Probability of rolling a sum of 6 before a 7 or 9

  • #1
Having a lot of trouble with this question.

So first I tried making an equation, and I wrote that the probability = P(rolling a 6)+P(rolling a 6 and not a 7 on the first roll and not a 9 on the first roll) + P(rolling a 6 and not a 7 on the first roll and not a 9 on the first roll and not a 7 on the second roll and not a 9 on the second roll) + ...

At this point I said let P(6) = x, P(not a 7) = y, P(not a 9) = z.

So I have P= x+xyz+xy^2z^2+... Factor out the x;
x(1+yz+y^2z^2+...)

Then I rewrote that as x*the sum of (y^i)(z^i) from i = 0 to infinity. Using an identity I got this to be x*[1/(1-yz)]. I subbed the numbers in to that, and got .5357 as an answer. Now, I don't know what the correct answer is but that seems wrong to me.

Can anyone help?

Thanks
 

Answers and Replies

  • #2
144
0
I assume you are throwing two dice, and that the game ends once the sum is 6, 7 or 9, and that you seek the probability for the game to end with a 6.

If the game ends after the first throw, can you find the probability for a 6?

What can be said about the next throw, and the ones thereafter? (the same)

If you need to be further convinced, write down, and compute, the corresponding infinite sum of probabilities, and you will again get the same answer.
 
  • #3
So I do have the right answer?
 
  • #4
315
1
You first mistake is that the events of not having a 9, and of not having a 7, are not independent, so you can't multiply them. Your second mistake is that the the event of getting the 6 on the second roll requires no 9, 7, OR 6 on the first roll.

If instead we have the single event: w = "the probability the roll is not a 9 or a 7 or a 6" then the probability is
(x + xw + xw^2 + xw^3 + ...) = x (1 / (1-w))
substituting x = 5/36, w = (1 - (3/36 + 6/36 + 5/36)), the probability is 0.35.
 
  • #5
144
0
So I do have the right answer?

Sir, I had to make a guess at what your question was, so I suggest you first clear that up. Assuming for a moment that my guess was correct, and that we are throwing two dice, and that you are aware that throwing 7 is the most likely, I do not understand how you can possibly consider the chance of getting a 6 first to be more than 53%.
 
  • #6
144
0
You first mistake is that the events of not having a 9, and of not having a 7, are not independent, so you can't multiply them. Your second mistake is that the the event of getting the 6 on the second roll requires no 9, 7, OR 6 on the first roll.

If instead we have the single event: w = "the probability the roll is not a 9 or a 7 or a 6" then the probability is
(x + xw + xw^2 + xw^3 + ...) = x (1 / (1-w))
substituting x = 5/36, w = (1 - (3/36 + 6/36 + 5/36)), the probability is 0.35.

You should have another go at correcting your answer. But you are getting closer. (Whats the chance of throwing 9?)
 
  • #7
Okay, that makes sense now (mostly). Wouldn't x and w be dependent as well?
 
  • #8
2,017
291
Okay, that makes sense now (mostly). Wouldn't x and w be dependent as well?

no. in the expression (x + xw + xw^2 + xw^3 + ...)

w in xw is the probability 6,7 or 9 wasn't thrown on the first turn, and x is the probability 6 is thrown on the second turn. These are independent.

in xw^2, w^2 is the probabiltiy 6,7 or 9 wasn't thrown the first two turns, and x is the probability 6 is thrown the 3rd turn etc.

The addition of x,xw,xw^2 etc. is all right too, as these are the probablities of the mutually exclusive events that a 6 is thrown first on the 1st, 2nd, 3rd,..... turns
 

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