Question

You are going to play a game where you bet a dollar and get to flip a coin ten times. If you get four heads in a row, you win. If you make the tenth flip without getting four heads in a row, you lose. Run this game ten thousand times. Approximately what is the probability that you will win?

Answer 1

seems to me like there are 10-4 ways to get 4 heads in a row and the rest tails

subtract 10-4-2 ways to get ht. 4h rest tails

dunno how simple this is to brute math tho

Answer 2

P(4h| 6t n 0h)
+P(4h | 5t n 1h)
+P(4h | 4t n 2h)
+P(4h | 3t n 3h)
+P(4h | 2t n 4h)
+P(4h | 1t n 5h)
+P(4h | 0t n 6h)

its a thought but i dont think its all that accurate

Answer 3

let hhhh = a

the number of ways something can happen is a formula: n!/(a!b!c!...k!) for the number of duplications of a term

a tttttt 7!/(6!1!0!)
a ttttt h 7!/(5!1!1!)
a tttt hh 7!/(4!2!1!)
a ttt hhh 7!/(3!3!1!)
a tt hhhh 7!/(2!4!1!)
a t hhhhh 7!/(5!1!1!)
a hhhhhh 7!/(6!1!0!)

each of these arrangements is a winning outcome

out of 2^(10) possible outcomes

Answer 4

I was looking up a question I asked a LONG time ago:
You flip a coin 10 times and if it lands "heads" 4 or more times in a row is considered a success. What is the probability of a success?
(Someone asked this a few hours ago and I worked out a computer simulation and quite recently I used Excel to get the exact probability.
Of the 2^10 (or 1,024) possible ways that 10 coin tosses can occur, 251 are successes (so p=0.2451171875).

Answer 5

7
+ 42
+105
+140
+105
+ 42
+ 7

2(7+42+105) + 140

2(154) + 140

308 + 140

448/2^10 is the probability of getting a winning toss out of all possible tosses is what im thinking

but im sure thers duplicates in it