Answer to the birthday riddle from last Tuesday:
If there are 23 people in the room there is a greater than 50% chance that two of them have the same birthday (as Squeak hinted at last year “If 23 people walk into a room…”).
For those interested, here’s how it can be solved in Excel:
People Odds Against Odds in Favor
1 =365/365 =1-B2
2 =B2*((365-(A3-1))/365) =1-B3
3 =B3*((365-(A4-1))/365) =1-B4
4 =B4*((365-(A5-1))/365) =1-B5
5 =B5*((365-(A6-1))/365) =1-B6
6 =B6*((365-(A7-1))/365) =1-B7
7 =B7*((365-(A8-1))/365) =1-B8
8 =B8*((365-(A9-1))/365) =1-B9
9 =B9*((365-(A10-1))/365) =1-B10
10 =B10*((365-(A11-1))/365) =1-B11
11 =B11*((365-(A12-1))/365) =1-B12
12 =B12*((365-(A13-1))/365) =1-B13
13 =B13*((365-(A14-1))/365) =1-B14
14 =B14*((365-(A15-1))/365) =1-B15
15 =B15*((365-(A16-1))/365) =1-B16
16 =B16*((365-(A17-1))/365) =1-B17
17 =B17*((365-(A18-1))/365) =1-B18
18 =B18*((365-(A19-1))/365) =1-B19
19 =B19*((365-(A20-1))/365) =1-B20
20 =B20*((365-(A21-1))/365) =1-B21
21 =B21*((365-(A22-1))/365) =1-B22
22 =B22*((365-(A23-1))/365) =1-B23
23 =B23*((365-(A24-1))/365) =1-B24
24 =B24*((365-(A25-1))/365) =1-B25
25 =B25*((365-(A26-1))/365) =1-B26
26 =B26*((365-(A27-1))/365) =1-B27
27 =B27*((365-(A28-1))/365) =1-B28
28 =B28*((365-(A29-1))/365) =1-B29
29 =B29*((365-(A30-1))/365) =1-B30
30 =B30*((365-(A31-1))/365) =1-B31Once there are 23 people the odds are greater than 50%. Interestingly, if there are 57 people the odss are greater than 99%.