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.

1≠21\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