Alternating dice roll game Imagine we played a game using a dice. I start, then you get a turn, then me again and so on. The first person to get a 6 wins. What is my probability of winning if I start?
Now consider we did the same but with a coin. Without calculation, would this probability increase or decrease?

Here's what I was thinking:
On average, you get 6 every $6$ rolls. So on average, the $6$th person to roll will win. So you going first doesn't change the probability (three rolls each), which must therefore be $\frac 12$.
Using this logic for a two-sided die, we'd get the same probability: $\frac 12$.
These results are extremely unintuitive to me and and almost certainly incorrect. I'd like to know (1) the flaw in the reasoning, and (2) how to get the correct solution.
 A: Let $P$ be the probability that the person about to roll wins the game.    
Let $p$ be the probability that a particular roll is a winning result.
Therefore $P= p+ (1-p)(1-P)$, because either the person about to roll immediately obtains a winning result, or they don't and then their opponent is about to roll next.   With a little arithmentic rearrangement :
$$P= \frac{1}{2-p}$$
This is also the probability that whoever starts the game wins the game.


On average, you get 6 every $6$ rolls. So on average, the $6$th person to roll will win. So you going first doesn't change the probability (three rolls each), which must therefore be $1/2$.

Sure, the average count for rolls until winning is $6$, but this sais nothing about the probability for who wins.   The probability for obtaining this expected value is only about $0.067$, and it is but one outcome in the event for you not winning.
You want the probability that the count for rolls until winning is an odd number.$$P~{=\sum_{k=0}^\infty \mathsf P(X=2k+1)\\=\sum_{k=0}^\infty (1-p)^{2k}p\\ = p+(1-p)^2p+(1-p)^4p+\cdots}\qquad\text{Where }X\sim\mathcal {Geo}_1(p)$$
A: If you both were to throw at the same time, then yes, you would both have an equal chance of winning. But, since you go first, you have an advantage, so you should have more than a 50% chance of winning.
Moreover, with a coin you have a pretty good chance of winning any time it is your turn, so that advantage should be greater than the advantage you have with throwing a die.
Indeed, consider the extreme case where we each would only have a 1 in 1000 chance of winning whenever it is out turn. Since the game would probably last many turns, the advantage will now only be very small, and we're close to being back to a 50/50 game (though you will always keep a slight advantage!)
A: I like to treat such problems as finite state machines. My answer isn't as simple as Graham Kemp's approved answer, but this approach can be used for more complicated scenarii.
There are two states, $A$ (you're about to roll) and $B$ (your opponent is about to roll), not counting terminal states (you've won/lost).
Define $W_A$ and $W_B$ as your probability of winning in each state.
$$W_A = \frac{1}{6} + \frac{5}{6}W_B$$
because, one time in six, you win, and the rest of the time the new state is $B$.
$$W_B = \frac{5}{6}W_A$$
with no constant term because you can't win (immediately) from state $B$.
Put these together and we have
$$W_A = \frac{1}{6} + \frac{25}{36}W_A$$
$$(1-\frac{25}{36})W_A = \frac{11}{36}W_A = \frac{1}{6}$$
$$W_A = \frac{6}{11}$$
So the first player wins six times out of eleven.
A: Consider your dice game. You go first:
Probability(you win) = 1/6
P(we keep playing) = 5/6
My go:
P(I win) = P(we're still playing) * 1/6 = 5/6 * 1/6
P(we keep playing) = P(we're still playing) * 5/6 = 5/6 * 5/6
Your go:
P(you win on this roll) = P(we're still playing) * 1/6 = 5/6 * 5/6 * 1/6
P(we keep playing) = P(we're still playing) * 5/6 = 5/6 * 5/6 * 5/6
My go:
P(I win on this roll) = P(we're still playing) * 1/6 = 5/6 *5/6 * 5/6 * 1/6
P(we keep playing) = P(we're still playing) * 5/6 = 5/6 * 5/6 * 5/6 * 5/6
Your go:
P(you win on this roll) = P(we're still playing) * 1/6 = 5/6 *5/6 * 5/6 * 5/6 
* 1/6
P(we keep playing) = P(we're still playing) * 5/6 = 5/6 * 5/6 * 5/6 * 5/6 *5/6
That's far enough and you should be able to see the pattern emerging. So the probability that you win is the sum of all the probabilities that you win when it's your turn to roll the dice:
P(you win) = 1/6 + 1/6 * 5/6 * 5/6 + 1/6 * 5/6 *5/6 * 5/6 * 5/6 + ...
= 1/6 (25/36 + (25/36)^2 + (25/36)^3 + ...)
A geometric progression with a = 1/6 and r = 25/36. So the probability that you win = a/(1-r) = (1/6)/(1-25/36) = 0.545454 = 54/99
Your chance of winning is skewed to be greater than 50% by virtue of the fact that you go first! You can work out the exact probability for your coin toss game using the same reasoning, but the chances of you winning will be greater than 54/99 as the increased chances of you winning on the first toss will skew the results further in your favour. 
A: In the dice game, it's clear that the probability of winning depends on whether we go first.  Let's call that probability $p$.  If we lose on the first attempt, our opponent will now be going first in the "rest of the game", so his chance of winning in that scenario will also be $p$ and ours will be $1-p$.
We have a $\frac 1 6$ chance of winning on that first go, so our overall winning probability is $\frac 1 6+\frac 5 6 \times (1-p)$ and by definition this is equal to $p$. 
Solving $p=\frac 1 6+\frac 5 6 \times (1-p)$ gives $p=\frac 6 {11}$.
In general, if our game has $n$ outcomes, the winning probability for the player going first is $\frac {n} {2n-1}$. For the coin-toss game where there are only 2 outcomes (and, say, first "Heads" is the winner), the advantage to playing first increases to $\frac 23$, the most for any game of this first-to-a-winner type.
If we had a game with lots of outcomes such as "keep drawing (with replacement) from a deck of cards until someone draws the Ace of Spades", then the advantage of playing first will be very small: our winning probability would be just $\frac {52}{103}$ which is barely more than 0.50 (even chance).
A: Let's say your opponent gets to throw after you regardless of what you throw, without affecting who wins (so if you get a six, he knows he loses, but he still gets to throw one last time). In that case, the game can be divided into rounds, where in a round you both throw the die once. A single round can go one of four ways:


*

*None of you get a six. In this case, you play another turn.

*You get a six, but not your opponent. In this case, you win.

*Your opponent gets a six, and you don't. In this case, your opponent wins.

*You both get a six. In this case, you win, since you threw first.


Note that 1. is completely symmetric, and 2. and 3. are evenly matched. Together they comprise $\frac{35}{36}$ of all rounds that will ever be played, and they are fair. However, in one round out of $36$, we will get into case 4, which is strictly in your favour.
As for calculation of probability, each game ends in exactly one of the three cases 2., 3. or 4. Case 2. and 3. each account for $5$ out of the $36$ possible die throw results, and case 4. accounts for $1$. Thus there are $11$ equally probable die throws that end the game, and six of them are in your favour. The probability of you winning is therefore $\frac{6}{11}$.
If you flip a coin instead, the probability of case 4. is drastically increased, so you have an even higher chance of winning.
