calculation of probablity: 7 houses.....

thinking

are you sure that the house winner still participates in the next draw

so that in theory one person could win all seven houses?

for the first draw the chance is 7/70

but as each house can only have one owner (but apparantly one person can own more than one house) the next draw only has 6 houses for 70 people

i.e. 6/70

third draw is 5/70 etc

so total chance of winning at least one house is

(7+6+5+4+3+2+1)/70 = 28/70 = 0.4 or 40%

If the winner of the first house didnt participate in the next draw it would be 6 houses for 69 people so the probability of getting ONE house would be

7/70 + 6/69 + 5/68 + 4/67 + 3/66 + 2/65 + 1/64 = 0.412 or 41.2%

which feels about right because one would intuitively expect the chance of winning only one house would be greater than the chance of winning at least one house.

I'm editing this because I suspect its fundamentally flawed somewhere but I leave my calculation for others to point where I went wrong

dammit

you are right martjn

for the first house the chance is 1/70

and that is repeated total 7 times

so its 7/70 or 10%

STEVE, YOU THOUGHT TOOO MUCH IN UR FIRST REPLY

AND ABOUT 2ND REPLY, WELL THOUGHT. :wink:

thats my problem vins, too much thinking, not enough drinking

btw you would think Martjin might stop by to say thanks for the homework help :wink:

With that logic, you'd have a 71/70 chance (greater than one) of getting a house if there were 71 houses.

The probablility of not winning any houses is (69/70)^7. Therefore, the probability of winning at least one house is 1-(69/70)^7, or 0.0958.

well done top markrs