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September 20th, 2019, 08:15 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88  Probability When Four People Flip Coins
One at a time, four people flip coins. Each person flips until they get one wrong, and their score is how many they get correct. What is the probability that the four people combine to score at least 5? When the game is done, everybody will have been wrong once for a total of being wrong four times. Is the answer the same as the probability of one person scoring at least 5 if he or she flipped until he or she was wrong four times?

September 20th, 2019, 10:12 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,644 Thanks: 1476 
If I understand your question the underlying experiment doesn't matter. As long as being a different person has no effect on the outcome of the experiment it doesn't matter if your samples come from 4 people or 1 person repeating the experiment 4 times. 
September 20th, 2019, 10:40 AM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 208 
Assuming the probability of being right is 50 %, the odds of one person scoring 0 is 1/2. The odds of scoring 1 (first right, then wrong), is 1/4. The odds of scoring S is 1/2^(S1). To answer your question, it is easier to calculate the probability of four people scoring 4 or less. Let $T = S_1+S_2+S_3+S_4$ be the combined score. T = 0 means all four people scored 0. (1/2)*(1/2)*(1/2)*(1/2) = 1/16 T = 1 means the scores were 1,0,0,0 (odds 1/4*1/2*1/2*1/2 = 1/32) or 0,1,0,0 (odds 1/32) or 0,0,1,0 (odds 1/32) or 0,0,0,1 (odds 1/32). Odds of T = 1: 4*(1/32) = 1/8 Keep going for T = 2, T = 3, and T = 4, then add them together. If you know how to calculate numbers of combinations, you can simplify the counting somewhat. 
September 20th, 2019, 04:47 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88 
What's wrong with this? For T = 2, there are ten possibilities: 2000: (1/8)(1/2)(1/2)(1/2) = 1/64 0200: (1/2)(1/8)(1/2)(1/2) = 1/64 0020: (1/2)(1/2)(1/8)(1/2) = 1/64 0002: (1/2)(1/2)(1/2)(1/8) = 1/64 Subtotal: 4/64 1100: (1/4)(1/4)(1/2)(1/2) = 1/64 1010: (1/4)(1/2)(1/4)(1/2) = 1/64 1001: (1/4)(1/2)(1/2)(1/4) = 1/64 0110: (1/2)(1/4)(1/4)(1/2) = 1/64 0011: (1/2)(1/2)(1/4)(1/4) = 1/64 0101: (1/2)(1/4)(1/2)(1/4) = 1/64 Subtotal: 6/64 The sum is 10/64, which is 5/32. For T = 3, there are 24 possibilities: 3000: (1/16)(1/2)(1/2)(1/2) = 1/128 0300: (1/16)(1/2)(1/2)(1/2) = 1/128 0030: (1/16)(1/2)(1/2)(1/2) = 1/128 0003: (1/16)(1/2)(1/2)(1/2) = 1/128 Subtotal: 4/128 2100: (1/8)(1/4)(1/2)(1/2) = 1/128 2010: (1/8)(1/2)(1/4)(1/2) = 1/128 2001: (1/8)(1/2)(1/2)(1/4) = 1/128 1200: (1/4)(1/8)(1/2)(1/2) = 1/128 1020: (1/4)(1/2)(1/8)(1/2) = 1/128 1002: (1/4)(1/2)(1/2)(1/8) = 1/128 0210: (1/2)(1/8)(1/4)(1/2) = 1/128 0201: (1/2)(1/8)(1/2)(1/4) = 1/128 0021: (1/2)(1/2)(1/8)(1/4) = 1/128 0012: (1/2)(1/2)(1/4)(1/8) = 1/128 0120: (1/2)(1/4)(1/8)(1/2) = 1/128 0102: (1/2)(1/4)(1/2)(1/8) = 1/128 Subtotal: 12/128 1110: (1/4)(1/4)(1/4)(1/2) = 1/128 1101: (1/4)(1/4)(1/2)(1/4) = 1/128 1011: (1/4)(1/2)(1/4)(1/4) = 1/128 0111: (1/2)(1/4)(1/4)(1/4) = 1/128 Subtotal: 4/128 The total is 20/128 = 5/32 The total of 0, 1, 2, and 3 is 1/16 + 1/8 + 5/32 + 5/32 = 16/32 = 1/2. That looks wrong because the expected value is 4, and I don't think the probability of being less than the expected value would be 1/2. For example, if you flip an even amount of coins so that the expected value is a whole number, the probability of more heads than the expected value and of fewer heads than the expected value will both be under 1/2. This is different because it's not based on a fixed number of trials, but I'd still be surprised if the total of 0, 1, 2, and 3 is 1/2. If what I'm doing is correct, I can do the probability for 4 on my own, but I want to know if I'm correct. 
September 21st, 2019, 02:47 PM  #5 
Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 208 
No, you were right. With 4 people playing, there is a 50 % chance of getting a score of 4 or higher. Validated against 10^6 simulations. 4 is not the most likely score, however. 
September 21st, 2019, 03:18 PM  #6 
Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 208 
Additional fun facts (which would probably be obvious to mathematicians if they expressed the probabilities in equation form): * If N people are playing, there is a 50 % chance of the total score being N or higher (not surprising  50 % chance of getting a score of zero). * If N people are playing, the most probable total score is tied between N1 and N2 (except if N = 1, then the most probable score is N1 = 0). Now a question for the statisticians: As $N \rightarrow \infty$ (with the scores appropriately scaled), does this distribution approach a Poisson distribution, a lognormal distribution, or something else? 
September 21st, 2019, 05:25 PM  #7  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle  Quote:
Each person flips until they get Tails, and every time one gets Heads earns a 1 point. Technically, there is no difference between 4 people tossing a coin, each or one person tossing 4 coins, given he (or she) can toss coins which came Heads, again. # What is the probability of getting 0 points? $${[Pr(Tails)]}^4=1/16=0.0625$$ # What is the probability of getting 1 point? You get 1 Heads on the first round and 0 on the second round. $$Pr( \text{1 Heads && 3 Tails}) \times Pr( \text{1 Tails}) = \frac{4}{16} \times \frac{1}{2} = 1/8$$ # What is the probability of getting 2 points? Either you get 2 points on the first round and 0 on the second OR get 1 point on the first and second round and 0 on the third. $$[Pr( \text{2 points on first round}) \times Pr( \text{0 points on second round})] + [Pr( \text{1 point on first round}) \times Pr( \text{1 points on second round}) \times Pr( \text{0 points on third round})] = \frac{6}{16} \times \frac{1}{4} + \frac{4}{16} \times \frac{1}{2} \times \frac{1}{2}=\frac{5}{32}$$ Do it for 3 points and 4 points. $$Pr( \text{At least 5 points}) = 1 Pr( \text{At most 4 points})$$ $$ Pr( \text{At most 4 points}) = Pr( \text{1 point}) + Pr( \text{2 points})+ Pr( \text{3 points}) + Pr( \text{4 points})$$ NOTE: You shouldn't use it as the solution for the problem. It is rather an explanation of how you can approach such problems. Now for the "Is the answer the same as the probability of one person scoring at least 5 if he or she flipped until he or she was wrong four times?" No, it is not.  
September 27th, 2019, 06:44 PM  #8  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 675 Thanks: 88  Quote:
romsek said the answer is the same.  
September 27th, 2019, 08:42 PM  #9 
Senior Member Joined: Jun 2019 From: USA Posts: 383 Thanks: 208 
I think EvanJ is right. Regardless who's flipping the coins, we're flipping coins until we get four "wrong" flips, and counting the "correct" ones. I don't see a difference in the probabilities.


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