why is craps probability so difficult?

I took physical chemistry in college which is thermodynamics and quantum mechanics. Those complicated differential equations were way easier than craps math.

Anyway,

Can someone explain to me like I am 10 years old - how come the probability of rolling a 10 before a 7 is 1/3?

That makes no sense to me. Because the chance of rolling a 7 is .166666666666, and the chance to rolling a 10 is 0.0833333333333 (exactly 1/2).

So how come I can't say....I am twice as likely to roll a 7 then I am a 10? How come I can't say, I am half as likely to role a 10 than I am a 7?

I found this somewhere and don't understand it at all. I don't understand while it is an infinite equation.

"Let pr(x) stand for the probability of event x happening on any given roll. The answer is:

pr(4) +
pr(anything other than 4 and 7) * pr(4) +
pr(anything other than 4 and 7)2 * pr(4) +
pr(anything other than 4 and 7)3 * pr(4) +
pr(anything other than 4 and 7)4 * pr(4) +
+ ...

Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.

pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)2 * 1/12) + ((3/4)3 * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)i
= 1/12 * (1/(1-3/4))
= 1/12 * 4 = 1/3.
"

Quote:

So how come I can't say....I am twice as likely to roll a 7 then I am a 10? How come I can't say, I am half as likely to role a 10 than I am a 7?

What's wrong with that?

If you have 3 red marbles and 6 blue marbles and 50 white marbles and the white marbles don't matter.....there are only 9 marbles that matter. Chance of a red marble being picked before a blue is 3/9 or 1/3. Chance of blue before red is 6/9 or 2/3. The difference (or rather, ratio between the two) is 1:2.

Quote: RS

Quote:

So how come I can't say....I am twice as likely to roll a 7 then I am a 10? How come I can't say, I am half as likely to role a 10 than I am a 7?

What's wrong with that?

If you have 3 red marbles and 6 blue marbles and 50 white marbles and the white marbles don't matter.....there are only 9 marbles that matter. Chance of a red marble being picked before a blue is 3/9 or 1/3. Chance of blue before red is 6/9 or 2/3. The difference (or rather, ratio between the two) is 1:2.

Thanks. The marble analogy helps.

however, I'm still struggling here - because I can't resolve this idea that I am half as likely to roll a 10 than I am a 7, yet the chance of that happening is only a 33%.

I'll think about it.

Okay RS -

Thanks for the analogy. I think I get it.

But can you help me take this a little further?

If I want to know the probability of drawing a red before a blue, 5 times in a row - it is (1/3)^5 = about 0.4%

Now, let's say the number 4,5,6,8,9 represent yellow marbles. So the chance of drawing a yellow (21 marbles) before a red is 87.5% (and a 12.5% chance of drawing red).

Okay - now lets say I want to know the probability of drawing a yellow marble 7 times before a red. I assume that is 0.875^7 - or about 39% chance.

Now if I want to double that, can I just do the ratio? For example, now I want to say - what is the probability of drawing 14 yellow before drawing 2 red - what is the math or ratio for that?

Yes, it'd be (1/3)^5, for drawing at least 5 reds in a row before a blue. If you want to do 5 reds THEN a blue without a 6th red, then you'd do:

[(1/3)^5] * 2/3


But, close enough.



Okay, 21 yellow marbles, for 4-9 [no 7], is 21/(21+3) = 21/24 = 7/8 = 0.875 = 87.5%.


And yeah, 0.875^7 (or [7/8]^7) would be 7 yellows before a red.

Quote:

Now if I want to double that, can I just do the ratio? For example, now I want to say - what is the probability of drawing 14 yellow before drawing 2 red - what is the math or ratio for that?

Sorry, can't help here. :( Although I'm almost certain the formula would be different altogether (ie: can't multiply it by 2, divide by 2, or use the same formula).

Quote: slackyhacky

however, I'm still struggling here - because I can't resolve this idea that I am half as likely to roll a 10 than I am a 7, yet the chance of that happening is only a 33%.

Physical chemistry, you say? Think about it this way:
1/3 of the atoms in a volume of water are oxygen, and 2/3 are hydrogen.
There are twice as many hydrogen atoms as there are oxygen, but that doesn't mean that 50% of the atoms are oxygen.

One key concept of probability: the sum of the probabilities of each result of a particular event happening = 1.
Let p denote the probability that a 10 is rolled before a 7; you already know that 2p is the probability of a 7 being rolled before a 10. These are the only two possible results of "will a 7 or a 10 be rolled before the other number?", so p + 2p = 1, which means p = 1/3.

Be warned; it's not always that simple. For example, calculating the probability of winning a pass line bet involves determining the probabilities of each come-out roll, then, for each one, multiplying that probability by the probability of winning from there (for point numbers, calculating the probability that the number will be rolled again before a 7; for other numbers, the winning probability is 1 for a 7 or 11, and 0 for a 2, 3, or 12), and finally, adding up the numbers. Some multi-roll bets, like the "fire bet," require some serious hoops to jump through.

Quote: slackyhacky

however, I'm still struggling here - because I can't resolve this idea that I am half as likely to roll a 10 than I am a 7, yet the chance of that happening is only a 33%.

I'll think about it.

33% IS half as likely as 66%.

Quote: slackyhacky

Quote: RS

Quote:

So how come I can't say....I am twice as likely to roll a 7 then I am a 10? How come I can't say, I am half as likely to role a 10 than I am a 7?

What's wrong with that?

If you have 3 red marbles and 6 blue marbles and 50 white marbles and the white marbles don't matter.....there are only 9 marbles that matter. Chance of a red marble being picked before a blue is 3/9 or 1/3. Chance of blue before red is 6/9 or 2/3. The difference (or rather, ratio between the two) is 1:2.

Thanks. The marble analogy helps.

however, I'm still struggling here - because I can't resolve this idea that I am half as likely to roll a 10 than I am a 7, yet the chance of that happening is only a 33%.

I'll think about it.

Is it the odds part that's giving you a hurdle, or is it the infinite series reduction? If you have 3 white marbles in an urn, one with "10" written on it and two with "7" written on it, the chances of drawing the 10 are 1 in 3, also known as 2-to-1 against. Odds and probabilities (chances) are related but not identical. The probability of a discrete event occurring is stated as "number of ways to succeed" / "number of total ways", but the odds against the same discrete event are stated as "number of ways to fail" to "number of ways to succeed". Flip the odds over for the "odds of" an event.
Here's a link with some visuals:
http://stats.seandolinar.com/statistics-probability-vs-odds/

For the infinite series reduction, a good analogy is if you take the urn above, with the three white labelled marbles and dump in a bunch of black ones. If you pull one of those, it doesn't count as a result either way so you have to keep drawing. Eventually you'll pull one of the white ones, and the odds of doing that are the same as when there were no black marbles at all. Mathematically, here's how you get to the infinite geometric series SUM(r^i) = 1/(1-r):
http://mathworld.wolfram.com/GeometricSeries.html

Thanks for the link - it helped too.

I think I'm getting somewhere.


Okay, so -

For my problem, let me try to work it out here - and see if it is correct.

Definitions first
P(A) = probability of a 10 before a 7 = 1/3 = .33333
P(B) = probability of a 7 = 6/36 = .166
P(C) = probability that a 4,5,6,8 or 9 rolls before a 10 = (21/36)/(24/36) = 0.875

What is the P(A) 5 times before a 7? (1/3)^5 = .004115 (lets call this even P(Q))
What is the P(C) 7 times before a 10? 0.875^7 = .393 (lets call this event P(Y))

So if I say

P(x) = probability of event A happening/Probability of all events -

and my question is - What is the probability of event event Y or P(B) happening before event Q?

Is the formula then P(x) = [P(Y)+P(B)]/[P(B)+P(Q)+P(Y)] ? This calculates to 0.559/.563 = .992

What if I ask, what is the chance of rolling a 4,5,6,8 or 9 fourteen times, or rolling a 7 once, before rolling a 10 3 times in a row?


To roll a 10 three times in a row = (1/3)^3 = 0.037
To roll 4,5,6,8, or 9 before a 10 fourteen times = 0.875^14 = 0.154

Can I then do the same thing? P(x) = (0.1666+0.154)/(0.16666+0.154+0.037) = 0.321/0.358 = 0.897

You're over thinking it. You're comparing each numbers probability on being rolled when you should be comparing the 7 vs 10/4 or 7 vs 5/9 or 7 vs 6/8.

The first thing is to look at the 2-dice WAYS table. The odds of an individual come point hitting before a seven is 1/2 (4 and 10 points), 2/3 (5 and 9 points), and 5/6 (6 and 8 points). This is simply the count of ways to make a 7 (always 6) vs. the counts of ways to make the points;

There are 3 ways to make the 4 or 10 point vs. 6 ways to make a 7... 6 to 3 = 2 to 1.

(so that is 6 ways out of 9 to lose, 3 ways out of 9 to win, that's where the 1/2 - 1/3 confusion is)

There are 4 ways to make the 5 or 9 point vs. 6 ways to make a 7... 6 to 4 = 3 to 2.
There are 5 ways to make the 6 or 8 point vs. 6 ways to make a 7... 6 to 5.

Also, the odds of the come bets are different from the place bets... 9 to 5 on the 4 or 10,, 7 to 5 on the 5 or 9, and 7 to 6 on the 6 or 8.

Much harder is being able to calculate the odds of winning/losing the pass line bet. Here is a link to a .txt file that I generated that discusses the process of calculating the odds of winning/losing a pass-line bet in the games of CraplessCraps and then again in the game of TraditionalCraps...

http://spikersystems.com/FlashNet_Pointer/www/downloads/SpikerSystems/e-Gamers/AAA_Information/Pass_line_odds_output.txt
(see sections 1 through 4)

I have a lame YouTube video that discusses this .txt file...
https://youtu.be/KBn9DMH4g-k?t=170
(watch from 2:50 to 9:30)