Kieth1,( post 98) is even more wrong than TruthSeeker (post 95)or Green Destiny(post 9). 8 eggs is a more probable answer than either 10 or 11 to James' question in the OP. Instead of just guessing, take a second look by analysis. I.e. first assume that there are 10 eggs in the original 100 boxes. You found 1 and have now 9 eggs in 99 boxes. What is the chance the next box you open will be empty? Obviously, 90/ 99 as 90 boxes are empty. Then assuming second box opened was empty as James states, then there are then 9 eggs in 98 boxes of which 89 are empty. Etc. So here is the way to compute probability of what happed (Only one egg found in 10 boxes) assuming there are 10 eggs in the 100 boxes. Chance box 2 found to be empty is 90/99. Chance box 3 found to be empty is 89/98. Chance box 4 found to be empty is 88/97. Chance box 5 found to be empty is 87/96. Chance box 6 found to be empty is 86/95. Chance box 7 found to be empty is 85/94. Chance box 8 found to be empty is 84/93. Chance box 9 found to be empty is 83/92. Chance box 10 found to be empty is 82/91. So if there are 10 eggs the probability of what James told happening is the product of these nine different fractions. I get that to be less than 0.408 for the probability that you will find only 1 egg in first 10 boxes you open. To make James’s story typical it should have near a 50/50 chance of being what happened. Thus there must be less than 10 eggs initially in the 100 boxes. Lets change the assumption to be that there were initially 9, not 10, eggs are in the 100 boxes and see how probable it is that you only find one in the first ten tries. Then after finding first egg is found there are 8 left in 99 boxes or 91 empty boxes. So expected probability of box 2 being empty is 91/99 and assuming box 2 was found to be empty, etc. the probability that box 10 was also found to be empty is 83/91. I.e. all the prior numerators are one higher than before and dominators are the same. So probability of Jame’s event happening if there are 9 eggs initially is: 91/99 x 90/98 x 89/97 x 88/96 x 87/95 x 86/94 x 85/93 x 84/92 x 83/91 which I get to be less than 0.453 For 8 egg case, again using fact all the prior numerators are one higher than before and dominators are the same, the probability of Jame’s event happening if there are 8 eggs initially is: 92/99 x 91/98 x 90/97 x 89/96 x 88/95 x 87/94 x 86/93 x 85/92 x 84/91 for which I get 0.50187 Nice problem James. From the OP I get the impression you invented it. Intuition can often be wrong. I felt sure 10 was too many as to avoid finding even one more egg when there are nine left seemed unlikely to me, so I decided to do it correctly. Hope you appreciate my effort - it is very rare that I calculate this much at my age.