16 replies to this topic

[bonanova]

By his own admission Alex spent too much
time in the back corner at Morty's throwing
darts. But the results could not be ignored: 
he cold hit the bullseye reproducibly 97% 
of the time.

Davey, on the other hand could hit only
90% of his attempts, and Ian's accuracy
was worse, at 80%.  Neither if them would
play Alex for that reason.

Last night Alex proposed that the three
of them play: Alex's single dart against
the better of his opponents' two darts.

Davey scribbled some 1's and 0's on a
napkin, trying to work out the odds. He
wasn't certain, but he figured that the
two of them just might be able finally to
win some beer money from their nemesis,
and he persuaded Ian to give it a try. 

Had Alex finally made a bad bet?

[MikeD]

Spoiler for i hope the logic is right

If Alex threw 200 darts, and Davey and Ian threw 100 each.

Alex would hit the bullseye 194 time on average
Davey would hit it 90 and Ian would hit it 80 time on average or 170 times together.

Since it would be the better of either Davey or Ian darts, then at best its still 90% chance to hit the bulleye verus 97% of Alex.
So Alex still would be in a better position to win the bet then Davey and Ian.

[mrzinks]

Yes :-)

[psykomakia]

I'm sure there is something wrong with this logic, but...

Spoiler for eh?

Davey will hit the target 90% of the time.
In the 10% of the time that he misses, Ian will hit the target 80% of the time.
So if Davey throws 100 darts, he will hit 90 times.
In the 10 times he misses, Ian will (likely) hit the target 8 times.
They will combine for a score of 98 times.
Alex, of course, would hit the target 97 times.

[MikeD]

@psykomakia

Spoiler for the flaw in your logic (i think)

Davey throws 100 darts hitting 90, Ian throws 10 darts and hitting 8 = 98 hits (as you said)
But to be fair Alex need to throw 110 darts and he would then hit 106.7 times. (he is still in a better position)

[curr3nt]

Spoiler for ...

There are eight possibilities:

D I A
0 0 0 p0 = 10% * 20% * 3%  =  0.06%
0 0 1 p1 = 10% * 20% * 97% =  1.94%
0 1 0 p2 = 10% * 80% * 3%  =  0.24%
0 1 1 p3 = 10% * 80% * 97% =  7.76%
1 0 0 p4 = 90% * 20% * 3%  =  0.54%
1 0 1 p5 = 90% * 20% * 97% = 17.46%
1 1 0 p6 = 90% * 80% * 3%  =  2.16%
1 1 1 p7 = 90% * 80% * 97% = 69.84%

Alex loses = p2 + p4 + p6 = 0.24% + 0.54% + 2.16% = 2.94%
Alex wins = p1 = 1.94%
Everything else is a draw.

Edited by curr3nt, 23 July 2012 - 02:58 PM.

[Rockmond]

Seems pretty simple. The odds that Davey and Ian both miss is 2% (.2*.1). The odds that Alex will miss is 3%, so he has made a bad bet. Is there something more complicated that I am missing?

[psykomakia]

@psykomakia

Spoiler for the flaw in your logic (i think)

Davey throws 100 darts hitting 90, Ian throws 10 darts and hitting 8 = 98 hits (as you said)
But to be fair Alex need to throw 110 darts and he would then hit 106.7 times. (he is still in a better position)

Spoiler for Nah

Davey throws 100 darts. Alex throws 100 darts. Ian throws 100 darts. Say Davey misses the last ten. Ian has also thrown a "last ten" and will miss 20% of them. 90 Davey made + 8 Ian makes in the last ten = 98.

[jim]

It seems very unlikely that dart players in a pub would play a game where they simply count the numner of bullseyes.

[MikeD]

I see where my mistake was. I like the way you worked it out curr3nt.