# Pretty tricky probability "dice" question

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• 01-20-2018, 02:37 PM
bennyJ
Pretty tricky probability "dice" question
 Suppose a modified version of the dice game craps is played with two regular (i.e., perfectly symmetrical) dodecahedra. Each die has its sides numbered from 1 to 12 so that after each throw of the dice the sum of the numbers on the top two surfaces of the dice would range from 2 to 24. If a player gets the sum 13 or 23 on his first throw (a natural), he wins. If he gets 2, 3, or 24 on his first throw (craps), he loses. If he gets any other sum (his point), he must throw the dice again. On this or any subsequent throw the player loses if he gets the sum 13 and wins if he gets his point but must throw both dice again if any other sum occurs. The player continues until he either wins or loses. To the nearest percent, what is the probability at the start of any game that a dice thrower will win?
• 01-21-2018, 12:15 AM
tkhunny
Not that tricky. Draw yourself a 12x12 grid and start counting!
• 01-21-2018, 09:51 AM
bennyJ
Does anyone have any ideas on how to approach this problem?
• 01-21-2018, 11:02 AM
Subhotosh Khan
Quote:

Originally Posted by bennyJ
Does anyone have any ideas on how to approach this problem?

Did you apply the "idea" provided above?
• 01-21-2018, 01:05 PM
j-astron
Quote:

Originally Posted by bennyJ
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.
• 01-21-2018, 05:13 PM
Denis
Silly question:
why not use 2 stacks of 12 cards, each labelled 1 to 12,
and pick a card at random from each?
• 01-21-2018, 06:23 PM
j-astron
Quote:

Originally Posted by Denis
Silly question:
why not use 2 stacks of 12 cards, each labelled 1 to 12,
and pick a card at random from each?

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?
• 01-22-2018, 12:13 AM
bennyJ
Quote:

Originally Posted by Denis
Silly question:
why not use 2 stacks of 12 cards, each labelled 1 to 12,
and pick a card at random from each?

Does that take into account the probability of drawing again (potentially infinitely) if you don't win on the first try?
• 01-22-2018, 09:24 AM
Denis
Quote:

Originally Posted by bennyJ
Does that take into account the probability of drawing again (potentially infinitely) if you don't win on the first try?

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.
• 01-25-2018, 09:07 AM
bennyJ
Is this question beyond everyone's powers?
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