Calvin and Dan's coin game

Since you are also asking about a generalization, I'll do the general derivation, and then look at the case $n=4$ (which means $2n=8$ We first note that the coin is fair, so $P(heads)=P(tails)=\frac{1}{2}$ . Since we are tossing it $2n$ times, the probability of getting any given sequence will be $\frac{1}{2^ {2n} }$ (by the multiplication principle). So what we need to figure out is how many ways you can get at least $n$ heads. Since $x$ heads can occur in ${2n \choose x}$ ways, we want to find the sum (A): $\sum^{2n}_{k=n} {2n \choose k}$ . Now consider the sum (B): $\sum^{2n}_{j=0} {2n \choose j}=2^{2n}$ (which can be justified using the binomial expansion for $(1+1)^{2n}$ ). Also note that ${a \choose b}={a \choose a-b}$ . Using this symmetry, the sum (B) can be written: $\sum^{2n}_{j=0} {2n \choose j}={2n\choose n}+2 \sum^{2n}_{i=n+1} {2n \choose i}=-{2n\choose n}+2 \sum^{2n}_{k=n} {2n \choose k}$ which gives: $-{2n\choose n}+2\sum^{2n}_{k=n} {2n \choose k}=2^{2n}$ , $\sum^{2n}_{k=n} {2n \choose k}=\frac{1}{2} (2^{2n}+{2n\choose n})$ which is the number of ways you can get at least $n$ heads). Now all we have to do is multiply by the probability of each case (which is $\frac{1}{2^ {2n} }$ for all cases). If $N$ is the number of heads: $P(N\geq n)=\frac{1}{2^ {2n} }\times \frac{1}{2} (2^{2n}+{2n\choose n}) =\frac{1}{2^ {2n+1} }(2^{2n}+{2n\choose n}) = \frac {1}{2} + \frac {{2n\choose n}}{2^{2n+1}}$ So, in the case $n=4$ , we get $P(N\geq 4)=\frac{1}{2^ {9} }(2^{8}+{8\choose 4})=\frac {256+70}{512}=\frac{163}{256}$ . And since $gcd(163,256)=1$ , $\frac ab=\frac{163}{256}$ , and $a+b=163+256=419$ .

We wish to compute (8C4+8C5+...+8C8)/(8C0+8C1+...+8C8). The denominator is equivalent to 2^8=256. Now, to compute the numerator, we look at the denominator divided by two (think in half), which is the equivalent of 8C8+8C7+...(8C4)/2 (This is true by the symetric sums). This also must be 256/2=128. Therefore, to "complete" our numerator, we must add and extra 8C4/2 to 128, giving us 163. Therefore our answer is 163+256=419.

The classical theory of probability states that in a sample space $S$ with $n(S)$ possible outcomes, if the number of possible outcomes of an event $E$ occurring is $n(E)$ , the probability of event $E$ occurring $P(E)$ can be calculated by $P(E) = \frac {n(E)}{n(S)}$ .

In this scenario, the sample space $S$ is the set of all possible outcomes of the result of tossing 8 fair coins. Using the rule of product , the total number of possible outcomes $n(S) = 2^8 = 256$ , since firstly, there are 2 possible results for the tossing of each coin, and secondly, the result of tossing each of the 8 coins is independent of each other.

Now, letting $E$ be the event that Calvin wins, $n(E)$ would the number of possible outcomes of tossing 8 fair coins with a result of at least 4 heads. $n(E)$ can also be expressed as the number of possible outcomes of tossing 8 fair coins with a result of exactly 4, 5, 6, 7 or 8 heads. In order to calculate $n(E)$ , we find the number of possible outcomes for each possible number of heads, before using the rule of sum to find the total number of possible outcomes.

Firstly, the number of possible outcomes of tossing 8 fair coins with a result of exactly 4 heads is given by ${8 \choose 4} = 70$ . In other words, just like how there are 70 ways for us to pick 4 coins out of 8, there are 70 ways for 4 of the 8 coins to return heads. We repeat this by finding the number of possible outcomes of tossing the coins with a result of exactly 5, 6, 7 or 8 heads. ${8 \choose 5} = 56$ ; ${8 \choose 6} = 28$ ; ${8 \choose 7} = 8$ ; ${8 \choose 8} = 1$ .

