Dicier Probability...

We know that we can a roll a 7 with two dice in 6 ways. While a 6 and an 8 both can be rolled in 5 ways. Yet the probability of rolling both a 6 and an 8 before throwing two 7's is higher. Lets see why...

The basic reason is that the order for getting a 6 or an 8 does not matter. The 6 - 8 combo is a favorite at the outset because there are 5+ 5 = 10 ways of getting either a 6 or an 8, whilst there are only 6 ways of getting a 7.

There are three ways that 6 and a 8 appear before two 7's.

1) Both 6 and a 8 appear before even one 7.

Now there are 10 out of 16 ways that either a 6 or an 8 appears first before a 7 because you don't mind either of them appearing.

Having got either of them, the chance of success, that is of getting the other number is 5/11 because now we specifically need either 6 or 8, whatever the case may be. (Remember the other numbers are disregarded when they are thrown)

Probability = 50/176

2) Either a 6 or 8 appears, then we hit a 7, then we throw the other number(6/8) before a second 7 appears.

10/16 chances of getting either 6 or 8 first.

6/11 chances that now a 7 comes up.

5/11 chance that the required relevant number, either 6 or 8 whichever is needed.

Probability = 300/1936

3) Finally 6, 8 appears after we throw the first 7.

6/16 chance of the first number being 7

10/16 chance of either 6 or 8

5/11 chance of the other required number of the two.

Probability = 300/2816

Adding up all these probabilities = 54.558 % !

Lets also check if the converse is correct.

Satvik wins in the 3 scenarios. Lets call rolling a 6 or 8 as Failure F

1) 77F

6/16 chances of rolling a 7

6/16 chances of rolling a 7

Probability = 36/256

2) 7F7

6/16 chances of getting a 7

10/16 chances of getting either 6 or 8

6/11 chances of getting a 7

Probability = 360/2816

3) F77

10/16 chances of of getting a 6/8

6/11 chances of a 7

6/11 chances of a 7

Probability = 360/1936

Adding them all together Percentage probability = 45.442 %

Combined total = 54.558 + 45.442 =100 %

This problem can be easily pictured by structuring it as a Markov Chain, as per the illustration.

S represents the start, W a win for Agnishom, L a loss, T a current tie and W' and L' a definitive win/loss, respectively. Note that at first the chances of winning start at 10/36, which later decreases to 5/36 when one of his numbers is struck down.

We want to calculate the absorption probability of S with regards to W'. Thus, a(W') is 1 and a(L') is 0. We can derive then the remaining absorption probabilities:

a(T) is equal to $\frac{5 \times a(W')}{36} + \frac{25 \times a(T)}{36} + \frac{6 \times a(L')}{36}$ Plugging the corresponding values for a(W') and a(L'), we get that a(T) = 5/11.

a(W) is equal to $\frac{5 \times a(W')}{36} + \frac{25 \times a(W)}{36} + \frac{6 \times a(T)}{36}$ Plugging the corresponding values for a(W') and a(T), we get that a(W) = 85/121.

a(L) is equal to $\frac{10 \times a(T)}{36} + \frac{20 \times a(L)}{36} + \frac{6 \times a(L')}{36}$ Plugging the corresponding values for a(L') and a(T), we get that a(L) = 25/88.

To top it off, we get the information from a(W) and a(L) to discover our desired a(S), which must correspond to $\frac{10 \times a(W)}{36} + \frac{20 \times a(S)}{36} + \frac{6 \times a(L)}{36}$ This gives us, by plugging the apropriate values, a final result of 4225/7744 or 54.558%

.we call the chance of Ag winning is S His chance of winning after rolling 6/8 is x His chance of winning after rolling 7 is y His chance of winning after rolling 6/8 AND a 7 is z we have these equations: z=5/36 + 25/36z y=10/36z + 20/36y x=5/36 + 6/36z + 25/36x S = 10/36x + 6/36y + 20/36 S solving these yields: z=5/11 y=25/88 x=85/121 S=4225/7744~54.558%