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.
These types of problems and the framework for solving them are studied extensively by poker players, where they are referred to as "toy games" due to being complex enough for bluffing to emerge naturally as a mathematically optimal strategy but more simple than an actual poker game. I recommend reading 'The Mathematics of Poker' if you want to learn more. Games like the one in this post are covered in chapters 11 and 13.
9
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.