another probability question

[DrVanNostrin]

Instead of having overtime in a basketball game every player on both teams shoots a free throw. The probability of any player making their free throw is p. Whichever team hits shoots the best % wins. Team A has 5 players, team B has 6. For what values of p will team A have an advantage?

[mykey1961]

Off the top of my head I'd guess any p less than 0.5 but greater than 0.

If the shots were interleaved, where team B shoots once, then A, etc.

After each have shot 5 times, on average they should be still be tied.

So it all comes down to the 6th shooter for team B.

If it's made less than 50% of the time, Team A has an advantage. With the exception of p = 0%. Then it's a lock tie.

No advantage at p = 1, p = 0.5, and p = 0.

Advantage for B when p < 1, p > 0.5
Advantage for A when p < 0.5, p > 0

I'm guessing this is wrong because it doesn't make for a good puzzle.

[DrVanNostrin]

Consider a few extreme examples:
p = 0.9999999, who would have an advantage? Team A
p = 0.0000001, who would have an advantage? Team B

[bigpooch]

Obviously, there is symmetry about p=1/2, so no team has an
advantage at p=1/2, p=0 and p=1 and it is sufficient to look
at values of p between 0 and 1/2. [p=1/2 reminds me of the
n coins and (n+1) coins question posted earilier]

I would guess (and I hate guessing) that the critical values
between 0 and 1/2 are 1/5, 2/5, 1/6 and 1/3 and that the
advantage swings from B (at very low values) to A just above
1/6 to just below 1/5 and then back to B between 1/5 and
1/3, etc.

I would like to be more rigorous and use an epsilon argument
around these values, but maybe somebody else can find a
more accessible and easy solution.

If I am right, the more general question when Team A has
m players and Team B has n ( > m ) players is also solved.

[PairTheBoard]

The extreme example I would look at to see the principle at work would be Team A with 1 player and Team B with 1 million players. If p is say .25 team A loses close to 75% of the time. If p is .75 team A wins about 75% of the time.

PairTheBoard

[mykey1961]

Assume q = 1 - p

PctA is the probability that Team A wins for a given p
PctB is the probability that Team B wins for a given p

<font class="small">Code:</font><hr /><pre>
PctA := p*p*p*p*p * 6*p*p*p*p*p*q+ // 5 vs 5
p*p*p*p*p * 15*p*p*p*p*q*q+ // 5 vs 4
p*p*p*p*p * 20*p*p*p*q*q*q+ // 5 vs 3
p*p*p*p*p * 15*p*p*q*q*q*q+ // 5 vs 2
p*p*p*p*p * 6*p*q*q*q*q*q+ // 5 vs 1
p*p*p*p*p * q*q*q*q*q*q+ // 5 vs 0
5*p*p*p*p*q * 15*p*p*p*p*q*q+ // 4 vs 4
5*p*p*p*p*q * 20*p*p*p*q*q*q+ // 4 vs 3
5*p*p*p*p*q * 15*p*p*q*q*q*q+ // 4 vs 2
5*p*p*p*p*q * 6*p*q*q*q*q*q+ // 4 vs 1
5*p*p*p*p*q * q*q*q*q*q*q+ // 4 vs 0
10*p*p*p*q*q * 20*p*p*p*q*q*q+ // 3 vs 3
10*p*p*p*q*q * 15*p*p*q*q*q*q+ // 3 vs 2
10*p*p*p*q*q * 6*p*q*q*q*q*q+ // 3 vs 1
10*p*p*p*q*q * q*q*q*q*q*q+ // 3 vs 0
10*p*p*q*q*q * 15*p*p*q*q*q*q+ // 2 vs 2
10*p*p*q*q*q * 6*p*q*q*q*q*q+ // 2 vs 1
10*p*p*q*q*q * q*q*q*q*q*q+ // 2 vs 0
5*p*q*q*q*q * 6*p*q*q*q*q*q+ // 1 vs 1
5*p*q*q*q*q * q*q*q*q*q*q; // 1 vs 0

