Some Coin Stats

Rare time I get to use my stats degree for something useful.

It takes 1/p flips for a head, on avg (e.g 2 flips with a 50% coin and 4 flips with a 25% coin)

The expected number of flips to get 10 heads (let’s call that E[10]) in a row is the same as the expected number for 9 in a row + 1, multiplied by how many times it would take to make that next flip successful (which is 1/p): E[10]=(E[9]+1)/p.

We can do some fancy math (not shown) from this fact to get the expected flips to get 10 in a row: E[10]=(1-p^10)/((1-p)p^n)

If the last flip has a p/10 probability instead of the normal p, this whole thing just multiplies by ten. This is because it would take, on avg, 10 tries to be “successful” on the last flip, given it would normally have won.

So this gives these estimates in num expected flips (and work time assuming 40hrs/wk): 60%: 4,110 flips (1.14 hours) 55%: 8,751 flips (2.43 hours) 50%: 20,460 flips (5.68 hours) 40%: 158,929 flips (1.10 work weeks) 30%: 2,419,284 flips (16.80 work weeks) 25%: 13,981,000 flips (1.87 work years)

This means unfair flips probably until late August 2027, which is lame because then I won’t have anything good to watch over the holidays in 2028.