need help solving dice question odds/probability.

scruffy

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hello new to forum. could someone please help me solve this question and im not sure if I am asking for odds/probability. thanks in advance for any input. WHAT ARE THE CHANCES/ODDS/PROBABILITY OF THE #7 APPEARING SIX TIMES IN A ROW (NOT CONSECUTIVE) BEFORE A #4 WOULD APPEAR USING 2 DICE AS IN A CRAPS GAME.
 
hello new to forum. could someone please help me solve this question and im not sure if I am asking for odds/probability. thanks in advance for any input. WHAT ARE THE CHANCES/ODDS/PROBABILITY OF THE #7 APPEARING SIX TIMES IN A ROW (NOT CONSECUTIVE) BEFORE A #4 WOULD APPEAR USING 2 DICE AS IN A CRAPS GAME.
P(7) = 1/6
P(4) = 1/12
P(other) = 3/4

Think about how many throws (N) are made before a 4 shows up on throw number (N+1):
P(N) = P(not4)^N * P(4) = 11^N * 1 / 12^(N+1)

Then for a given value of N, what is the probability that 6 (or more) of those throws are 7s? Remember that you know none is a 4, so the probability of a 7 on any throw is
p = (1/6) / (11/12) = 2/11
For every value of N > 6, use a binomial distribution to find P(x>6 | N).

The probability you are looking for is
Sum[N=6, inf] { P(N) * P(x>6 | N) }

If you work out the first few terms (N=6, N=7, ...), you should find them decreasing with N, so keep as many as you need.
 
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to dr phil

Dr phil, thanks for your reply. I am not good at math nor do I understand anything about what you wrote. If its possible can you just give me an answer in laymans tems. I play craps and use a system that I have done quite well with and was just trying to figure out what I am looking at as far as risk goes with my system. Can you possibly just give me a straight up answer like 64-1 or 100-1 chance of a 7 showing six times in a row before a 4. thanks for your valuable time in advance.
 
hello new to forum. could someone please help me solve this question and im not sure if I am asking for odds/probability. thanks in advance for any input. WHAT ARE THE CHANCES/ODDS/PROBABILITY OF THE #7 APPEARING SIX TIMES IN A ROW (NOT CONSECUTIVE) BEFORE A #4 WOULD APPEAR USING 2 DICE AS IN A CRAPS GAME.
Have a look at this webpage. The coefficients of each tell you how many ways to get the sum equal to the exponent.
The \(\displaystyle 6x^7 \) says there are six ways to have a sum of seven. So the probability is \(\displaystyle \dfrac{6}{36}=\dfrac{1}{6} \) of a sum of seven. Likewise, the probability is \(\displaystyle \dfrac{3}{36}=\dfrac{1}{12} \) of a sum of four. The probability is \(\displaystyle \dfrac{27}{36}=\dfrac{3}{4} \) of any other sum .

Now we are set up to find your answer. The probability that you throw six sevens before the first four appears is \(\displaystyle \binom{N - 1}{6}{\left( {\frac{1}{6}} \right)^6}{\left( {\frac{3}{4}} \right)^{N - 7}}\left( {\frac{1}{{12}}} \right)\), where \(\displaystyle N\) denotes the number of the throw upon which the first four appears.

NOW NOTE that \(\displaystyle N\ge 7 \) and also that is exactly six sevens before the first four.
 
to w.h. auden

Thanks for your time. So is it correct to say that it would happen 1 in 12 times so therefore a 8.33% chance of it happening?
 
Thanks for your time. So is it correct to say that it would happen 1 in 12 times so therefore a 8.33% chance of it happening?
W H Auden has been dead for many years. He was a poet of note.

No that is not the correct answer.
If the first four appears on the tenth throw the answer can be found here .

I see you asked for a simple layman's answer. Well there is not one.
However you can use that webpage and change the value of N.
 
to: pka

Thank you for the link. I honestly don't know how to use it. Based on what you typed in for values am I reading correctly that it would happen 7 times in 110592
 
Thanks for your time. So is it correct to say that it would happen 1 in 12 times so therefore a 8.33% chance of it happening?
NO, the probability 1/12 refers to throwing a 4 on any particular roll. I expected that to be something a craps player would understand inherently.

The probability of NOT throwing a 4 is 11/12,
So the probability of N throws before the first 4 is
(11/12)*(11/12)*. . . *(1/12)= (11/12)^N * (1/12)
That tells you how long a run you can expect without a 4 being thrown. [Note that I have defined N as the number of not-4s, while the form used by pka is N= the first throw that is a 4.]
Code:
 N     P(N)    P(6|N)     P(7|N)     P(8|N)  . . .
 0    0.0833
 1    0.0764
 2    0.0700
. . .
 6    0.0494  0.0000214
 7    0.0453  0.0001125  0.0000036
 8    0.0415  0.0003376  0.0000214  0.0000006  
. . .
For example, the probability is 0.0415 that there will be exactly 8 throws before the first 4. That is one chance in 24. If that is the value of N, then the probability that 6, 7, or 8 of those are 7s is 0.0003596. There is a chance of 1/67000 that there will be 8 throws preceding the 4, AND that at least 6 of the 8 are 7s.

To get the total probability, you have to add up ALL values of N. The best way to do that would be a spreadsheet.

If you don't understand probability, you are the kind of guy the house loves to see at the craps table. Most "systems" work most of the time, IF you have a big enough bankroll to cover adverse runs. You wind up risking a LOT of money to expect a small positive return. You can continue to make small gains, right up to the day your bank goes bust. It is likely your bankroll is not as large as the house's.
 
to: dr phil

Thanks for time again.. yes I understand fully that there are 36 combinations of dice and that only 3 combinations add up to 4 and that 6 combinations add up to 7. let me explain the way I play a little and perhaps I can get an answer to my question/dilema in a way that I can understand so I can be more aware of my risk the way I play. I play 5 units on the pass line on the come out roll and 4 units on the don't pass line. In my calculations I break even on that bet on the come out over millions of rolls. I then play 5 units if the point is 4,5,9,10 on the pass line odds (true odds). I track each number as a separate profit center and will press into it 6 times until I get a positive outcome (hit) on that number. My press bets are staggered so I end up with the same positive units on a favorable outcome no matter if its the first bet or the consecutive press bets. So my dilemma or question is trying to find out exactly how many times/frequency/odds I would actually lose all 6 bets in a row when pressing into the 4,5,9,10 in this manner in a way I can understand it. Yes I am somewhat educated but cant figure this out.
 
to: dr.phil

Here is an example of units I play on the pass line odds chasing a 4 as the point. 4 6 9 14 21 31 if they 7 out 6 times in a row I am - 91 units between the pass line and pass line odds. If they make the point in one of those 6 attempts I make +8 units and start over again.
 
to: denis

denis, thanks for input and time. are you saying that it would take 800000+ attempts to make 6 number 7s in a row before a 4 would appear or am I missing something? I don't have the same vocabulary as all the brains in this forum.
 
to denis

denis in terms of the 7:80 am I correct in saying that if I attempted it 87 times I would lose 7 times and win 80 times?
 
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