What is the formula for determining the odds of winning exactly X coin flips out of Y number of tosses?

Question

For example What are the odds of me winning exactly 9 fair coin tosses out of 10?

Answer 1

The probability of winning exactly X coin flips out of Y tosses is:

C(Y, X) / 2^Y

C(Y, X) is the number of combinations of Y things taken X at a time:
C(Y, X) = Y! / [X! · (Y - X)!]

In your example, the probability of winning 9 flips out of 10 is:
C(10, 9) / 2^10 = 10 / 1024 = 5 / 512.

The probability of winning 8 of 10 is:
C(10, 8) / 2^10 = 45 / 1024.

The probability of winning 5 of 10 is:
C(10, 5) / 2^10 = 252 / 1024 = 63 / 256, or about 24.6%.

Answer 2

No. All the answers so far are wrong. For one thing, you want to win EXACTLY 9 times, which means you must win 9 times and lose once.

First, the probability of winning the first nine tosses and losing the last is

(1/2)^9 * 1/2

But in the original problem, we don't care which of the ten tosses is the losing toss. The losing toss can be any of the ten, so we must multiply the above probability by 10. The answer is

10 * (1/2)^10

The general case, of winning X tosses out of Y, is more complicated. Louise's answer (below) gives the general formula. Do you know what the "C" notation means?

Answer 3

The intuitive and correct answer in both cases is that the fewer the number of possible outcomes, the higher the chance of any particular outcome being realized. With 10 tosses, for example, the only options are 1, 2, ...10, so the probability is of the order of 0.1 (1/10); with 100 there are ten times the number of possible results with the probability of each around 0.01 (1/100).

Answer 4

it depend you have x coin or x coins
anyways
if you flip x coin for y times the odds are
(x) x (y)=2 or
xy/2 as coin has only two probability either heads or tails

as per your example
9 coins tossed for 10 times then the odds of winning would be
9 x 10/2
90/2.......
45 times.

Answer 5

P=(1/2)^9

P = Probability of winning 9 consecutive times.

P = 1/512

Answer 6

u must loose x-y tosses and win x tosses of coin

so, choose x positions out of y tosses... this is done in yCx ways...

For both head or tail the prob is 1/2... so in y tosses you have (1/2)^y irrespective of whether u take heads or tails on each toss

so your prob of winning exactly x times out of y tosses is

yCx (1/2)^y

so, for 9 tosses out of 10, the prob is
10C9 (1/2)^10

Note that this prob is the same as winning only once... 10C9 = 10C1
Thats because getting heads 9 times and tails once is same as getting heads once and tails 9 times out of ten (in terms of probablility).

Answer 7

Garypopkin has the correct answer. The best way to visually understand it is to build a probability tree. Since the order in which you lost the 1 toss is irrelevant to you, there are 10 combinations in which you will win 9 times and lose once:

1st combination: you lose the first toss, but win the next nine

2nd combination: you lose the second toss, but win the first and last eight.

etc...

Then each branch of your tree (each toss) has a probability of "success" = 0.5 = probability of "failure", since each toss is either heads or tails.

Each of the 10 combinations described above has a ramification of branches. For example:

"win-lose-win-win-win-win-win-win-win-win"

would be combination No.2. For this combination to occur you need (^="and"):

win^lose^win^win^win^win^win^win ^win^win.

The probability of it occurring is therefore:

"0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5*0.5 = 0.5^10".

Since there are ten combinations, the probability you are looking for is indeed:

10 * (0.5^10).

Answer 8

its 50/50
X/Y=Z

Answer 9

50%