# Pretty tricky probability "dice" question

#### bennyJ

##### New member
 Suppose.

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#### tkhunny

##### Moderator
Staff member
Not that tricky. Draw yourself a 12x12 grid and start counting!

#### Subhotosh Khan

##### Super Moderator
Staff member
Does anyone have any ideas on how to approach this problem?
Did you apply the "idea" provided above?

#### j-astron

##### Junior Member
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.

#### j-astron

##### Junior Member
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?

#### mmm4444bot

##### Super Moderator
Staff member
Is this question beyond everyone's powers?
Hmm? Are you simply waiting for somebody to complete it for you?

You were given a suggestion, in post #2.

You were asked if you tried that suggestion, in post #4. You haven't responded to that question, yet.

There's also a lengthy reply, in post #5. Did you see it?

#### Subhotosh Khan

##### Super Moderator
Staff member
Is this question beyond everyone's powers?
Nope... we are waiting for you to show your work....

#### j-astron

##### Junior Member
Well, the odds of getting a point on the first throw appear to be 7/72 since there's a 1/12 chance of getting a 13 (12/144) and a (2/144) chance of getting a 23. The chances of losing on the first throw by getting a 2, 3, or 24 appear to be 4/144, or 1/36.
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.

#### j-astron

##### Junior Member
Forever sounds daunting, but I would imagine that you're alluding to a converging infinite series.
I was!

Did you write out P(win on 2nd throw), P(win on third throw), and P(win on fourth throw)?

It should be enough to establish a pattern

#### j-astron

##### Junior Member
Since I'm not completely confident I wouldn't fudge one of the small calculations along the way (or even set them up properly), do you think there would be any way to calculate the win probability at the start of the game knowing the probability of making a point on the first throw (87.5%) and the probability of losing on any one of the subsequent throws (8.333333% per throw, 1/12, or the chances of getting 13 on each subsequent throw of the die)? I guess, by proxy, you know that the chances of winning on a subsequent throw or having to throw again would be 91.666666%).
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.

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#### j-astron

##### Junior Member
I must have made some mistakes in calculating the following outcomes because my numerator is more than 144 when you add them all up:

2 - 2/144
3 - 2/144
4 - 4/144
5 - 4/144
6 - 8/144
7 - 7/144
8 - 8/144
9 - 8/144
10 - 10/144
11 - 10/144
12 - 10/144
13 - 12/144
14 - 12/144
15 - 12/144
16 - 12/144
17 - 8/144
18 - 8/144
19 - 6/144
20 - 6/144
21 - 4/144
22 - 4/144
23 - 2/144
24 - 2/144

Whatever your point would be, I would guess you would take 1 - (probability of particular point) - 12/144 (lose on 13)
It looks like you are writing out the probability of rolling each sum, which is a prerequisite for a solution, but is not a solution itself. And yes, there are some mistakes. For example, there is only ONE way to roll a 24. For that matter, there is also only one way to roll a 2.

You also don't need all of these. You only need to know the probabilities of rolling a 13, 23, 2, 3, or 24.

Let's focus on the actual solution:

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...
I want you to actually write these out like this

P(win on second throw) = P(continue on 1st) * P(his point on second) = (____) * (_____)

P(win on third throw) = P(continue on 1st) * P(continue on second) * P(his point on third) = (____) * (_____) * (_____)

So for each subsequent throw, you write the equivalent of everything that is on the lines above, and you fill in the numbers in the blanks, and you explain where/how you got those numbers. Then we might start to get somewhere...

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