Heads Me!
Ben and Sarah are each trying to get exactly 2 heads from a throw of coins.
- Ben is going to throw 3 coins.
- Sarah is going to throw 4 coins.
Who will have a higher chance of accomplishing their task?
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3 solutions
falling a head is anyway AN INDEPENDENT EVENT to the no of throws as long as there are enough total no of chances given i.e., >=2, no. of chances given(any) has equal likelihood of giving exactly two heads .
I don't understand what you mean by that.
For example, if you only throw 2 coins, you only have a 25% chance of getting exactly 2 heads.
But if you throw 3 coins, your chances increase to 37.5%.
If you are given more number of throws, the chances of falling a head is affected.
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excuse me for the false generalization. yes you are spot on. the number of successes.the chances of falling a head do get affected . I think , only for the particular case of 3 and 4 trials , the probability is the same.
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Yes, and that's what makes this problem so special! xD
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@Winston Choo – and there seems to be a underlying pattern too. say 2n is the number of trials , then n and n-1 seem to be the no of successes for which there is a same chance of occurrence. Is this unique to the distribution we are talking about?
Using binomial probability:
Ben has a probability of ( 2 3 ) ( 2 1 ) 2 ( 2 1 ) 1 . Sarah has a probability of ( 2 4 ) ( 2 1 ) 2 ( 2 1 ) 2 .
However, ( 2 4 ) is double that of ( 2 3 ) , but Sarah has another factor of 2 1 . Therefore Ben and Sarah have the same chance of getting 2 heads.
There are 2 3 possible ways that Ben can throw the coins, and there are 2 4 possible ways that Sarah can throw the coins. In Ben's case, 3 out of the 8 total outcomes result in exactly 2 heads (HHT, HTH, THH), thus the probability is 8 3 . In Sarah's case, 6 out of the 1 6 total outcomes result in exactly 2 heads (HHTT, HTHT, HTTH, THHT, THTH, TTHH), thus the probability is 1 6 6 = 8 3 . As you can see, the probabilities are equal.