Revisiting Prior Odds

Remember our earlier examples of deciding whether we have a six or twenty sided die?

When I glibly divide 1/6 by 1/20 I assume that I am equally likely to choose a six sided or a twenty sided die. Let's instead assume a case where 101 dice are placed in a bag. 1 is a six sided die. The remainder are 20 sided. The odds of pulling a six sided die out of the bag is 1 in 100:

P of six sided die: 1/101

P of 20 sided die: 100/101

odds = (1/101)/(100/101) = 1 in 100.

We pull one die out of the bag and start rolling it to see what results we get and to determine how many sides it has. To be confident that we've pulled a six sided die out of our bag, we'd want the result of our rolls to overcome our 100 to one odds of being wrong. For one roll where the number is six or less, our calculation would be:

1/100 * 1/(6/20) = 1/100 * 3.333 = 0.03333

Which is a roughly 1 in 30 chance or 3.2%

For two rolls:

1/100 * 20/6 * 20/6 = 1/100 * 11.111 = 0.11111

About a 1 in 10 chance or 9.9%.

For three rolls:

1/100 * 20/6 * 20/6 * 20/6 = 1/100 * 37.037 = 0.37037

slightly less than 1 in 3 or 27%.

For four rolls:

1/100 * 20/6 * 20/6 * 20/6 * 20/6 = 1/100*123.456 = 1.23456

About 1.23 to one or about 55%.

For five rolls:

1/100 * (20/6)*(20/6)*(20/6)*(20/6)*(20/6) = 4.11

About 4 to 1 or 80%.

So in four rolls with numbers of six or less we could be 80% certain that we had a six sided die. If we rolled even one seven, we would be 0% certain that we had a six sided die.

If there were equal numbers of six and twenty sided dies in the bag, our numbers would be different. The prior probability would be 1/2, the prior odds would be 1 in 1. Therefore our result above would be 411 to one, with a probability of 99.75%.

No matter how compelling our scientific data, the particulars of our situation change the resulting probability.