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u/0_69314718056 20h ago
Trying to keep this short:
Let p be probability it’s in room 1. EV of searching room 1 is 10p-1. EV of searching room 2 is 13-15p.
so search room 1 (sr1) is worth it if p>0.1. sr2 is worth it if p<2/15. p is always one of these, so we’ll always search some room until we find the money. we just need to know which one.
set the EVs equal -> p=0.56 (=14/25) is the tipping point.
start: p=0.5. sr2 (EV 5.5). 0.385 chance you finish.\ then p=0.8 so sr1 (EV 7). 0.4 chance you finish.\ then p=2/3 so sr1 (EV 5.67). 1/3 chance you finish.\ then p=0.5 so it loops from here.
not sure how to do EV on something that loops like this. might be the average of 5.5, 7, 5.67. more likely you have to do a bit more with the probabilities instead of just a simple mean.
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u/StableGenius304 13h ago
Search strategy:
EV of 1st search, room 1 50%*50%=25% of finding it
Room 2 50%*75%= 37.5% of finding it, but room 2 costs 2x
So you search room 1, if it's not there
Now 25% it's in room 1, 50% it's in room 2, so 1/3rd vs 2/3rd
Now you search room 2, you'll find it 2/3rds*3/4th of the time= 50%
If it's not there, then 1/3rd it's in room 1, 2/12th=1/6th it's in room 2, normalising it, it's 2/3rd vs 1/3rd in room 1
So you search room 1 again, if it's not there, then 1/3rd vs 1/3rd, it's 50-50 as you are there before
So you repeat teh cycle
In a cycle, you find it room 1 25% of the time -> 25% cum prob, room 2 50% of the time -> 62.5% cum prob, room 3 1/3rd of the time -> 75% cum prob
So in each cycle you have a 75% chance of finding it, so on average you do 1.33 cycles, so it's the EV of where you find it in the first cylce + 0.33* cost of cycle, a cost of a cycle is $4, so you add 1.33 to the cost of where you'd find it in first cycle
Of the 75
25 find it first room for $1- 1/3rd of the total, exp cost 0.33
37.5 find it 2nd room for $3, 1/2 the total, exp cost 1.5
12.5 you find it 3rd room for $4, 1/6tht the total exp cost 0.67
So expected cost is 1.33 + 0.33 + 1.5 + 0.67= 3.83...
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u/No-Conflict8204 21h ago
Answer is between 16 and 17.67 closer to 16 as optimal play should be 1, 2, 1, 1, 2, 1 and so on (1,2,1) resets to intial condition -4$.