Probability Question: Flipping coins to determine who pays for lunch at work (elimination game) I am trying to figure out how to determine the probability of someone "winning this game" (having to pay for everyone's lunch that day. The game is played by everyone flipping a coin. Whoever is in the minority flip has to then continue flipping. i.e. if 6 people flip heads and 4 flip tails, the 4 who flipped tails have to then flip again until 1 person is left. When there are 2 people whoever hits heads first wins (a tie would be a reflip). How do I calculate the probability I would be the last person remaining? I can then figure out the cost of lunch and my EV based on it to see if it is worth flipping that day. Sorry if this doesn't make sense (software engineer on a trading desk; hence these work shenanigans haha). Feel free to ask any additional questions if I did not provide enough information.
 A: As others have pointed out, everyone has an equal chance of being the one paying the bill. But here are a couple of other considerations:
(1) If flips occur before lunch orders are placed, the non-payers may order more expensive lunches than they would have otherwise.
(2) Also if your lunch order costs more than the average lunch order for the group, this is a good game for you to play (from the point of view purely of expected value); but if it costs less than average, it is a bad game for you to play.
A: Bafs’ answer is correct, but I want to add a little bit more information and context.
As both Bafs and Gerry mention, since the game is symmetric (every player is interchangeable) each player must have the same probability of winning. To make this slightly more intuitive, suppose that Alice and Bob are both playing this game. Ahead of time, they agree that if Alice wins the game then Bob has to pay for lunch, and if Bob wins the game then Alice has to pay for lunch. Notice that this doesn’t actually change the likelihood of Alice of Bob having to pay! Indeed, we could “permute” the players in any way and everyone would still have the same probability of having to pay.
It immediately follows that with $N$ players, there is a $\frac{1}{N}$ probability of having to pay for any individual day.

Now that we understand that, we can do a little financial analysis of your lunch game. For ease of analysis, let’s suppose we’re always playing the lunch game with the same $N$ people, and everyone always gets the same thing for lunch (hence a total bill of $B$ every game).
Index the days we play the game by $1,2, \cdots$. let $P_n$ be the amount you have to pay on day $n$. We immediately see that:
$$\mathbb{E}[P_n] = \frac{B}{N}$$
That is, on average, we pay one $N$’th of the lunch bill. We may also compute variance:
$$\text{Var}(P_n) = \frac{B \cdot (N-1)}{N^2}$$
Now suppose we have play the game $K$ total days. The total amount you will have to pay is:
$$P := \sum_{n=1}^K P_n$$
It immediately follows that $\mathbb{E}[P] = \frac{KB}{N}$. Since the games day to day are independent of one another, we also see that:
$$\text{Var}(P) = \frac{KB(N-1)}{N^2}$$
Explicitly, $P$ is $B$ times a binomial random variable with $K$ trials and a probability of $1/N$. This will introduce a lot of variance on the total amount you have to pay. For example, suppose that you play this game with in a group of five people ($N=5$) and the total bill is always 50 dollars ($B = 50$), and we play five rounds.
If you simply bought your lunch everyday, you would have a total bill of 50 dollars. But if we play the game and you happen to win twice (or more), you will immediately have at least doubled your total bill! What’s the probability of winning twice? We may immediately calculate that there is about a $25\%$ of having to pay $\geq 100$ dollars. Now of course, there’s also a $33 \%$ chance that you have to pay nothing at all!
Is this a good bet? It will depend on how risk averse you are. If you like sure-fire bets, I wouldn’t play the game. One other consideration is that playing means you will be unexpectedly required to make large payments. This kind of non-smooth consumption can be difficult for people!

One last consideration comes from relaxing the “everyone gets the same lunch” constraint I imposed. Since the cost of your individual lunch is “on average” being distributed across the group, this incentivizes people to order more expensive things than they would if they were guaranteed to be paying for their own meals. This may result in everyone paying more overall than they would have had they not payed the game.
A: I agree with the comments above. The more people, the less chance of paying, but the higher the bill. The question then becomes how much you can afford, versus a lower chance of paying. For any number of people n, the calculations will simplify to $p = 1/n$ for each person. The expected value is the same for all n.
$E[C] = (1/n)(c)(n)$
where c is the same cost of a meal for each person, every day.
