Winning the Dice Game

Shawn can only roll a number from 1-9.

If Shawn rolls a 1, Bret must roll a number from 12-20 to win. The probability of Shawn rolling a 1 is $\frac{1}{20}$ and the probability of Bret rolling a number from 12-20 is $\frac{9}{20}$ . Which gives us a probability of $\frac{1}{20}$ * $\frac{9}{20}$ .

If Shawn rolls a 2, Bret must roll a number from 13-20 to win. The probability of Shawn rolling a 2 is $\frac{1}{20}$ and the probability of Bret rolling a number from 13-20 is $\frac{8}{20}$ . Which gives us a probability of $\frac{1}{20}$ * $\frac{8}{20}$ .

If Shawn rolls a 3, Bret must roll a number from 14-20 to win. The probability of Shawn rolling a 3 is $\frac{1}{20}$ and the probability of Bret rolling a number from 14-20 is $\frac{7}{20}$ . Which gives us a probability of $\frac{1}{20}$ * $\frac{7}{20}$ . . . . If Shawn rolls a 9, Bret must roll a 20 to win. The probability of Shawn rolling a 1 is $\frac{1}{20}$ and the probability of Bret rolling a 20 is $\frac{1}{20}$ . Which gives us a probability of $\frac{1}{20}$ * $\frac{1}{20}$

So adding all of this we get: $\frac{1}{20}$ * $\frac{9}{20}$ + $\frac{1}{20}$ * $\frac{8}{20}$ + $\frac{1}{20}$ * $\frac{7}{20}$ +...+ $\frac{1}{20}$ * $\frac{1}{20}$

=( $\frac{1}{20}$ )( $\frac{9}{20}$ + $\frac{8}{20}$ + $\frac{7}{20}$ +...+ $\frac{1}{20}$ )

= ( $\frac{1}{20}$ )( $\frac{45}{20}$ )

= $\frac{45}{400}$

= $\frac{9}{80}$

9+80=89

For Bret to win the game, he must gain at least 10 + 1 = 11 points to have a sum total greater than Shawn's.

In other words, his die roll must have a value greater than Shawn's by at least 11.

Since the lowest die roll Shawn can get is 1, the minimum die roll Bret must get is 12.

If Bret gets a 12, for him to win, Shawn must get a 1. The chances of this happening is 1/20 (which is the chance of getting a 12 for Bret) x 1/20 (which is the chance of getting a 1 for Shawn) = 1/20 x 1/20

Similarly, if Bret gets a 13, Shawn must get either a 1 or 2. And the chance of this happening is 1/20 x 2/20.

Extra: We can generalise this probability to 1/20 x n/20 = n/400 where n is the value of (Bret's die roll - 11) and n>0

The overall probability is the sum of all the probabilities where Bret wins when he gets a die roll greater than 11.

So we can represent it like this:

Overall probability = (1/20 x 1/20) + (1/20 x 2/20) + ... + (1/20 x 8/20) + (1/20 x 9/20)

= 1/20 x (1/20 + 2/20 + ... + 8/20 + 9/20)

= 1/20 x 1/20 x (1 + 2 + ... + 8 + 9)

= 1/20 x 1/20 x (no. of terms/2) x (1st term+last term)

= 1/20 x 1/20 x ((9-1+1)/2) x (1+9)

= 1/400 x 45

= 45/400

= 9/80

a = 9

b = 80

a + b = 89

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Extra:

We can generalise the overall probability into a summation, using the aforementioned n/400 as the probability that Bret will win for a certain die roll from 12 to 20, and n = Bret's die roll - 11.

\displaystyle \sum_{n=1}^9 (n/400)

or if we take n to be Bret's die roll:

\displaystyle \sum_{n=12}^20 ((n-11)/400)

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Since Shawn is winning the game by 10 points and each person has a roll on the last turn, in order for Bret to win, Shawn needs to roll something less than or equal to 9, otherwise there is no way for Bret to win.

We can then write the probabilities for Bret to win. If Shawn rolls a 1 on the die, then Bret needs anything greater than 11 on the die to win. We can express the probability of this happening as

(frac {1}{20}) \times \frac {9}{20}

If Shawn rolls a 2, then Bret needs to roll anything greater than 12 on the die to win, and we can express the probability as

\frac {1}{20} \times \ frac {8}{20}

and so on.