PctB := 5*p*p*p*p*q * p*p*p*p*p*p+ // 4 vs 6
5*p*p*p*p*q * 6*p*p*p*p*p*q+ // 4 vs 5
10*p*p*p*q*q * p*p*p*p*p*p+ // 3 vs 6
10*p*p*p*q*q * 6*p*p*p*p*p*q+ // 3 vs 5
10*p*p*p*q*q * 15*p*p*p*p*q*q+ // 3 vs 4
10*p*p*q*q*q * p*p*p*p*p*p+ // 2 vs 6
10*p*p*q*q*q * 6*p*p*p*p*p*q+ // 2 vs 5
10*p*p*q*q*q * 15*p*p*p*p*q*q+ // 2 vs 4
10*p*p*q*q*q * 20*p*p*p*q*q*q+ // 2 vs 3
5*p*q*q*q*q * p*p*p*p*p*p+ // 1 vs 6
5*p*q*q*q*q * 6*p*p*p*p*p*q+ // 1 vs 5
5*p*q*q*q*q * 15*p*p*p*p*q*q+ // 1 vs 4
5*p*q*q*q*q * 20*p*p*p*q*q*q+ // 1 vs 3
5*p*q*q*q*q * 15*p*p*q*q*q*q; // 1 vs 2

Ties := p*p*p*p*p * p*p*p*p*p*p+ // 5 vs 6
q*q*q*q*q * q*q*q*q*q*q; // 0 vs 0
</pre><hr />


I guess a person could assume PctA = PctB and then solve for p, but not me.

Using a binary search for values of p, we find the value for p where PctA = PctB is approximately 0.542310724269960960

PctA is greater than PctB for values of p where

0 &lt; p &lt; 0.542310724269960960

[pzhon]

Team A will have the advantage when p is p_0 to 1/2, and 1-p_0 to 1, where p_0 is 0.039794942508919688556..., a root of an 8th degree polynomial.

Mathematica code:
<ul type="square">
f[p_, m_, n_] := Sum[
If[i/m &lt; j/n, -1, If[i/m &gt; j/n , 1, 0]]
Binomial[m, i] Binomial[n, j]p^(i + j)
(1 - p)^(m + n - i - j),
{i, 0, m}, {j, 0, n}]

Plot[f[p, 5, 6], {p, 0, 1}]

Solve[f[p, 5, 6] == 0, p][/list]

[mykey1961]

I missed the other point where PctA = PctB at p = 0.960204909275475293

so that makes the range where Team A has an advantage at

0 &lt; p &lt; 0.542310724269960960
and
0.960204909275475293 &lt; p &lt; 1

[pzhon]

[ QUOTE ]
Team A has an advantage at

0 &lt; p &lt; 0.542310724269960960
and
0.960204909275475293 &lt; p &lt; 1

[/ QUOTE ]
[img]/images/graemlins/diamond.gif[/img] The range has to be the reflection about 50% of the complement. If p gives side A an advantage, then 1-p gives side B the same advantage by reversing the result of each throw, but you concluded that probabilities around 50% give A an advantage as do both very high and very low probabilities.
[img]/images/graemlins/diamond.gif[/img] You left out a few terms. There should have been 6*7=42, as there are 6 scores in {0,1,2,3,4,5} and 7 in {0,...,6}. You left out most scores of the form 0 to n. You'll find that your probabilities for win, lose, and tie don't add up to 1.

The surprising thing to me is that your calculated start of the upper interval disagrees with mine, 0.96020505749108031144, only by about 1/10 million, even though the disagreements are much larger for lower values. The reason is that the omitted terms are all very small when p is large, since they contain (1-p)^5, but the derivative of the advantage is relatively large.

[mykey1961]

The ommitted terms made all the difference.

I hate when my brain fades like that.
While it seems to be happening more and more lately, I haven't yet reached the point where I expect it to happen.