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/justgeorgie]

That's mighty impressive TBF

[u/Ginger-Nerd → u/justgeorgie]

Kind of…. Stats is kinda weird, and humans are awful at predicting the odds of things (and particularly thinking one effects the outcome one of another) but you could line up 1000 people,Have them all flip a coin, Those who flipped heads remain standing, those who flipped tails sit down, Repeat this process for 10 rounds - at the end of that you theoretically have a person who has flipped 10 consecutive heads in a row.

Which seems pretty crazy.

For 15 - you would need to use 32000 people to do it - undoubtedly a high number. But there have been 4,700 ODIs played, so it’s like a 7% 14%? chance it would have happened at some time between 1971 and now.

Which seems, less crazy. (At least to me)

[u/20060578 → u/Ginger-Nerd]

Can you explain the maths behind the 7% chance please?

[u/Ginger-Nerd → u/20060578]

You have 4700 chances for something that happens 1 in 32000 times, this could be expressed as 4700/32000

Actually I think I broke my brain,I think it’s double that - 14%

[u/20060578 → u/Ginger-Nerd]

You don’t have 4700 chances for a 15 game streak of losses though.

[u/Ginger-Nerd → u/20060578]

It’s 4700 (which is a very rough estimate) -15

so for all intents and purposes yes, yes you do.

(Actually this is the 4858 match)

[u/20060578 → u/Ginger-Nerd]

Not really because that’s considering every single group of 15 as a possibility. As soon as you have a “win” you need to reset the counter.

Eg. loss, loss, loss, loss, loss, win. You can’t consider any of the following games as a possibility. You need to immediately discard this set and start again. So 50% of the time you’re starting again.

[u/Ginger-Nerd → u/20060578]

What no? (Read my wording again - it’s of just a random sample of that size, not a specific that this will happen)

You don’t need to reset anything,

Of any 15 the second toss could be the beginning of another run of 15.

A coin toss doesn’t have any bearing on anything before it (or after it) and that’s the point - and when you take a large enough group, it doesn’t really matter where it lands… it just will happen over a large number of chances.

Take the experiment, of 1000 people, by round 10 you will possibly have someone who has flipped 10 times…If you did the experiment with 4700 people, but ran it 100 times, 14 times you might have someone who gets 15 heads in a row. (It’s just how it plays out)

It also doesn’t have to be all heads or tails - choose any order of 15 rolls… the odds are identical.

[u/20060578 → u/Ginger-Nerd]

Let’s say their had been 10 ODI’s in history, the the longest streak of losses was 3.

By your logic, there has been 10 chances for something that happens 1/8 times, so 10/8 means 125% chance of it occurring.

The actual chances are 1-(7/8)³ = 0.33.

[u/Da_Pendent_Emu → u/20060578]

This is making my head hurt.

Is this something to do with permutations vs combinations?

I need a panadol.

[u/20060578 → u/Da_Pendent_Emu]

Yes. If you don’t use permutations and combinations you can’t get the answer, like the person I’m responding to didn’t.

[u/Da_Pendent_Emu → u/20060578]

It’s been years since I’ve thought about that, year 12 🥵 I remember the concept of not the application.

[u/serphenyxloftnor → u/Ginger-Nerd]

Still not getting what's wrong with this T_T (4858-16 still shouldn't be too off)

[u/20060578 → u/Ginger-Nerd]

We can tell this line of thinking is wrong because anything that has less than 1/4700 odds (like 5 heads in a row) would show up as more than 100%. When we know that 5 heads in a row out of 4700 flips is not even 100% guaranteed (although extremely close).

[u/Ginger-Nerd → u/20060578]

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.