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Jun 23, 2011

Flipping Results from Discovery 2008 Talk

I didn't get any guesses on the odds of flipping various items from my Flipping Odds post. Instead, the commenters went directly to the question of interest, which is how does the chance for one flip affect the chance for two flips being the same? That turns out to be a straightforward formula. If we call the odds of landing up p, then the odds of landing the same on two flips turn out to be p*p + (1-p)*(1-p). That's a quadratic with a minimum of 0.5, so it's always better to call "same" if you're in the two-flip situation, such as at the start of an Ultimate Frisbee game.

Here's the JMP-made graph of the results that I showed in my Discovery 2008 session.

Notice that though the two-flip odds aren't 50%, they do get closer to 50% than each single-flip case. In the disc case, the odds go from 57% to 51%, so the double flip seems like a reasonable thing to do, even if it's not perfectly even odds.

For completeness, here are the numbers for landing in the "up" position. For the cat and the coin, I show theoretical values. The disc results represent 400 flips, and the others represent 100 flips each.

ObjectUp Chance
Disc57%
Lid28%
Bowl32%
Funnel12%

I'm still not sure what the exact probabilities are, but JMP tells me they're not likely to be 50%. For instance, the probability of getting my disc results from a fair disc are less than 1%.

Community Member

Xan Gregg wrote:

I made a copy of the graphics script from the overlay function plot and adding min() and max() arguments to the Y Function().

Pen Color( "dark gray" );

Pen Size( 3 );

Y Function(

::Up Chance_ ^ 2 + (1 - ::Up Chance_) ^ 2,

::Up Chance_,

Min( 0.5 ),

Max( 1 )

);

Community Member

Daniel wrote:

Can you remind us how to "cut" the connecting line so that it ends with the points? I forgot to write that down at your talk!

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