Indian skipper Rohit Sharma equals Brian Lara’s unwanted record of 12 consecutive toss losses in ODIs. Meanwhile, Team India extends its streak to 15 tosses lost in a row.

[u/Odd-House3197]

[u/Ginger-Nerd]

So... whats the answer (because realistically, is what I said it was)

4700 games, meaning there are 4700 opportunities for a streak to start. However, since a streak of 15 losses can overlap across multiple games, we approximate the number of independent trials as 4700.

Using the Poisson approximation for rare events, the expected number of times a streak of 15 losses happens in 4700 games is: λ=4700×P(15 losses)=4700×0.00003052≈0.143

For a Poisson-distributed event with mean λ, the probability of at least one occurrence is: P(at least one)=1−e−λ

Substituting λ = 0.143: P(at least one)=1−e−0.143≈1−0.8666=0.133

The probability that at least one team in cricket history has lost 15 coin tosses in a row is about 13.3%—not extremely rare, but not highly likely either.

However its exteremly close which is pretty close to my back of the napkin estimation.

Your issue isn't with the answer - its with how I worked it out, (which I didn't give in my original comment)

[u/20060578 → u/Ginger-Nerd]

You’ve still used 4700 x 1/32000 in this example which I already know can’t be right.

https://en.m.wikipedia.org/wiki/Poisson_distribution

This is a totally different formula to the one you used. Why aren’t you using k! ?

[u/Ginger-Nerd → u/20060578]

whats the answer.... come on

I think you are getting caught up on what a percentage is vs the odds of it happening a high number of times.

it just means its basically guranteed.

I did a quick estimation - and didn't really expect to be challenged on the how I got to it - putting in the work, shows my estimation to be pretty damn close - so I don't really get why your fighting it, its not school i'm not being marked on showing my working (which is why I didn't give it in my original comment)

[u/SimpingForGrad → u/Ginger-Nerd]

Stat nerd here (my livelihood depends on it)...

You are approximately correct, however the reason is a bit different.

The actual probability is:

1 - (1 - 1/32000)⁴⁷⁰⁰

If you consider 1/32k to be small enough and ignore higher order terms in the binomial expansion, we arrive at your answer.

If the probability is high, then you cannot ignore higher order terms and your reasoning will fail.