Scottish Cup Draw - Hearts v Hibs

[CropleyWasGod]

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You're correct except your 64 should read 32.

**** aye

[Hibbyradge]

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But is it not correct to look at them in isolation - after the fact anyway? I’ve seen black spin in 13+ times in a row on a roulette table. Happens.

Your roulette analogy helps prove the argument, actually.

Am I right in saying that betting on a particular number gets you 35/1 return? (Ignore the inbuilt advantage for the house because the real odds are 37/1.)

Ok, so you bet £1 on number 7 and it comes in, you get paid £35.

If you let your bet run, you'd get £35 x 35 and it comes in for a second time. That return £1225.

You let it run and if it comes in, bingo, £42875.

That's not even accurate cos of the stakes etc but it does show how the odds of it happening are calculated.

[Liberal Hibby]

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Your roulette analogy helps prove the argument, actually.

Am I right in saying that betting on a particular number gets you 35/1 return? (Ignore the inbuilt advantage for the house because the real odds are 37/1.)

Ok, so you bet £1 on number 7 and it comes in, you get paid £35.

If you let your bet run, you'd get £35 x 35 and it comes in for a second time. That return £1225.

You let it run and if it comes in, bingo, £42875.

That's not even accurate cos of the stakes etc but it does show how the odds of it happening are calculated.

And to add to that analogy if you play roulette every year for 140 odd years getting three sevens in an row isn't that odd.

[SRHibs]

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Your roulette analogy helps prove the argument, actually.

Am I right in saying that betting on a particular number gets you 35/1 return? (Ignore the inbuilt advantage for the house because the real odds are 37/1.)

Ok, so you bet £1 on number 7 and it comes in, you get paid £35.

If you let your bet run, you'd get £35 x 35 and it comes in for a second time. That return £1225.

You let it run and if it comes in, bingo, £42875.

That's not even accurate cos of the stakes etc but it does show how the odds of it happening are calculated.

I understand how the multiplicative nature works. Any combination of 3 draws would’ve had the same chance of happening in terms of the probability though, which just makes it pretty futile after the fact.

I think I agree and am just a confused person. Strange, but happy, coincidence anyway.

[Geo_1875]

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This indeed is the correct answer - one in 55 800.

Working out is:-

(1/2) x (1/15) x (1/2) x (1/15) x (1/31) = 1 / 55 800

The half is required to set the home team each year and the '1 in 15' etc is the probability of the away team being picked.

The general formula is:-

(1/2)^k × (1/(n1 - 1)) x (1/(n2 -1)) x (1/n3 -1)) x ..... x (1/(nk-1))

where k is the number of years and n1, n2, n3 ..., nk is the number of teams left in at each round in a given year.

Who said mathematics / statistics was boring.






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Very well explained and exactly as I understood it to be, only getting the number of teams in each draw wrong.

Except your calculation is wrong by a factor of 2.

[Geo_1875]

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I understand how the multiplicative nature works. Any combination of 3 draws would’ve had the same chance of happening in terms of the probability though, which just makes it pretty futile after the fact.

I think I agree and am just a confused person. Strange, but happy, coincidence anyway.

But what are the odds that a bookie would lay your bet after the fact?

[SRHibs]

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But what are the odds that a bookie would lay your bet after the fact?

Yes, exactly.

[Cardinal Hibernian]

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Very well explained and exactly as I understood it to be, only getting the number of teams in each draw wrong.

Except your calculation is wrong by a factor of 2.

Thank you - missed out the (1/2) in front of the (1/31) - apologies.

Not easy trying type into tapatalk / Hibs.net.

Possibly a case for an equation editor to be included in this discussion thread?


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[Cardinal Hibernian]

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This indeed is the correct answer - one in 55 800.

Working out is:-

(1/2) x (1/15) x (1/2) x (1/15) x (1/2) x (1/31) = 1 / 55 800

The half is required to set the home team each year and the '1 in 15' etc is the probability of the away team being picked.

The general formula is:-

(1/2)^k × (1/(n1 - 1)) x (1/(n2 -1)) x (1/n3 -1)) x ..... x (1/(nk-1))

where k is the number of years and n1, n2, n3 ..., nk is the number of teams left in at each round in a given year.

Who said mathematics / statistics was boring.






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Sent from my SM-T820 using Tapatalk

[Geo_1875]

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Thank you - missed out the (1/2) in front of the (1/31) - apologies.

Not easy trying type into tapatalk / Hibs.net.

Possibly a case for an equation editor to be included in this discussion thread?


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Should just put a link to an odds checker in every "What are the chances...." post

[Sioux]

Back on track....

We really shouldn't worry about the cup tie. The goons will have 3, 4 or 5 quality players in their side before then. We will lose. They will also be up to third or fourth behind Celtic and Aberdeen, qualifying for Europe. Money is not an issue. They are better off then they've been for years. Oh, and they're back at the dung heap.

I think that's about it.

Weapons grade knob ends!

[SON OF PADDY]

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I call their stadium the 'Minimum Wage' stadium. Doesn't get 20k.

Brilliant 😂
Hope you don't mind,but I'm having that.

[DickieDastardly]

If we could get to ten in a row against them that would be a fabulous achievement; any statos know when we last managed that?

[1van Sprou7e]

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If we could get to ten in a row against them that would be a fabulous achievement; any statos know when we last managed that?

1974-1978, 12 games undefeated

Before the loss in 1973 we went 10 undefeated, starting in 1968 .

1 defeat in 10 years, lets hope we can match that again