0

What is the probability of getting $8-6$ (Win-lose), when you bet 14 times in a win or lose a game with $60%$ winning percentage?

According to the answer, it must be $ 17.4$%.

I tried to solve this using binomial distribution but my answer is not $ 17.4$%. Here is my solution:

$B(x;n,p) = B(8;14,0.6)$ = $14 \choose 8$$(0.6)^8(0.4)^6$ $= (3003)(0.6)^8(0.4)^6$ $= 0.20659760529408$

Here is the exact problem

"Let’s say, based on your betting system’s 60% winning percentage, you want to know Harvard's most likely record over the next 14 bets. 60% of 14 is 8.4, so our record should be 8-6. Using the binomial distribution calculator, we learn that even though 8-6 is the most likely record, it will actually only occur 17.4% of the time. Show that it will actually only occur 17.4% of the time."

Please help me solve this problem. Any comments or suggestions will be much appreciated.

Alessio K
  • 10,599
AYA
  • 578
  • 2
    What answer did you get; and how did you get it? – Angina Seng Aug 31 '20 at 11:21
  • I get $0.206597605$, I can't come up with a way to get $.174$ – lulu Aug 31 '20 at 11:23
  • 2
    Can you print the problem exactly as it appears, with no edits? Or include a link to the source (if possible). As stated, the $20.7%$ answer appears to be correct. – lulu Aug 31 '20 at 11:29
  • @lulu I include here the actual text of the problem. Thank you! – AYA Aug 31 '20 at 11:32
  • Well, it's confusing. The problem says "your betting system" has a $60%$ winning percentage...but then the rest of the problem doesn't mention "your betting system". It all concerns "Harvard's record" which, for some reason, also appears to have a $60%$ winning percentage. Also, both $(8,6)$ and $(9,5)$ are equally probable though for some reason they fix on $(8,6)$. But I still can't guess where the $17.4%$ comes from. – lulu Aug 31 '20 at 11:37
  • In Bin(x;n,p) with x = 8 and n = 14 17.4% arises when one takes e.g. p = 0.488 or p = 0.652, but that does not explain how they get it here, since they state p = 0.6.. – delivery101 Aug 31 '20 at 11:53

1 Answers1

1

As you probabily already did, the probability to get 8 successes on 14 independent draws is

$$\binom{14}{8}0.6^8 0.4^6\approx 20.7\%$$

Edit: reading the text, the probability that there is an error (more than one error) is getting higher and higher.

Here is the graph of your 60% winning game (with 14 independent trials)

enter image description here

As you can see, the most likely events are 2, with same 20.7% probability

tommik
  • 32,733
  • 4
  • 15
  • 34