[Mr Rick]

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As an example, if I was deciding whether to make a call on the flop to fill a gut shot and while I was making that decision the dealer prematurely turned over one of my 4 outs for that gutshot, it changes the pot odds I would need to make that call. I now have 3 outs on the turn and 4 on the river. Before the fiasco I would have thought I had 4 outs on the turn (4/47) and then 4 outs on the river (4/46) to hit or: 1 - (43/47 * 42/46) = 16.5%. But after the fiasco my turn chances have been reduced to 3 outs (3/46) so by the river my chances are 1 - (43/46 * 42/46) = 14.7%.

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You chances of hitting on the river are better than you suggested because you have a 1 in 24 chance of hitting exactly the exposed card.

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I see your point but I think you made a typo. Assuming one other opponent the chances of hitting that one card would be 1 in 42 instead of the 1 in 46 I had mistakenly assumed. This takes into account the 3 flop cards, 1 turn card, 2 opponent cards, 2 of my own cards, and the 2 prior burn cards.

So the chances on the river of hitting the 4 outs are (3/46 + 1/42) and the total chances of hitting the hand after the fiasco would be: 1 - ((43/46) * (1 - (3/46 + 1/42)) = 14.74%

Rounding to two places the difference was 14.65% vs. 14.74%.

This doesn't significantly change the point of my argument - which is that using this procedure the player yet to act has an advantage - which could be significant depending on what card originally came out on the turn.

That advantage is much greater than the slight change in odds the player yet to act would have if the seen card was shuffled back in for the turn (roughly 0.1% change for the turn and 0.1% for the river).