Gambles in Your Life Andre Dabrowski Mathematics and Statistics

Pick the Prize!

One Chance in Three Prob[Winner] =#(winning choices) / #(all possible choices) = / #( ) = 1 / 3

Prob[event]

Gambles in your Life P[winner] =#(winning choices)/#(all choices) =1/#(all choices)

“Lottery 216” 1.Everyone has a sample ticket. 2.Every ticket has 3 numbers, each number chosen from {1,2,3,4,5,6}. E.G. 136 or 524 or 652, but not Is 222 more or less likely to win than 452? 4.What is your chance of winning?

Is 222 more or less likely to win than 452? Put one marker in the box for each ticket. Mix them up. Draw one out. All tickets have the same chance at winning! So 222 has the same chance as 452 of winning.

What is your chance of winning? P[winner]=1/#(all possible choices) #(all possible choices) = #(choices for first digit) X #(choices for second digit) X #(choices for third digit) = 6 X 6 X 6 = 216 P[winner]=1/216.

Lotto 6/49 P[win by matching all 6 numbers] =1/#(all possible choices) #(all possible choices) = 49 x 48 x 47 x 46 x 45 x 44 / in 13,983,816 chances!

Which is more likely? Matching all 6 numbers in a 6/49 lottery Being struck by lightning sometime during the year. 1/ 13,983,816About 1/ 1,000,000

UO Xmas Lottery! 1.Everyone has a ticket. 2.We will draw from a box to choose the winner. 3.P[winning]=1/216.

Now that we know HOW to calculate probabilities, we can look for interesting ones to compute.

The Birthday Problem There are 365 days in the year. The chance that any one person shares your birthday is 1/365. Pretty small! What is the chance at least two people in this room share birthdays?

P[no matching birthdays] P[no match for 2 people] = = 365 X 364 / 365 X 365 = 364/365. #(ways of choosing 2 without matching) #(ways of choosing 2 birthdays)

P[no match in 5 people]= 365 X 364 X 363 X 362 X X 365 X 365 X 365 X 365 =.97 approximately = #(ways of choosing 5 without matching) #(ways of choosing 5 birthdays) =

P[no match in 25 people]= #(ways of choosing 25 without matching) #(ways of choosing 25 birthdays) = 365 X 364 X … X 342 X X 365 X … X 365 X 365 = = 0.43 approximately There is about a 57% chance a class of 25 will have at least two sharing a birthday.

P[birthday match in k people]

Gambles in your Life Small probabilities can become large if we do many simultaneous experiments. Coincidences are not really coincidences in large groups. Yell “Hey Pete” in a crowd and someone will answer! How reliable are complex systems? A system can survive one component failing, but what is the chance two fail at once?

UO Xmas Birthday Giveaway! First two birthdays to match win!

We know How to compute probabilities for simple games How do we compute probabilities for more complicated problems?

Simple --8 Heads in a Row in 8 tosses Chance of 8 heads in a row = ½  ½  …  ½  ½  ½ =1/256 Pretty small!

Harder -- 8 Heads in a Row somewhere in 100 tosses Toss a fair coin 100 times. What is the chance of at least 8 heads in a row somewhere in the string of 100? HHTHTTHTTTHTHHTHTHTH TTTHTHTHTHHTHHHHHHHH TTHHHTHTHTHHHTTHTHTH HHTHTHHTHTHTTTHTTTTHT THTTHTHTHHHTHTHTHTHTT

Monte Carlo Methods “Toss” a coin 100 times Find the longest string of H’s Repeat this 100,000 times ,000,000 tosses! P[at least 8 H’s in a row] is approximately #( at least 8 H’s in a row)/100,000

Monte Carlo Methods “Toss” a coin 100 times using a computer Find the longest string of H’s Repeat this 100,000 times ,000,000 tosses! using a computer P[at least 8 H’s in a row] is approximately #(at least 8 H’s in a row)/100,000

Monte Carlo Methods libname here 'h:/XmasLecture'; libname there 'c:/tmp'; %macro dupit; %do ii=1 %to 100; x_&ii=(ranuni(0)<.5); %end; %mend; data there.runs; do i=1 to by 1; output; end; data there.runs; set there.runs; %dupit; run; %macro runs; %do ii=2 %to 100; %let iii=%eval(&ii-1); a=0+run_&iii; b=0+x_&ii; run_&ii=a*(a>0)*(b=1)+(b=1); runmax=max(runmax,run_&ii); %end; %mend; data there.runs; set there.runs; run_1=0+(x_1=1); runmax=0; %runs; data here.runs; set there.runs; keep runmax; run; data there.runs; run; proc gchart data=here.runs; axis1 value=(height=10); vbar runmax / midpoints = 1 to 15 by 1 type=percent caxis=axis1; run; proc freq data=here.runs; table runmax / nofreq nocumulative; run; quit;

100,000 Simulations

The FREQ Procedure runmax Frequency Percent ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ > All P[ run of 8 H or more ] =.17 approx., = 1/6 >> 1/256. The chance of 4 or more heads in a row is about 97%. We can use this to pick out which sequences on the sheet are unlikely to really have been generated at random.

Gambles in your Life “good” days and “bad” days. Long lineups for no reason. Design of bridges, power plants. Weather prediction. Biological evolution.

Thanks for coming!