5 en Katherine wins then Katherine will win more frequently. And there is such a situation: Zyan gets 5 and Katherine gets 6. The answer is: Katherine wins more frequently!

Katherine has 3,3,3,3,3,6 Zyan has 2,2,2,5,5,5.

Just see individually 4 of Katherine's numbers 3,3,3,6 are greater than Zyan's 2,2,2,5. On the contrary, individually only 3 times Zyan get the greater number 5,5,5 than Katherine's 3,3,3. So, Katherine gets more win if we randomly roll the dice together for numerous times.

If Katherine's die has the number 3 on all sides, then both people would be equally likely to win. This is because Zyan has a 50% chance of rolling 2 and a 50% chance of rolling 5, and so wins half of the time and loses half of the time.

However, Katherine can roll the number 6, which guarantees a win. Therefore, she has >50% chance of winning every time she rolls the dice.