+0  
 
0
964
2
avatar

what are the odds of losing 6 coin flips in a row

Guest Dec 9, 2014

Best Answer 

 #2
avatar+85624 
+5

There's something interesting about this problem that I'd like to bring out.....many people think that there is some "righting" force in nature that tends to bring things "back to center"....in this case........many think that, if they haven't been successful after "N" flips, that there is a greater chance that they WILL be successful on the next flip......this isn't true as shown below...

The probability that a person has one success on any of 6 coin flips  = 1 minus the probaility that he isn't successful on any of them.....

Thus......before any flips are attempted.....the probabilty of success = 1 - (1/2)^6 = 0.984375

Let's suppose he isn't successful after the first toss......then he has five remaining in which to achieve a success and this probability = 1 - (1/2)^5 = 0.96875.....notice that this isn't as good as before the first flip

Look at the remaining probabilities, assuming no successes on previous attempts

After the second toss =  1 - (1/2)^4 = 0.9375

After the third toss =  1 - (1/2)^3 = 0.875

After the fourth toss =  1 - (1/2)^2 = .75

Notice, that as more unsuccessful flips are made, the probability of attaining just one success drops....

And, after the fifth toss......the probabilty is just....1 - (1/2)^1 = 1/2 = .50 ......just what we would expect it to be on one remaining toss !!!

This demonsrates something known as the Gambler's Fallacy......(it's why those big buildings exist in Las Vegas....!!!).......why does this happen???....because, as more "failures" pile up in the sequence of some N trials, the fewer chances there are remaining for a person to achieve a success........!!!

 

CPhill  Dec 9, 2014
Sort: 

2+0 Answers

 #1
avatar+92193 
+5

$$\frac{1}{2^6}=\frac{1}{64}$$

Melody  Dec 9, 2014
 #2
avatar+85624 
+5
Best Answer

There's something interesting about this problem that I'd like to bring out.....many people think that there is some "righting" force in nature that tends to bring things "back to center"....in this case........many think that, if they haven't been successful after "N" flips, that there is a greater chance that they WILL be successful on the next flip......this isn't true as shown below...

The probability that a person has one success on any of 6 coin flips  = 1 minus the probaility that he isn't successful on any of them.....

Thus......before any flips are attempted.....the probabilty of success = 1 - (1/2)^6 = 0.984375

Let's suppose he isn't successful after the first toss......then he has five remaining in which to achieve a success and this probability = 1 - (1/2)^5 = 0.96875.....notice that this isn't as good as before the first flip

Look at the remaining probabilities, assuming no successes on previous attempts

After the second toss =  1 - (1/2)^4 = 0.9375

After the third toss =  1 - (1/2)^3 = 0.875

After the fourth toss =  1 - (1/2)^2 = .75

Notice, that as more unsuccessful flips are made, the probability of attaining just one success drops....

And, after the fifth toss......the probabilty is just....1 - (1/2)^1 = 1/2 = .50 ......just what we would expect it to be on one remaining toss !!!

This demonsrates something known as the Gambler's Fallacy......(it's why those big buildings exist in Las Vegas....!!!).......why does this happen???....because, as more "failures" pile up in the sequence of some N trials, the fewer chances there are remaining for a person to achieve a success........!!!

 

CPhill  Dec 9, 2014

9 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details