Basic probability issue confusion: Please help

[Benga]

Hello, FMH. I hope that one of you may be able to help me with this simple problem.

So suppose you have a coin, where you win a dollar if it comes up heads, or you win nothing if it is tails, but you get to re-flip the coin one time if it is tails (you get a second attempt to get heads and win the dollar). If you got heads the first time, you take the dollar and you're done.

So the odds could be calculated (maybe correctly, maybe not) as follows:

Option 1: Heads comes up.
Option 2: Tails, then the re-flip is heads.
Option 3: Tails, then the re-flip is tails. So it would seem as though there is a 2/3 chance to win the dollar.

The same math should apply to a die roll, if 4-6 gets you the dollar, and 1-3 gets you a single re-roll. It should be the same exact odds as the coin flip, right? Each time you roll the die, you have a 50% chance of a 4 or higher.

But the options now are:
Option 1, 2 and 3: 6 comes up; 5 comes up; 4 comes up. You get the dollar.
Options 4 through 9: 3 comes up, and then your re-roll will give you 3 / 6 possibilities of you winning the dollar (the re-roll would be a 4 or higher in 3 of the 6 re-rolls).
Options 10 through 15: 2 comes up, and then the re-roll gives a 3 / 6 chance at the dollar (same math).

Options 16 through 21: 1 comes up, and then the re-roll gives a 3 / 6 chance at the dollar (same math).

Here, the total number of options is 21, and the total number of wins is 12 (3 + [3 * 3]). 12/21 is 4/7, not 2/3, even though it should be the same odds as the coin flip.


So maybe the correct way to do this is that I have to account for the hidden "options" for when I flip/throw a successful coin/die, since otherwise I'm skewing the result by over-sampling the first-toss misses. In other words, maybe the actual dynamic is "you always get 2 tosses to win, it's just that you're automatically guaranteed a win if your first toss is a win, but you still have to throw the second toss for probability's sake."
In this case, there are 4 options for the coin:

Heads-Heads; Heads-Tails; Tails-Heads; Tails-Tails.

So 3 of the 4 combinations win the dollar, or a 3/4 chance.


So which is it: Do I have a 2/3, 4/7 or 3/4 chance? And why can't I find a way to re-produce this experiment from one medium (the coin) to the other (the die) without coming up with different numbers? What am I doing wrong?

 

[ecoo]

Your mistake is assuming, for example in the coin flip game, that all of the 3 options have equal probabilities of occurring.

All probabilities of a game should add up to 1.

For the coin flip, winning is considered gaining a dollar or more, so winning encompasses two different scenarios (heads, or tails then heads). The probability of getting a heads is \(\displaystyle \frac{1}{2}\), and the probability of getting a tails then heads is \(\displaystyle \frac{1}{2}*\frac{1}{2}=\frac{1}{4}\), meaning the probability of winning is \(\displaystyle \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\).

Losing encompasses one scenario, which is to receive a tails and then another tails. The probability of this occuring is \(\displaystyle \frac{1}{2}*\frac{1}{2}=\frac{1}{4}\).

The same reasoning occurs with the dice game.

 

[pka]

So suppose you have a coin, where you win a dollar if it comes up heads, or you win nothing if it is tails, but you get to re-flip the coin one time if it is tails (you get a second attempt to get heads and win the dollar). If you got heads the first time, you take the dollar and you're done.

So the odds could be calculated (maybe correctly, maybe not) as follows:

Option 1: Heads comes up.
Option 2: Tails, then the re-flip is heads.
Option 3: Tails, then the re-flip is tails. So it would seem as though there is a 2/3 chance to win the dollar.

The same math should apply to a die roll, if 4-6 gets you the dollar, and 1-3 gets you a single re-roll. It should be the same exact odds as the coin flip, right? Each time you roll the die, you have a 50% chance of a 4 or higher.

So maybe the correct way to do this is
Heads-Heads; Heads-Tails; Tails-Heads; Tails-Tails.

So 3 of the 4 combinations win the dollar, or a 3/4 chance. CORRECT!

In effect the coin is flipped just twice. The outcomes are:
\(\displaystyle \begin{array}{*{20}{c}}
H&H\\
H&T\\
T&H\\
T&T
\end{array}\)
Clearly you win in three of the four possible outcomes.
BTW. This is a most difficult lesson for a beginner to learn:
List all possible outcomes regardless.

 

[Benga]

Thank you for your help, PKA.