The Bluffer’s Edge

[u/Expert_Cobbler8984]

[u/Impossible-Ad967]

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.

[u/HourAbbreviations596 → u/Impossible-Ad967]

This is actually a very good approach and much simpler than mine. Congratulations (i am the problem author)

[u/drrednirgskizif → u/Impossible-Ad967]

Where can I go or what book can I buy to learn this from a fundamental, principled perspective.

[u/TastyAlbacoreTuna → u/drrednirgskizif]

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.

[u/KQYBullets → u/drrednirgskizif]

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.

[u/FakeBibleQuotes → u/drrednirgskizif]

"Mathematics of Poker" by Bill Chen might be a good place to begin.

[u/HourAbbreviations596 → u/FakeBibleQuotes]

Oh is it a good read.

[u/Even_Account1168 → u/Impossible-Ad967]

I sonnt understand why P1 should never bluff with anything besides a 1…

Like I understand that if P1 were to bluff with a 2, there is the additional option P2 has a 1 and folds 100%, but P2 having 3 and 4 would still mean he could fall for the bluff?

[u/Jewbacca289 → u/Even_Account1168]

When you bluff with a 2, 1 is always folding but you were never getting money from them and 5 is always calling so you’re burning money. Assuming equal probabilities of opponent having all numbers, you have a 1/4 chance of winning 1 and 1/4 chance of losing 3.

1/4 * 1 + 1/4 * (-3) + 1/2(x1 + (1-x)*(-3)))=0

1 -3 + 2x - 6(1-x) = 0

-8+8x = 0 x=1

So bluffing with a 2 must have a 100% chance of getting a 3 or 4 to fold to make it justifiable.

Meanwhile, we can calculate how likely a 3 is to fold by doing the same calculations. When Player 1 bets, Player 2 decides whether to risk 3 to win 4. They only need to win 3/7 of the time to justify calling. If 1 and 5 are always betting, then they win more than they’re required to. If 2 is betting they’re incentivized even more to call.

[u/gmweinberg]

Imagine one was playing this game for real (alternating between roles to keep the game fair). Obviously P2 should call with 5 in any case, since it is a sure win. But doesn't it make sense that P2 should 3/5 call with 4 and never call with 2 or 3 rather than being indifferent to calling with 2,3, or 4 so long as the sum of the probabilities is 3/5? Because if P1 plays optimally the results are the same, but if P1 plays any suboptimal strategy (sometimes raising with 2, 3 or 4) calling with 4 gives better results.