Two toy models can have the same visible correct count while assigned true-label probabilities give different exact bit losses.

highlighted = computed this step

Both toy models show the same correct count

Model A and Model B both have correct count 2/2 in this assigned toy comparison. That count is visible before looking at the loss bits.

Model A correct=2/2,Model B correct=2/2\text{Model A correct}=2/2,\quad \text{Model B correct}=2/2
Same correct count can have different lossExact bit loss changes when assigned true-label probabilities change.same correct count, different lossmodelcorrect counttrue-label pbitssummeanModel A2/2[1/2, 1/2][1, 1]21Model B2/2[1/2, 1/8][1, 3]42same correct count 2/2; different exact losscorrect-count is assigned for this toy comparison; assigned toy probabilities; log base 2 chosen for exact bits; lossarithmetic; NOT training; NOT generalization; NOT a probability calibration claim; NOT calibration proofcorrect-count is assigned for this toy comparison; assigned toy probabilities; logbase 2 chosen for exact bits; loss arithmetic; NOT training; NOT generalization;NOT a probability calibration claim; NOT calibration proof

Model A has two one-bit rows

Model A assigns true-label probabilities 1/2 and 1/2. Those are 1 bit and 1 bit, so the sum is 2 and the mean is 1.

1+1=2,2/2=11 + 1 = 2,\quad 2/2=1
Same correct count can have different lossExact bit loss changes when assigned true-label probabilities change.same correct count, different lossmodelcorrect counttrue-label pbitssummeanModel A2/2[1/2, 1/2][1, 1]21Model B2/2[1/2, 1/8][1, 3]42same correct count 2/2; different exact losscorrect-count is assigned for this toy comparison; assigned toy probabilities; log base 2 chosen for exact bits; lossarithmetic; NOT training; NOT generalization; NOT a probability calibration claim; NOT calibration proofcorrect-count is assigned for this toy comparison; assigned toy probabilities; logbase 2 chosen for exact bits; loss arithmetic; NOT training; NOT generalization;NOT a probability calibration claim; NOT calibration proof

Model B has one larger bit row

Model B assigns true-label probabilities 1/2 and 1/8. Those are 1 bit and 3 bits, so the sum is 4 and the mean is 2.

1+3=4,4/2=21 + 3 = 4,\quad 4/2=2
Same correct count can have different lossExact bit loss changes when assigned true-label probabilities change.same correct count, different lossmodelcorrect counttrue-label pbitssummeanModel A2/2[1/2, 1/2][1, 1]21Model B2/2[1/2, 1/8][1, 3]42same correct count 2/2; different exact losscorrect-count is assigned for this toy comparison; assigned toy probabilities; log base 2 chosen for exact bits; lossarithmetic; NOT training; NOT generalization; NOT a probability calibration claim; NOT calibration proofcorrect-count is assigned for this toy comparison; assigned toy probabilities; logbase 2 chosen for exact bits; loss arithmetic; NOT training; NOT generalization;NOT a probability calibration claim; NOT calibration proof

The count matched, but the loss did not

The visible correct count is the same, but the exact mean losses differ: 1 bit for Model A and 2 bits for Model B. This is assigned loss arithmetic only. No claim about training, generalization, or calibration is made.

121\ne2
Same correct count can have different lossExact bit loss changes when assigned true-label probabilities change.same correct count, different lossmodelcorrect counttrue-label pbitssummeanModel A2/2[1/2, 1/2][1, 1]21Model B2/2[1/2, 1/8][1, 3]42same correct count 2/2; different exact losscorrect-count is assigned for this toy comparison; assigned toy probabilities; log base 2 chosen for exact bits; lossarithmetic; NOT training; NOT generalization; NOT a probability calibration claim; NOT calibration proofcorrect-count is assigned for this toy comparison; assigned toy probabilities; logbase 2 chosen for exact bits; loss arithmetic; NOT training; NOT generalization;NOT a probability calibration claim; NOT calibration proof