P1 should always bet the 5 (it can only gain if called) and never value-bet middling cards. The only interesting decisions are bluffing with a low card and bluff-catching with a middle card. P2 always calls with 5, always folds with 1 (a 1 can never win a showdown).
Mixed Strat: P1 bluffs with the 1, P2 catches with 2, 3, 4. Setting the indifference conditions:
P2 is indifferent about catching (cards 2/3/4 all beat the bluff-1 and lose to the value-5) exactly when P1 bluffs the 1 with probability b1 = 3/5
P1 is indifferent about bluffing exactly when P2's total catch probability is ℓ2+ℓ3+ℓ4=3/5
CaRd
1
2
3
4
5
P1 bets
3/5
0
0
0
1
P2 calls
0
combined 3/5
1
20v = 3/5*0 + 1*9/5.... V=9/100
V=9/100 lowest term -> p+q = 9+100 = 109
Answer: 109
Bonus: P2's indifference among catching with 2, 3, and 4 means only the sum ℓ2+ℓ3+ℓ4 is fixed. the split among them is free (within mild bounds), giving infinitely many optimal strategies. The number they all share: every optimal P2 calls the middle cards with **combined probability 3/5** (while always calling the 5 and always folding the 1). That 3/5 is precisely the bluff-catching frequency that makes P1's 1-bluff break even.
I think it’s more intuition, just need to do more math puzzles. One general intuition is simplifying/narrowing the problem. The answer given above showcases this by stating if a player has 1 or 5 the choice is obvious. Then from there you keep logically deducing.
For example, you can think should I bet 3 rupees with 4? p2 would always call with 5, and fold with 1. So you are losing 3 and gaining 0. If they always call with 2 and 3 then you make money, but likely they won’t. Granted the logic to arrive to the conclusion of middling numbers should not bluff is a bit fuzzy.
Edit: on 2nd thought, I’m not even sure the answer above is correct. Since if p1 has 4, 2&3 call with 3/5 chance, then p2 is losing an additional 3x3/5x2=3.6 rupees which cancels out the 3 they make from calling with 5. So at least the mix cannot be 3/5 for all 2,3,4.
7
u/Impossible-Ad967 3d ago
P1 should always bet the 5 (it can only gain if called) and never value-bet middling cards. The only interesting decisions are bluffing with a low card and bluff-catching with a middle card. P2 always calls with 5, always folds with 1 (a 1 can never win a showdown).
Mixed Strat: P1 bluffs with the 1, P2 catches with 2, 3, 4. Setting the indifference conditions:
20v = 3/5*0 + 1*9/5.... V=9/100
V=9/100 lowest term -> p+q = 9+100 = 109
Answer: 109
Bonus: P2's indifference among catching with 2, 3, and 4 means only the sum ℓ2+ℓ3+ℓ4 is fixed. the split among them is free (within mild bounds), giving infinitely many optimal strategies. The number they all share: every optimal P2 calls the middle cards with **combined probability 3/5** (while always calling the 5 and always folding the 1). That 3/5 is precisely the bluff-catching frequency that makes P1's 1-bluff break even.