Darts at Morty's

[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]

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...

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

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]

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 July 23, 2012 by curr3nt

[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

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)

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.

[EventHorizon]

and I came up with the same probabilities as curr3nt.

I thought perhaps combining the two darts would still have a larger variance and that would change things. I worked up formulas for the binomial distribution, combined them, and ended up with another binomial distribution of the new probability .98% (which really should have been obvious from the Bernoulli trial).

There surely was something I was missing.

Since then I've decided that....

I noticed the word reproducibly, and that changes things since if he never gets less than 97% we are no longer dealing with the binomial distribution for him.

Let B(k;n,p) be the probability of getting k successes in a Binomial distribution with n trials and a probability p of success.

Assuming he always gets exactly 97 out of 100, he will still lose since the probability the duo get 98, 99, or 100 is (B(100;100,.98) + B(99;100,.98) + B(98;100,.98)) which is slightly over 67%.

But will Alex ever get more than 97 out of 100?

Assuming Alex's distribution is uniform (so he gets 97, 98, 99, and 100 darts out of 100 with equal probability)...

The probability the duo will win is:

.25 * ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) ) +

.25 * ( B(100;100,.98) + B(99;100,.98) ) +

.25 * ( B(100;100,.98) ) +

.25 * 0

= about .303144

The probability they tie is :

.25* ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) + B(97;100,.98) )

= about .214740

This leaves the probability Alex win at .482116.

But that only works if Alex's distribution is uniform between 97 and 100 when throwing 100 darts.

But what if they only throw 33 darts at a time? Alex must always get all 33 or he will not have reproduced the result of a percentage greater than or equal to 97%. In this case he will never lose.

So that's my guess. They always throw darts in groups of 33. Alex always gets 97% or more, so he cannot miss. Alex gets some more beer money.

Though perhaps I'm reading too much into it...

[ujjagrawal]

When Davey and Ian playing together:

Combined probability of hitting a bull's eye will be -

90% + 10% x 80% = 98% (If Davey fails to hit, when going first)

or

80% + 20% x 90% = 98% (If Ian fails to hit, when going first)

Clearly, more than 97% probability by Alex, so Alex definitely made a BAD bet.

[mrzinks]

Wouldn't Davey and Ian's average accuracy be 95%, or 2% less than Alex's ??

[Rockmond]

Curr3nt did a lot of figuring to come up with the same conclusion I got by multiplying .1 by .2 and comparing it to .03. On 100 throws Alex loses a net 1 time.

[Rockmond]

When Davey and Ian playing together:

Combined probability of hitting a bull's eye will be -

90% + 10% x 80% = 98% (If Davey fails to hit, when going first)

or

80% + 20% x 90% = 98% (If Ian fails to hit, when going first)

Clearly, more than 97% probability by Alex, so Alex definitely made a BAD bet.

Easier to just calculate odds of neither Davey or Ian hitting bullseye: .1 * .2 = .02 or 2%

[jim]

Suppose they each throw 100 times with 97 bullseyes for Alex while Davey gets 90 and Ian gets 80. What would the team of Davey and Ian score? Suppose Davey throws first and misses 10 times. Ian lets the pressure get to him and misses every time Davye does.lus 10 additional times. In this case the team scores 90. On the other hand, if Davey is good under pressure and makes all 10 of those throws while missing on 20 other occassions, then the team scores 100. If the correlation coefficent for Davey and Ian is zero, then the team scores 98 and they win. With a negative correlation they do even better, but with a large enoiugh positive correlation the bet will be a good onwe for Alex.

[TheChad08]

Suppose they each throw 100 times with 97 bullseyes for Alex while Davey gets 90 and Ian gets 80. What would the team of Davey and Ian score? Suppose Davey throws first and misses 10 times. Ian lets the pressure get to him and misses every time Davye does.lus 10 additional times. In this case the team scores 90. On the other hand, if Davey is good under pressure and makes all 10 of those throws while missing on 20 other occassions, then the team scores 100. If the correlation coefficent for Davey and Ian is zero, then the team scores 98 and they win. With a negative correlation they do even better, but with a large enoiugh positive correlation the bet will be a good onwe for Alex.

You're using factors that aren't given to us and over examining the question.

What is 1 + 1?

Answer is 2

But what if I add to the question (like you) and turn it into 1 half + 1 half.

Then the answer would be 1.