linusakesson.net

A lateral-thinking puzzle

I came up with a nice little lateral-thinking exercise.

Note that puzzles like these are inherently unfair, because you have to think outside the box, without breaking out of the (ill-defined and completely unrealistic) world model. But usually, once you see the solution, you'll know that it's the correct one.

Beware of spoilers in the comments.

Here we go:

You are a medical doctor working under rather chaotic conditions. Somebody turns up with three unconscious patients and their papers, and promptly disappears again. The patients have lost a lot of blood, and you realise that they're all going to die soon unless they get blood transfusions. From the papers, you can see that one patient has blood type A, another has blood type B, and the last one has blood type O. However, the papers are all mixed up, and you cannot tell which patient has which blood type.

If you do nothing, all three are going to die. Thus, you find it morally justifiable to take any risk that improves the odds of survival of at least one patient.

To clarify the rules of the game:

  1. Patient A will live after receiving a transfusion of Patient O's blood.
  2. Patient B will live after receiving a transfusion of Patient O's blood.
  3. Patient A's blood will instantly kill Patient B and/or Patient O.
  4. Patient B's blood will instantly kill Patient A and/or Patient O.
  5. The only way you can learn something about the blood type of a patient is by giving them a transfusion and seeing whether they die.

For instance, if we refer to the patients as 1, 2 and 3, one strategy might be to perform the transfusions 1→2 and 1→3. With 1/3 probability, Patient 1 has blood type O, and Patients 2 and 3 live. Otherwise, everybody dies. The expected number of people saved is thus 1/3 * 2 + 2/3 * 0 = 2/3.

Your own blood is of type AB and therefore useless.

Find the best strategy.

(If your strategy is expected to save 1 person, then good job, but there's an even better approach!)

Posted Saturday 19-Sep-2015 09:35

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Anonymous
Sat 19-Sep-2015 14:22
(No spoilers) I've got 4/3. Is a better solution possible?
lft
Linus Åkesson
Sun 20-Sep-2015 14:03
The intended solution is indeed 4/3. Well done! I haven't found a better way, but who knows.
Anonymous
Mon 21-Sep-2015 14:12
I'll give the first "hint". For my solution to give an expected value of 4/3, I assumed that a patient doesn't immediately die when donating blood.
Anonymous
Tue 22-Sep-2015 03:00
Draw a small amount of blood from two of them and give the blood to the other. ⅓ of the time you will have drawn from A and B and everyone will die. Otherwise, you will have drawn from O and one other. O will die but the other will still be alive, and you can use O's blood for the other two. This assumes that the small amount mixed into O's bloodstream now won't kill the third one upon receiving more blood from O though.
Vale
Valentino Miazzo
Tue 29-Sep-2015 08:09
I assume Hr doesn't matter and the amount of blood removed and necessary to save one patient doesn't kill immediately the donor.

Solution: Pick two patients randomly. Spill from the receiver a transfusion and keep the blood aside. Transfuse from donor to receiver. 1/3 of times the donor is O and receiver doesn't die immediately. Restore the receiver blood and do another transfusion from the donor to the other patient. You saved 2 people.
2/3 of times the donor is A or B and the receiver dies immediately. Reverse the transfusion.
In 1/2 of times the receiver was O. The corpe contains compatible blood and the reverse transfusion saves the original donor. The blood aside is O and useful to save the other patient.You saved 2 people.
In the other 1/2 of times the receiver is not O and all the patients die.
Therefore: 1/3*2 + 2/3*1/2*2 + 2/3*1/2*0=4/3

By the way, your projects are marvelous and very inspiring. Thanks!
lft
Linus Åkesson
Thu 1-Oct-2015 21:13

Vale wrote:

Solution: Pick two patients randomly. Spill from the receiver a transfusion and keep the blood aside. Transfuse from donor to receiver. 1/3 of times the donor is O and receiver doesn't die immediately. Restore the receiver blood and do another transfusion from the donor to the other patient. You saved 2 people.
2/3 of times the donor is A or B and the receiver dies immediately. Reverse the transfusion.
In 1/2 of times the receiver was O. The corpe contains compatible blood and the reverse transfusion saves the original donor. The blood aside is O and useful to save the other patient.You saved 2 people.
In the other 1/2 of times the receiver is not O and all the patients die.
Therefore: 1/3*2 + 2/3*1/2*2 + 2/3*1/2*0=4/3

That is exactly the solution I had in mind! Well done!

By the way, your projects are marvelous and very inspiring. Thanks!

Thank you!
Anonymous
Mon 30-Nov-2015 20:37
Is the assumption that other blood is not available? That is not spelt out. Why would you not give just O blood to everyone. DOes not matter who has what blood type. If the requirement is to transfuse from one to the other then that should be spelt out in the question. Otherwise my solution will save all three.
lft
Linus Åkesson
Tue 1-Dec-2015 23:00
A fair point. Yes, that assumption should have been spelt out.
Anonymous
Thu 10-Dec-2015 16:49
I reached 5/3, assuming that drawing blood enough to give transfusion to 2 persons does not kill patient immediately.

Define patients 1, 2 and 3. Draw and store blood from each of them. Give blood of patient 1 to 2, and if he survives, also to patient 3. If patient 2 does not survive, give his stored blood to patient 3. If patient 3 does not survive, give his stored blood to patient 1, but if patient 3 survives, give patient 2's blood also to patient 1.

List of scenarios with (1,2,3) marking blood types of each patient:

(o,a,b),(o,b,a): Patients 2 and 3 both survive, 1/3 cases.
(a,b,o),(b,a,o): Patient 1 survives, 1/3 cases.
(a,o,b),(b,o,a): Patients 1 and 3 both survive, 1/3 cases.

1/3*2+1/3*1+1/3*2=5/3
lft
Linus Åkesson
Fri 11-Dec-2015 14:38
That's an interesting approach! But I'm not entirely convinced about the third case. Let's assume (a,o,b). In the beginning, each patient lacks 1 transfusion's worth of blood. After the initial drawing, each patient lacks 3 units of blood. You use one of the stored units of Type A blood, giving it to Patient 2. Then you give one unit of Type O blood to Patient 3. Finally, you give the remaining unit of Type O blood to Patient 1. At this point, I assume that any leftovers are given back to the respective patient, so the two stored units of Type B blood are given to Patient 3. But you only have one unit left of Type A blood, so Patient 1 would still end up lacking one unit of blood. You'd have to transfuse some of the blood from Corpse 2 back to Patient 1. That may or may not work in real life (please don't try it), but it's lateral thinking all right. If it was part of the intended solution, then I think it would have been good to spell it out. Otherwise; did I miss something?
Anonymous
Fri 11-Dec-2015 14:54
Hi Linus,

You're right about giving leftover blood back to donor, and the needed assumption I missed to state is actually based on original rules:
Patient A's blood will instantly kill Patient B and/or Patient O.
Patient B's blood will instantly kill Patient A and/or Patient O.

So, we can assume that virtually a drop of blood is enough to see the potential negative effect, and transfusion can be stopped without consuming any significant amount of blood. We need to draw blood from everyone just to avoid contaminating patient O's blood with even a single drop of wrong type.

-Pets
lft
Linus Åkesson
Fri 11-Dec-2015 16:40
All right, then I follow you. Good work there!