calculate the probability that is tossing coin 10 times we get all heads up
Answers
Answer:
20000000
Explanation:
Explanation:
Jungsun Choi and Junho Oh from Nanjing International School sent some wonderful thoughts to each part of this problem. In this solution I give each of their comments for comparison, along with some of my own thoughts and comments.
A key fact in this problem is that it takes, on average, 2046 flips to achieve 10 heads in a row. Analysis of this problem is fascinating because it draws together a fascinating mix of theoretical and numerical probability along with estimates and approximations.
1. Jungsun: There is an 1/2 chance to get a head of a coin each time. To get 10 heads in a row, an 1/2 chance has to be multiplied for 10 times. So, the formula to complete the coin scam on the first attempt is (1/2)10. As a result, the chance of DB completing the coin scam on the first attempt is 1/1024.
Junho: The chance of DB completing the coin scam on the first attempt, which is to toss a coin and get 10 heads in a row, is very unlikely. When calculated, the probability of this happening is 1/1024 which is about 0.000967.
2. Jungsun: The chance to complete the coin scam on the first attempt is 1/1024, and it means that statistically, among 1024 trials (of 10 flips in a row), 1 trial may succeed to get 10 heads in a row.
Junho: According to probability, there is a 1/1024 chance of getting 10 consecutive heads (in a run of 10 flips in a row). However, this does not mean that it will be exactly that number. It might take one person less throws to get 10 consecutive heads. Also, one might spend the whole day trying to get it, but not succeed.
Steve: The actual exact expectation calculation is complicated because DB did not do sequences of 10 coin flips: he carried on until 10 in a row were seen; this is not the same as doing several individual trials of 10 flips. The full computation involves conditional probability, and works out to be
E(10H)=+=E(10H|10H)P(10H)+E(10H|9HT)P(9HT)E(10H|8HT)P(8HT)+⋯+E(10H|0HT)P(0HT)211−2=2046
The problem points to the fact that both the average and spread are relevant in probability calculations. I ran a simulation to determine the time of completion of trials in 2000 cases. You can view the data in this spreadsheet. The average for my trials was a little over 2053 (a little more because I terminated the trials at 10000 runs). The cumulative frequency chart was as follows:
3. Jungsun: Because his 5000 trials were all failed, he has to do the experiments again.
Junho: In order to make 10 consecutive heads, it is just a matter of chance. There is no exact number of flips that one can throw to get 10 consecutive coins; that is just a number of probability. There is no guarantee that x more flips will make 10 consecutive heads. Theoretically, one is supposed to get it after flipping 2046 times. But since this person did it already 5000 times and didn't get 10 consecutive heads, do it at least 2046 times again. (on average)
Steve: Coin flips are memoryless: regardless of how many previous flips have failed, I will still expect to have to make, on average, 2046 more flips. This is one of the most confusing aspects about probability to many people. Being able to think clearly 'What is the situation NOW? Do any PAST (i.e. happened) events effect any FUTURE (i.e. yet to happen) events?' really helps with probability.