25% probability that there was a chance of avoiding injury $\quad$ vs. $\quad$ 25% chance of avoiding injury
I ask about merely the math behind the last sentence of footnote 71 quoted below. I quote the legalistic sentences thereinbefore for context, but they may be immaterial.
How does "a 25% probability that there was a chance of avoiding injury" differ from "25% chance of avoiding injury"? Alas, my mind is conflating these 2 chances.
71 Though some academics do insist that damages for loss of a chance could have been awarded in Hotson [v East Berkshire Health Authority [1987] AC 750 268]: see Peel 2003b, 627, and references contained therein. An amazing number of academics argue that the fact that the House of Lords awarded the claimant in Hotson nothing means that Hotson is authority for the proposition that damages for loss of a chance of avoiding physical injury cannot be claimed in negligence: see Porat & Stein 2003, 679; Weir 2004, 214–15. As the majority of the Court of Appeal recognised in Gregg v Scott [2002] EWCA Civ 1471 (at [39], per Latham LJ and at [78], per Mance LJ) this is incorrect – the facts of the case in Hotson were such that the claimant simply could not bring a claim for loss of a chance against the defendants. See, to the same effect, Reece 1996; also Hill 1991. Fleming 1997 puts the point quite well (at 69): ‘[A] 25% probability that there was a chance [of avoiding injury cannot] be conflated into a 25% chance [of avoiding injury].’
N.J. McBride and R. Bagshaw, Tort Law, 6th edn (2018), page 280, footnote 71.
In the last sentence, "Fleming 1997" refers to John G. Fleming, “Preventive Damages” in N J Mullany (ed), Torts in the Nineties (LBC Information Services, Sydney, 1997), pp 56-71.
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My two cents ... for what they're worth.
Imagine someone places 4 identical boxes (A, B, C, D) in front of you. All of the boxes contain 4 identical cups of coffee each (4 × 4 = 16 cups of coffee that are indistinguishable from each other). In 3 of the boxes all the cups are laced with poison (injures, doesn't kill). In one box, only 3 cups contain poison.
You are to select a box and then select a cup of coffee which you must drink. Remember all the boxes look identical and the cups of coffee too are identical i.e. you'll have to guess.
The probability that you have a chance of avoiding injury = 1/4 = 25% (you choose the box that contains 1 cup of coffee that isn't poisoned; call this box C)
Once you've selected this particular box C, you'll have to pick the coffee that isn't poisoned. The probability of avoiding injury = 1/4 = 25%.
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| DNB | (no comment) | May 10, 2023 at 06:55 |
One is a chance of doing something. The other is a chance of having a chance of doing something.
Here are some analogies that may or may not help, depending on your familiarity with the kinds of situations described.
Analogy 1: A tabletop roleplaying game (RPG).
Your character is sneaking down a dungeon corridor. Unfortunately, you neglected to check for traps, and a poison dart flies out of a concealed hole. You roll the dice to avoid it, but you only have a 25% chance of succeeding at dodging it.
This is the “25% chance of avoiding injury”. $P\left(\mathrm{uninjured}\right)=0.25$.
Having learned your lesson, you start checking for traps as you go. Good thing too, as there’s a pitfall trap straight after the dart trap. The game master secretly rolls your search result; there is a 25% probability that you see it. If you don’t, you automatically fall in. If you do see it, you still have to roll your chance of avoiding it.
This is the “25% probability that there was a chance of avoiding injury”. $P\left(\mathrm{uninjured}\right)=0.25x$, where $x$ is your chance of succeeding on that second roll.
Analogy 2: The lottery ticket
You and three friends buy a lottery ticket together. For some reason, you don’t want to split the prize (if any). Instead, you randomly choose which of you gets any winnings. There is a 25% chance that the recipient is you.
This parallels the “25% chance of avoiding injury”. $P\left(\mathrm{recipient}\right)=0.25$.
But of course, you don’t know if there will be any prize coming. There’s a 25% probability that you’re the chosen recipient, but you still have to wait until the lottery is drawn before you know what you could receive. Your ticket probably only has a small chance of winning a prize.
This parallels the “25% probability that there was a chance of avoiding injury”. $P\left(\mathrm{winner}\right)=0.25x$, where $x$ is the chance of the lottery ticket winning something.
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