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I am new here and I am not sure if this is the correct place to ask this question.

It seems like a simple probability question but I am stuck on it. It is a problem I am trying to solve.

Lets say I have 43 bets. The probability of a bet winning is 0.64. The probability of a bet losing is 0.36.

What is the probability that 29 of those 43 bets will be losing/losers?

I worked it out like this:

29 / 43 = 67% 0.36 * 67 = 24.12% chance that 29 of those 43 bets will be losing/losers.

But this seems very simplistic and incorrect to me. Because for example - when the number of bets goes up e.g. 43 goes up to 100 bets. The probability that 29 of those 100 bets losing goes down to say 10.44%. Which seems wrong to me. Because I think if the number of bets goes up, then the probability of losing a fixed number of 29 should also go up, as there are higher "chances" of losing.

Could you list the correct answer, working out and possibly a link where I can read to learn more.

I did read up on this, but there were so many ways of achieving this. I read about Binomial distributions etc, but the answers I kept calculating just seemed very wrong to me.

Many Thanks!

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Let's indicate with $1$ a victory and with $0$ a loss. We can create so a bijection between all the possible results of a series of $43$ bet and a binary number of $43$ ciphers. In particular, we are searching all of the binary number in this form:

$$\underbrace{1...1}_{14 \text{ times}}\underbrace{0...0}_{29 \text{ times}}$$

All of the possible permutations of this binary number are:

$$P_{(43,14,29)}=\frac{43!}{14!29!}$$

The probability of each permutation is:

$$p(P)=(0,64)^{14}(0,36)^{29}$$

This must be multiplied for the number of permutation. So the searched probability $p(E)$ is:

$$p(E)=p(P)P_{(43,14,29)}=(0,64)^{14}(0,36)^{29}\frac{43!}{14!29!}=0.002 \%$$

:)

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  • $\begingroup$ Hi Eureka, Thanks for your answer. I read your comment above that the above answer switched the probabilities. The funny thing is...If I switch the above answers probabilities. {43 \choose 29} .36^{29} (1-.36)^{43-29} = 0.002% Which is the exact same answer as yours! Is this correct then? (please excuse if my math formatting is incorrect, as I am still learning how to use formatting on the site) $\endgroup$ – Nathan Mar 27 at 19:39
  • $\begingroup$ Hi :) , I meant that he switched the exponents because clearly $\binom{n}{n-k}=\binom{n}{k}$. Indeed our results are different as you see. $\endgroup$ – Eureka Mar 27 at 19:43
  • $\begingroup$ So are both the answers correct? or just yours above? thanks :-) $\endgroup$ – Nathan Mar 27 at 19:47
  • $\begingroup$ For sure they aren't both correct: math is just one ahaha. I think that he's wrong because he wrote $(0.64)^{29}$ , but you requested 29 losses not 29 winnings. $\endgroup$ – Eureka Mar 27 at 19:49
  • $\begingroup$ Yes true! haha, I will try to build your answer in Excel over next few days, thanks! $\endgroup$ – Nathan Mar 27 at 20:49
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$${43 \choose 29} (1-.64)^{29} .64^{43-29} = 0.0000205822$$

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  • $\begingroup$ You switched the two probabilities. $\endgroup$ – Eureka Mar 27 at 19:26
  • $\begingroup$ Hi David, thanks for your answer. I guess it is just binomial coefficient * pk(1-p)(n-k) Which is similiar to what is explained here mathsisfun.com/data/binomial-distribution.html $\endgroup$ – Nathan Mar 27 at 19:28
  • $\begingroup$ I managed to replicate your answer in Excel using the COMBIN for the binomial coefficient calculation, thanks! $\endgroup$ – Nathan Mar 27 at 19:29
  • $\begingroup$ Thanks for the correction David! thanks for the answer $\endgroup$ – Nathan Mar 27 at 20:50

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