This continues until we have hit the maximum case at which Shawn will always win, which is when Shawn rolls anything greater than 9. We can add all the individual cases to find the total probability of Bret winning:

\frac{1}{20} (\frac{1}{20} + \frac{2}{20} + \frac{3}{20} + \frac{4}{20} + \frac{5}{20} + \frac{6}{20} + \frac{7}{20} + \frac{8}{20} + \frac{9}{20}) = \frac {9}{80}

The total probability of Bret coming back is \frac{9}{80}.

(If someone could fix my LaTeX, that'd be awesome! I'm a newbie at this)

In the problem it states that Shawn is winning by 10 points and Brett need 10 or more to win. Also that it could be a tie. So you can say that in order for Brett to win, he has to roll one of the numbers that is higher than 11. The numbers are 12,13,14,15,16,17,18,19, and 20. With that we can see that Brett has a 9/20 chance to win. Shawn has a 1/20 chance to win because he can get any number and end up winning against Brett. Calculating the possible probabilities of Brett winning against Shawn:

If Shawn roles a 1 Brett has a 9/20 probability to win. ( 12,13,14,15,16,17,18,19,20)

If Shawn roles a 2 Brett has a 8/20 probility to win (13,14,15,16,17,18,19,20)

And this continues untill Shawn roles a 10. At that point Brett only has a 1/20 chance to win or tie.

With all the probabilities of Brett winning calculated we can use the probity addition rule to find the total probability of winning for Brett.

9/20+ 8/20+ 7/20+ 6/20+ 5/20+ 4/20+ 3/20+ 2/20+ 1/20=

45/20

Now we multiply the total calculated probability by the probability that Shawn can come back to win also.

Doing so we get

45/20 x1/20= 45/400

Simplifying:

9/80

So A= 9 and B=80 (9 and 80 are coprime because they have only one greatest common factor, that being 1)

A+B= 89

Answer = 89

Here,we see that Bret needs to win by at least 11 points on the last round,that means the difference between Shawn and Bret's roll is >=11. Let,shawn roll the number y Bret roll the number x then we get x-y>=11 here,we see that since y>=1,then x>=12.Now comes the part where counting is done cleverly.Here,we see that If x=12 then y=1 only.Again,if x=13 then y=1 or 2.so there are 2 possible outcomes.Counting in this way,till x=20 we see that the possible outcomes where bret can win is

9(9+1)/2=45. Again,total outcomes=20*20=400 taking the ratio,we get,45/400=9/80 so,a+b=9+80=89.

Bret needs to get > 10 more points than Shawn. Therefore, Bret could only win when Shawn gets < 10 in the last turn. When Shawn get 1 pt (1/20), Bret has 12-20 (9/20) chances to win. This probability is therefore 1/20*9/20.

So, considering all cases, the probability for Shawn to win is: 1/20(9/20+8/20+7/20+...+1/20) = 45/400 = 9/80

Thus a=9, b=80, a+b = 89

I made pairs through which bret can win if bret scores 19 points then he wins if shawn scores 1,2,3,4,5,6,7,8. if bret scores 18 points then he wins if shawn scores 1,2,3,4,5,6,7 similarly we can find other pairs,add them and divide them by total cases 45/400=9/80 9+80=89 which is the answer.

Solution 1: There are $20 \times 20 = 400$ different possible outcomes. Let's count the number of outcomes where Bret wins. If Bret rolls from 1 to 11, then since Shawn rolls at least 1, Bret cannot win. If Bret rolls $11+ k$ for $1 \leq k \leq 9$ , then there are $k$ possibilities for Shawn to roll in order for Bret to win. Hence, there are $\sum_{k=1}^9 k = 45$ possible outcomes. Thus, the probability that Bret wins is $\frac {45}{400} = \frac {9}{80}$ , which gives $a + b = 9 + 80 = 89$ .

Solution 2: Notice that if Bret rolls a number between 1 and 10, he can't win, no matter what Shawn rolls. Also, if Shawn rolls a number between 11 and 20, Bret can't win no matter what number he rolls. The probability of neither of these happening is $\left(\frac{10}{20}\right)^2 = \frac{1}{4}$ . We are left to consider the situation where Bret rolls a number between 11 and 20, and Shawn rolls a number between 1 and 10. Notice that this is equivalent to simply adding 10 to Bret's score and assuming that he also rolls a number between 1 and 10. In this case, the probability of Bret and Shawn rolling the same number is $\frac{1}{10}$ . Since they each have the same possible rolls with equal probability, if they do not roll the same number, they will each win half the time. Thus, In this situation, Bret wins with probability $\frac{9}{20}$ as does Shawn. Since this only occurs $\frac{1}{4}$ of the time, the probability of Bret winning is $\frac{9}{80}$ . So $a + b = 9 + 80 = 89$ .