Next, we add up the total number of possible outcomes $n(E)$ to get $n(E)=70+56+28+8+1=163$ . Using the classical theory of probability, $P(E) = \frac {n(E)}{n(S)} = \frac {163}{256}$ . To answer the question, we add up the values of the numerator and denominator to obtain $163+256=419$ .

the probability a coin showing a head is 0,5. so in every tosses, the probabilty a coin showing head is 0,5. Calvin has to obtains at least 4 heads which mean that (1) the probabilty the coins show 4 heads is 〖0,5〗^4x〖0,5〗^4 and there are (█(8@4)) ways. so the probability of 4 heads is 70/256 (2) the probabilty the coins show 5 heads is 〖0,5〗^5x〖0,5〗^3 and there are (█(8@5)) ways. so the probability of 5 heads is 56/256 (3) the probabilty the coins show 6 heads is 〖0,5〗^6x〖0,5〗^2 and there are (█(8@6)) ways. so the probability of 6 heads is 28/256 (4) the probabilty the coins show 7 heads is 〖0,5〗^7x〖0,5〗^1 and there are (█(8@7)) ways. so the probability of 7 heads is 8/256 (5) the probabilty the coins show 8 heads is 〖0,5〗^8x〖0,5〗^0 and there are (█(8@8)) ways. so the probability of 8 heads is 1/256. then count the probabilities so the probabilty of Calvin wins is 163/256 and a+b is 419

so the probabilities for 0 heads and 8 tails, 1 head, 7 tails... 8 heads, 0 tails are the 8th row of pascal's triangle.

that is of the form:

1,8 ,x,x,x,x,x, 8,1

0,1,2,3,4,5,6,7,8

now, the sum of the nth row of pascal's triangle is $2^n$ , so an exact half probability would be $\frac {2^7}{2^8}$ the middle term of the eighth row of the triangle is $8 \choose 4$ $= 70$ , so if we add one half of that to the numerator (because it was already divided half between winning and losing) we achieve the fraction.

$\frac {2^7 + 35}{2^8}=\frac {163}{256}$

$163 + 256 = 419$

it is asked that at least 4 heads therefore it can be seen other way around as 1 - probability of getting at most 3 heads let x denote the no of heads getting here possible value of x(possible no of heads getting) is 0,1,2,3 therefore solution goes as 1-(8c0+8c1+8c2+8c3)/256 i e 1-93/ 256 163/256 ans 163+256=419

Calvin wins if he obtains at least 4 heads; that is, if he does not obtain 0, 1, 2, or 3 heads.

The probability he obtains 0, 1, 2, or 3 heads is $\dfrac{{8 \choose 0} + {8 \choose 1} + {8 \choose 2} + {8 \choose 3}}{2^8}$ . Evaluating, this is equal to $\dfrac{93}{256}$ . Since we want the probability this event does NOT occur, we subtract from 1 to get $1-\dfrac{93}{256} = \dfrac{163}{256}$ as our answer.

• We have 9 general events, starting by ALL NOT HEAD and ending by ALL 8 HEADS , each one has its probability.

• So, for us to calculate the probability of at least 4 , its the sum of the probabilities of the events starting from 4 HEADS to 8 HEADS.

• The total number of possible results = 2^8 = 256

• The total number of 4 HEADS resulting = 8C4 = 70
• The total number of 5 HEADS resulting = 8C5 = 56
• The total number of 6 HEADS resulting = 8C6 = 28
• The total number of 7 HEADS resulting = 8C7 = 8

• The total number of 8 HEADS resulting = 8C8 = 1

• The total number of at least 4 HEADS = 163

• The probability of getting at least 4 HEADS = \frac {a}{b} = \frac {163}{256}

• So, a + b = 419

So, let's do complementary counting. Let's find out the probability of obtaining 3 heads, then 2, then 1, then none. So, the probability of obtaining 3 head is just ${8 \choose 3}$ which is 56. Next, the probability of obtaining 2 heads is just ${8 \choose 2}$ which is 28. Next, the probability of obtaining 1 head is ${8 \choose 1}$ or obviously 8. Finally, ${8 \choose 0}$ is 1, so add them all up and subtract from the total and we get 93, which you subtract from the total to get $\frac {2^8-93}{2^8}$ = $\frac {256-93}{256}$ = $\frac {163}{256}$ . Since they want the sum of the numerator and denominator, you take 163+256=419.

the probability of getting at least four heads is: p(4H) = 1 - [ p(0H) + p(1H) + p(2H) + p(3H) ] 1-[1/256+8/256+28/256+56/256]=1-93/256=163/256 It's easy to notice that 1, 8, 28 and 56, are the first four values in the 8th row of Pascal's triangle.

