56

It doesn't matter which option you choose, because Your probability of survival if you're one of n players left is as follows: Informal proof It was established in the question that if there are only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate both your opponents, which can only ...


36

The answers of rand al'thor and Callidus are great; I just want to give a different argument for the result. Claim: After each round, the number of surviving players is even. Proof: Let $f_i$ be the flip of player $i$, with tails $1$ and heads $0$ (it's arbitrary which is which). Let $s_i$ be $1$ if player $i$ survives the round and $0$ otherwise. Then, ...


33

They could toss it twice HT = first player wins; stop TH = second player wins; stop HH or TT = ignore these two tosses; repeat the procedure


26

Wendy wins on the first flip with probability $1/100$. Otherwise, the game keeps going and Sally has probability $p$ to win on the next flip, which has overall probability $99/100 \times p$. If not, the game returns to the start, which has no effect on fairness. So, to be fair, these two probabilities should be equal, which happens for $p=1/99$.


26

As Timmy has a "head" on his shoulders with probability of 1, he always wins at his turn. Then


23

Case 1: Waiting for HH $A$ = average nb. of flips to get HH after T or nothing. $B$ = average nb. of flips to get HH after H. A is 1 flip, plus B if you get H and continue, or plus A if you get T and restart: $A = 1 + {1\over2}B + {1\over2}A$ B is 1 flip, and you are done if you get H, or plus A if you get T and restart: $B = 1 + {1\over2}0 + {1\over2}A$ ...


18

The answer is to simply flip the coin once, but make sure you catch it without letting it bounce. The question makes the assertion that a weighted coin will be biased, but that only happens if the coin is allowed to bounce and/or spin (on a surface). For a flipped/caught coin, there is no significant bias. The law of conservation of angular momentum ...


16

If there are two players left, both lose. If there are three players left: If there are four players left: If there are more:


14

All numbers from 0 to 10 are equally likely. Here is one way to flip a coin with bias $p$. Choose a random number uniformly from $(0,1)$, and pick heads if that number is less than $p$. This gives us a way to simulate Bob's experiment. We choose $p$ uniformly from $(0,1)$, and then choose ten coin-flip numbers uniformly from $(0,1)$. The number of heads is ...


13

$p$ is a random variable, chosen with uniform probability over the interval [0,1]. The probability of two heads is $\int_0^1 p^2=\frac{1}{3}$, and by symmetry so is the probability of two tails.


12

The game is fair for Proof:


12

The answer is Proof


10

Lemma: On any given turn, an even number of players are eliminated. Proof: Go around the circle, skipping every other coin. Every time the coin you see changes from tails to heads or vice versa, the person you skipped over gets eliminated. You must see an even number of changes before returning to the start, so there are an even number of eliminations in ...


9

Lopsy answered 11. rand al'thor then asked to consider e.g. 9 and 13. I will give a recursive formula for the probability of winning for a game of $N$ players. According to Lopsy's Lemma, for $N$ even there is no chance of winning so I consider only odd $N$. First, let me define notation for two types of sequences of coins. I use $B(n_1)$ for a "block" ...


9

Frequentists won't be damned. This is a trick question. In looking at making a coin, you can't bias it by weighting. There is an entire class on such myths and methods of bias at Berkley (at least there was some years ago). I attached a link to a paper that describes actions such as taking a plastic ship and adding heavy putty to one side to see if you ...


8

The answer is 0.0101010101... Proof Say that $w$ is the chance of Wendy’s coin coming up heads; $s$ is the chance for Sally’s coin. Wendy will win with probability: $$p(W)= w + w's'w + w's'w's'w + w's'w's'w's'w ... = w ( 1 + (w's')^2 + (w's')^3 ... )$$ Sally wins with probability: $$p(S) = w's + w's'w's + w's'w's'w's ... = s ( w' + w'^2s'^1 + w'^3s'^2 + ...


8

Suppose instead that But in this new situation, For the game to be fair, the amount he wins should be


7

It is: because:


7

It's pretty easy to calculate the exact result using a computer. I haven't found an easy way to calculate it by hand (yet), but if someone wants to check his result before posting: Update: Optimized version, counting only matching combinations. It's now possible to verify, that the number of matches for all odd $n$ up to $11$ is listed in A007079. public ...


7

I have an idea. You take a normal coin. It is evenly weighted on each side (if you cut it exactly in half along the side the two halves would weigh the same). Now So However So Diagram of idea (apologies for bad freehand writing): It doesn't have to be scratch off ink or solder - it can be any material that can be eventually scratched off by a nail. ...


6

The probability is given by $P(fair|heads^m) = P(heads^m|fair) \cdot P(fair) / P(heads^m)$, where $P(heads^m) = P(heads^m|fair)\cdot P(fair) + P(heads^m|nonfair)\cdot P(nonfair) = 1/2^m \cdot K/N + 1^m \cdot (N-K)/N$. Hence, $P_m = \frac{1/2^m \cdot K/N}{1/2^m \cdot K/N + 1^m \cdot (N-K)/N)} .$ Solving for $K$, we find that it must be 88.8.


6

Making the events independent, we get a probabilty that is $p_{i,j} = \mathbb{P}(\text{player }i\text{ wins }j\text{ games}) = {6 \choose j}\cdot 2^{-6}$. (note: if six players wins 3 games, then the last player will also win 3 games) So, for all players, assuming independent events (which clearly is not true), we get $\prod_i p_{i,3} = \left({6 \choose ...


6

Unless I'm missing something This means


6

Yes, we can have a game in which So, And Thus, Assuming that the results from two coins tosses are independent. In simple words, Thus, there's no need of biased coins.


6

The number of possible outcomes for $100$ coin tosses is simply $2^{100}$, which will be our denominator. For the numerator we must count the number of outcomes without $2$ successive heads. Let us define 2 functions: $H(n)$ giving the number of outcomes for $n$ coin tosses without $2$ successive heads which end with heads $T(n)$ giving the number of ...


6



5

This seems to be mathematically impossible. The probability of two heads is $1/3$ and the probability of two tails is $1/3$. Since they're equal, this gives us $$ p^2 = (1-p)^2 $$ $$ p = 1-p $$ $$ p = 1/2 $$ But $(1/2)^2 = 1/4$, not $1/3$. If there really is a solution to this I'd be really interested to see it.


5

The answer: Reasoning: The probability of getting $k$ heads is $$p(k) = \int_0^1 \binom{10}k x^k(10-x)^{10-k}dx$$ Integrating $p(k+1)$ by parts yields $$p(k+1) = \int_0^1 \binom{10}{k+1}\frac{k+1}{10-k}x^{k+1}(1-x)^{10-k}dx - \left(\left.\frac 1 {10-k} x^{k+1}(1-x)^{10-k}\right|_0^1\right)$$ The last term vanishes for all k between 0 and 9 (it's undefined ...


5

Each person tosses the coin 3 times, whoever gets more heads wins, if tie then do it again. If ties keep occurring then toss it 5 times or more to decide OR each person tosses it once whoever gets tail wins. If tie or no one gets tail then each person tosses it again until someone wins.


5

HH Following the method outlined in this answer on Math.SE, we get the following: Let $e$ be the expected number of tosses. Start tossing. If we get a tail immediately (probability $\tfrac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\tfrac{1}{4}$), then the expected number is $e+2$. Finally, if our first $2$ ...


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