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.
Does anyone have any ideas on how to approach this problem?
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.
Silly question:
why not use 2 stacks of 12 cards, each labelled 1 to 12,
and pick a card at random from each?
I'm just an imagination of your figment !
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?
Sorry if that confused you Benny.
All I meant was throwing a 12sided die is equivalent
to picking at random a card from a 12card deck.
Disregard my post. Thanks.
I simply meant that my question was probably silly.
Clearer would have been "May I ask a silly question".
Sorry about the confusion I created.
Last edited by Denis; 01-22-2018 at 09:22 PM. Reason: added apology.
I'm just an imagination of your figment !
Is this question beyond everyone's powers?
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