Out of 8 fair coin tosses, atleast 4 heads can be selected in {8 \choose 4} + {8 \choose 5} + {8 \choose 6} + {8 \choose 7} + {8 \choose 8} ways as 'atleast heads' cover all cases from 4 heads to 8 heads. The total number of outcomes are 2^8 as every toss has 2 outcomes - heads or tails. Dividing them will give the answer \frac {163}{256} .

probaality of 8 heads 8C8/256=1/256 SIMLARLY 7 HEAD 8C7/256=8/256 6=28/256 5=56/256 4=70/256 TOTAL=163/256

We have 9 possible outcomes, (0,1,2,3,4,5,6,7,8 heads)

We want to work out the probability that we get 4,5,6,7,8 heads. Notice that the probability of getting 0 heads is the same as that of getting 8 heads as heads and tails are symmetric. Similarly the probability of getting 1 head is the same as getting 7 and so on.

So P(0,1,2,3 heads)=P(5,6,7,8 heads)=x

P(4 heads)= $\binom{8}{4} \times (\frac{1}{2})^8$ = $\frac{70}{256}$

We also have $\frac{70}{256}+ 2x=1$

So $x=93/256$

In total, probability of win $=\frac{70}{256}+\frac{93}{256}=\frac{163}{256}$

The probability that Calvin wins is equal to 1 minus the probability he loses, i.e., getting at most 3 heads. Denote P(k) to be the probability that he gets k heads on 8 tosses. Then, the probability that he wins is $1-\sum_{k=0}^3 P(k)$ = $1-{8 \choose 0}-{8 \choose 1} -{8 \choose 2}- {8 \choose 3}$ = $\frac{163}{256}$ . So $163+256=419$ .

There are 2 choices for every flip. so $2^8$ choices for 8 flip. now Calvin will win if he obtains 4,5,6,7 or 8 heads or 4,3,2,1,0 tails respectively. so we have ${8 \choose 4}+{8 \choose 3}+{8 \choose 2}+{8 \choose 1}+{8 \choose 0}=163$ choices for Calvin to win. So the the probability that Calvin wins is $163 / 256$ . so $a+b=419$ .

the sum of the values of the 8th row of pascal's triangle [1,8,28,56,70,56,28,8,1] from the 5th value is 163 163 + 128 = 291

Solution 1: Let $X$ be the event that we get at least 4 heads, $Y$ be the event that we get at least 4 tails. By symmetry, we get that $P(X)=P(Y)$ . Also, we clearly have $P(X \cup Y) = 1$ , since either of the events must occur.

Thus, $1 = P(X) + P(Y) - P(X\cap Y)$ . We know that $P(X \cap Y ) = \frac {8 \choose 4}{2^8}$ , so $P(X) = \frac {1}{2} \left( 1 + \frac {8 \choose 4}{2^8} \right) = \frac {163}{256}$ . Hence $a+b = 419$ .

Solution 2: To obtain at least 4 heads, there are ${8 \choose 4} + {8 \choose 5} + {8 \choose 6} + {8 \choose 7} + {8 \choose 8} = 70 + 56 + 28 + 8 + 1$ ways. Hence, the probability is $\frac {163}{256}$ , so $a+b = 163 + 256 = 419$ .

Firstly, find the probability that Dan will win.

By using the binomial probability function:

The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:

P(X=x) = nCx * H^x * T^(n-x)

then,

H=0.5; T=0.5; n=8; x={4,5,6,7,8}

Probabity that Dan will win = (8C4 0.5^4 0.5^4) + (8C5 0.5^5 0.5^3) + (8C6+0.5^6 0.5^2) + (8C7 0.5^7 0.5^1) + (8C8 0.5^8*0.5^0)

=163/256

Lastly, add the numerator and the denominator:

163+256=419

Thus, the answer is 419.