I just started probability class. We learned about independent events, and I'm a little confused about it, so I thought of problem and I'll appreciate it very much if anyone can help explain this question:

Question: Using a simple computer program to row a die with six sides (1 to 6). We're going to make the following bet:

Choice 1: You get only one chance to roll the die. If the die lands on the 6 side, I'll give you \$1 million, however if the die lands on any side other than 6, then you owe me \$1 million.

Choice 2: You get 500,000 attempts to roll the die (each attempt is independent event). If just one of those attempts lands on a 6, I'll give you \$1 million, else if no attempts lands on a 6, you owe me \$1 million.

Choice 3: It doesn't matter which choice 1 or choice 2, they are the same. I'll pick whatever choice 1 or 2.

My thoughts: Intuitively, I would pick choice 2! Since the probability of getting a 6 from rolling the die in a single attempt is $\frac{1}{6}$ which imo, is quite low for a \$1 mil gamble. However, if I had 500k attempts, most likely(but not surely, since each attempt is independent) I will get at least one attempt that lands on a 6. Since the probability of getting the 6 side is $\frac{1}{6}$ which means on average (but not always) I will land a 6 for every 6 attempts. So if I had 500k attempts, the chance of me getting the 6 side must be quite high.

But on the second thought, if each attempt is independent, then it doesn't matter how many attempts I have, the probability is still $\frac{1}{6}$. So choice 1 and choice 2 yield the same chance of winning $1 mil.

However, what if I get a trillion attempts rather than just 500k. Or better, what if I get near infinite (but still finite) attempts....most likely I will land a 6. So how does independent events really work in this situation?

Also, I recall there is a method so solve "what is the probability of event A happens in the first $n$ trials". So in this case: "What is the probability of not landing on a 6 from rolling a die in the first 500k attempts". Isn't the chance of this happening very low? Thus the probability of winning $1 mil from choice 2 is greater than choice 1?


1 Answer 1


You're mainly interested in the chance of not rolling a six, as that can be expressed as the occurring of several independent events (not rolling a six on the first throw, not rolling a six on the second throw and so forth). Denote not rolling a six on the $n$-th throw by $B_n$.

The chance of not rolling a six, through independence, is

$P_(B_1 \cap B_2 \cap \ldots \cap B_{500000}) = P(B_1) \cdot P(B_2) \cdot \ldots \cdot P(B_{500000}) = (\frac{5}{6})^{500000},$

which is extremely unlikely.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .