gambling math problem

crazygeryy

there were 92 soccer games on yesterday.29 of them were draws.

id like to know my odds on picking 4 of them to be draws.in other words picking 4 out of the 92 to be draws.can anyone work that out?just curiousity.
and im useless at maths.

UrbanSea

Could I be right in saying that if you wanted the % you'd put 4 over 92 and multiply it by one hundred. However,that would be assuming that all teams would be of the same quality and odds.
Maybe I'm way off there,which I probably am.

UrbanSea

Edit: That would be for four individual draws,not ones in particular.

jprender

118/1

clived2

119 to 1

clived2

jprender wrote: »

118/1

lols i rounded up

crazygeryy

wow high odds.so its hard to pick 4 draws lol

thankyou for solving it for me.

Im Only 71Kg

jprender wrote: »

118/1

how did you come up with this number? can you show us the math involved..thanks

clived2

There are 92 games, 29 draws,

Imagine a bowl with 92 balls in it, 29 blue and the rest black,

What are chances of picking a blue ball in your first go

29 out of 92 29/92

and the chance of a blue in your second 28/91 and so on


27/90
26/89

So you have

29/92= .315 = 1 divide by .315= 3.18 odds
28/91= .308 = 1 divide by .308= 3.25 odds
27/90= .3 = 1 divide by .3 = 3.33 odds
26/89= .29 = 1 divide by .29 =3.45 odds


3.18 x 3.25 x 3.33 x 3.45 = 118.7

crazygeryy

clived2 wrote: »

There are 92 games, 29 draws,

Imagine a bowl with 92 balls in it, 29 blue and the rest black,

What are chances of picking a blue ball in your first go

29 out of 92 29/92

and the chance of a blue in your second 28/91 and so on


27/90
26/89

So you have

29/92= .315 = 1 divide by .315= 3.18 odds
28/91= .308 = 1 divide by .308= 3.25 odds
27/90= .3 = 1 divide by .3 = 3.33 odds
26/89= .29 = 1 divide by .29 =3.45 odds


3.18 x 3.25 x 3.33 x 3.45 = 118.7

bravo! im impressed and thanks.