Suppose.
Suppose.
Last edited by bennyJ; 04-23-2018 at 01:53 PM.
Not that tricky. Draw yourself a 12x12 grid and start counting!
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
I would take an inventory of possible outcomes and see which ones lead to a win. What are the possible outcomes?
- win on first throw
- lose on first throw
- have to continue after first throw
These events are all mutually-exclusive, and cover all the possibilities, so their probabilities must add up to 1 (can you see why?)
P(win on 1st) + P(lose on 1st) + P(continue) = 1
What's the probability of winning? It's the sum of the probabilites of all the (mutually-exclusive) outcomes that lead to winning:
P(win) = P(win on 1st) + P(win if continuing)
Can you compute these probabilities? P(win on 1st) is straightforward, because it's just the probability of getting a couple of specific dice throws on your first try, and as Subhotosh said, that's just counting. Helpful things for that include
- what's the total number of outcomes for any given dice throw? 12 possibilities for the value of the first die, and 12 possibilities for the value of the second die. So the total number of outcomes = ?
- what Subhotosh said about constructing a 12x12 grid to count up the frequency of different sums more easily
P(win if continuing) is more interesting to compute...
I'm also wondering if it's easier to compute P(lose) and then take 1 - P(lose) to solve the problem. These are some thoughts on how to approach it.
Yeah, that would be a way to simulate the game, but how would it help solve the problem, short of playing the game thousands of times and seeing what the win ratio converges to?
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
Yeah, both of these probabilities you computed are correct. P(win on first throw) = 7/72
You are also correct that the probability of "his point" is 126/144 = 0.875.
What is P(win on second throw)?
What is P(win on third throw)?
What is P(win on fourth throw)?
You have to add all of these up...forever. Can you think of what kind of sequence you're going to end up with that has infinitely-many terms that add up to a finite sum?
Hint: P(win on second throw) = P(continue on 1st AND his point on second) = P(continue on 1st)*P(his point on second), where you take the product since these are independent events.
The Probability of winning on subsequent throws will consist of a similiar product of probabilities (continue n-1 times and get his point on the nth time). Except one more factor will be added on to that product in each subsequent term...
EDIT: This problem was a lot of fun! Slightly tricky until you see the pattern.
As I mentioned earlier, you can try to calculate the total probability of losing instead, and take 1 - P(lose). Either way, you have to set up "small calculations", one for each round. Just do it! Do it the way I suggested first, and post your work here. We can point out any mistakes and provide hints along the way. It just requires being a bit methodical. In the end, if computing the win probability directly, and excluding the first round, there are literally only two numbers that go into each term.
Last edited by j-astron; 02-19-2018 at 11:23 AM. Reason: typo
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