# Math Help - Magic Coin Toss aka: 3 sided die

1. ## Magic Coin Toss aka: 3 sided die

Ok, so, say we have a magic coin. The coin has a 30% chance to land heads up, a 30% chance to land tails up, and a 40% chance to land on its edge....

Or, a non-fair die that has a 30/30/40 chance of landing on 1,2 or 3 respectively...

If you flip the "Coin" 2 times, what are the chances you will get heads at least once? What are the chances you will get heads both times?

Either way, I was looking onto using the binomial distribution, but this doesn't fit. The criteria are (obviously) that the results:
• are dichotomous (nope)
• are mutually exclusive (yep)
• are independent and (yep)
• are randomly selected (yep)

So.... I was planning on using the equation

P(m of k in s from n) =

_____c(n,k)______

C(s,m)*C(n-s,k-m)

with m=1, k=2, s=30 & n=100

Unfortunately, i have no idea how to actually solve this equation as I am not in any sort of math classes. I know how to do basic algebra and geometry, but I have no idea what to do with a comma in an equation. Can someone please give me a crash course in how to actually use this equation.

Thx,
Cheers

2. Hello, Voydangel!

Ok, so say we have a magic coin.
The coin has a 30% chance to land Heads up, a 30% chance to land Tails up,
and a 40% chance to land on its Edge.

If you flip the coin twice:
(a) what is the probability that you get heads at least once?
(b) what is the probability that you get heads both times?
With only two flips, we don't need any fancy formulas . . .

We have: . $P(H) \:=\:0.3 \qquad P(\sim\!H) \:=\:0.7$

. . . $P(\text{no H}) \:=\:(0.7)(0.7) \:=\:0.49$

Therefore: . $(\text{at least one H}) \:=\:1 - 0.49 \:=\:0.51$

(b) $P(\text{HH}) \:=\:(0.3)(0.3) \:=\:0.09$

3. Ok, so that solves my immediate problem. But from a logistics standpoint, I am a little confused.

Given what you stated, if I were to extrapolate out 10 coin tosses, then the odds of getting at least one heads would be something close to 98%. (.7x.7x.7... etc). This seems odd to me. Logistically speaking, One would think that the more flips of the coin you did, you should get closer to the actual odds of getting heads (30%), but using your method, you get closer and closer to 100% with each toss.

However, I note that you said "With only two flips", which would lead me to understand that either: a) Once you go beyond 2 (or whatever), then you might actually need some sort of formula, or b) Math and logic don't always go together, or c) both a & b.

So I guess the question isn't IF my logic is flawed, but HOW...

So, lets say we do the same experiment, only this time we toss the "magic coin" 100 times. Then what?

4. Hello, Voydangel!

Given what you stated, if I were to extrapolate out 10 coin tosses,
then the odds of getting at least one heads would be something close to 98%. . Yes!
This seems odd to me.

Logistically speaking, one would think that the more flips of the coin you did,
you should get closer to the actual odds of getting heads (30%). .(1)

But using your method, you get closer and closer to 100% with each toss. .(2)
You are confusing two different problems.

(1) .You are correct.
If we flip the coin 1000 times, we would expect Heads 30% of the time.

(2) .It asks for the probability of at least one Head in, say, 1000 flips.
The probability is very close to 100% ... do you see why?

Imagine flipping this strange coin a million times
. . and praying that you do not get a Head.
Can you see that you are virtually certain to lose the bet?

5. Ah yes...it seems I am lacking the ability to turn the puzzle around in my head, as you have done twice now. (chances of NOT getting a heads...in both cases, is what solved it). I will have to practice turning things around more rather than just looking for an "in my face" number, as I am apparently prone to do.

Thanks very much
Cheers