Probability number comes up and then comes up again before another number

Question

I was trying to follow the logic in a similar question (Probability number comes up before another), but I can't seem to get it to work out.

Some craps games have a Repeater bet. You can bet on rolling aces twice before rolling a 7, rolling 3 three times before 7, etc. The patent for this game (https://patents.google.com/patent/US20140138911) says the odds for aces twice before 7 is 48:1. The wizard of odds (https://wizardofodds.com/games/craps/appendix/5/) says the probability is 0.020408 (which is 1/49).

I tried calculating this by multiplying the odds of the two events 1/36 for rolling aces and (1/36)/((1/36)+(1/6)) for rolling aces before 7. I got (1/36)*((1/36)/((1/36)+(1/6))) = 0.003968253968253969 which is like 1/252.

I'm obviously missing something, but can't see what.

Edit:...sorry...after typing this up i figured it out. The bet has to be made and then aces has to roll before 7 twice. So if 7 rolls before the first aces the bet loses, so I was wrong by using the 1/36 for the first aces.

((1/36)/((1/36)+(1/6)))*((1/36)/((1/36)+(1/6))) 0.020408163265306128

I still don't understand why one says 48:1 when its 1/49

So it seems you have solved your question. You might consider either deleting the question or posting an answer below. — Arctic Char, Aug 10 '20 at 20:07
Odds of $48:1$ are the same as a probability of $\frac 1{49}$. — lulu, Aug 10 '20 at 20:10
A friend explained to me 48:1 describes 49 choices 48 on one side and 1 on the other...hence 1/49. — shao.lo, Aug 10 '20 at 21:11

score 1 · Accepted Answer · edited Aug 11 '20 at 03:01

1

The bet has to be made and then aces has to roll before 7 twice. So if 7 rolls before the first aces the bet loses, so I was wrong by using the $1/36$ for the first aces.

$$\left(\frac{\frac{1}{36}}{\frac{1}{36}+\frac{1}{6}}\right)\cdot \left(\frac{\frac{1}{36}}{\frac{1}{36}+\frac{1}{6}}\right) =0.020408163265306128$$

edited Aug 11 '20 at 03:01

Arctic Char

16,007

answered Aug 10 '20 at 20:08

shao.lo

153

score 1 · Answer 2 · answered Aug 10 '20 at 20:26

Let $E_1$ denote the event that you roll snake eyes before a 7.
Let $E_2$ denote the event that you roll snake eyes before a 7, given that event E_1 has already occurred.

In fact, the chance of $E_2$ is the same as the chance of $E_1$.
I simply separated the events for clarity.

The key formula here is that if $A$ and $B$ are mutually exclusive events, and you are trying to compute the chance of $A$ happening before $B$, the probability is

$\frac{p(A)}{p(A) + p(B)}$,
where $p(A)$ refers to the chance of event A occurring and
$p(B)$ refers to the chance of event B occurring.

Here, there are 6 ways out of 36 to roll a 7 and 1 way out of 36 to roll snake eyes. Therefore, it is immediate that
$p(E_1) = \frac{1/36}{(1/36) + (6/36)} = \frac{1}{7}.$

Once event $E_1$ occurs the chance of event $E_2$ then occurring is similarly $= \frac{1}{7}.$

Thus the chance of events $E_1$ and then $E_2$ both occurring is
$\frac{1}{7} \times \frac{1}{7}.$

Probability number comes up and then comes up again before another number

2 Answers2