1. Viterbi once again! Model generates numbers –312453666641 1:1/6 2:1/6 3:1/6 4:1/6 5:1/6 6:1/6 Fair 1:1/10 2:1/10 3:1/10 4:1/10 5:1/10 6:1/2 Loaded 0.95 0.10 0.05 0.9 The unfair casino: Loaded dice p(6) = 0.5; switch fair to load:0.05; switch load to fair: 0.1

2. Model decoding (Viterby) Example: 566. What was the series of dice used to generate this output? 1:1/6 2:1/6 3:1/6 4:1/6 5:1/6 6:1/6 Fair 1:1/10 2:1/10 3:1/10 4:1/10 5:1/10 6:1/2 Loaded 0.95 0.10 0.05 0.9 FFF = 0.167*0.95*0.167*0.95*0.167 = 0.0042 FFL = 0.167*0.95*0.167*0.05*0.5 = 0.00666 FLF = 0.167*0.05*0.5*0.1*0.167 = 0.000070 FLL = 0.167*0.05*0.5*0.9*0.5 = 0.0019 LFF LFL LLF LLL

3. Model decoding (Viterby) Example: I now add one more say 5! After my first 3 throws I have ended up having either a loaded or a fair dice in my hand The most likely “path” to a loaded dice is FLL = 0.167*0.05*0.5*0.9*0.5 = 0.0019 The most likely path to a fair dice is FFF = 0.167*0.95*0.167*0.95*0.167 = 0.0042 Since the Markov model has no memory, the most likely path to having a fair dice after the 4th through is 1:1/6 2:1/6 3:1/6 4:1/6 5:1/6 6:1/6 Fair 1:1/10 2:1/10 3:1/10 4:1/10 5:1/10 6:1/2 Loaded 0.95 0.10 0.05 0.9

4. Model decoding (Viterby) 1:-0.78 2:-0.78 3:-0.78 4:-0.78 5:-0.78 6:-0-78 Fair 1:-1 2:-1 3:-1 4:-1 5:-1 6:-0.3 Loaded -0.02 -1.3 -0.05 Log model 1245666 F -0.78-1.58 L Null-3.08-3.88

5. Model decoding (Viterby) 1:-0.78 2:-0.78 3:-0.78 4:-0.78 5:-0.78 6:-0-78 Fair 1:-1 2:-1 3:-1 4:-1 5:-1 6:-0.3 Loaded -0.02 -1.3 -0.05 Log model 1245666 F -0.78-1.58 L Null-3.08-3.88 Identify what series of dice was used to generate this output?

6. Model decoding (Viterby) 1:-0.78 2:-0.78 3:-0.78 4:-0.78 5:-0.78 6:-0-78 Fair 1:-1 2:-1 3:-1 4:-1 5:-1 6:-0.3 Loaded -0.02 -1.3 -0.05 Log model 1245666 F -0.78-1.58-2.38-3.18-3.98-4.78-5.58 L Null-3.08-3.88-4.68-4.78-5.13-5.48 Series of dice is FFFFLLL