Hello! Let's say we 3-Bet bluff a player, there are 4 players behind who are yet to act. Every single of them will 4-Bet us with a 2% Range. Whats the probability that we will get 4-Bet? Is it simply 4*0.2, so 8%?
Hello! Let's say we 3-Bet bluff a player, there are 4 players behind who are yet to act. Every single of them will 4-Bet us with a 2% Range. Whats the probability that we will get 4-Bet? Is it simply 4*0.2, so 8%?
No. The easiest way to calculate it is 1 minus the probability they don't.
One way to see that NewOldGuy's formula is correct (or to remember it in the future) is to consider the question if each of the four players behind you has a 30% probability of 4-betting.
Of course, 4*30% = 120% which cannot be the correct overall probability of at least one player behind you 4-betting since probabilities cannot exceed 100%.
NewOldGuy's formula in this case gives the correct overall probability as:
One way to see that NewOldGuy's formula is correct (or to remember it in the future) is to consider the question if each of the four players behind you has a 30% probability of 4-betting.
I hate to correct such a helpful, nice guy, but I think it is more than a nitpick to point out that your first sentence should be "one way to see that your formula is not correct"
one more question about this: assuming we now know that the % of players behind 4-betting is, 5%. I we now assume that villain folds to our 3-Bet 50% of the time(whenever none of the players behind 4bets). Our total fold equity in that spot is 50% minus the 5%, so 45% total. Is that right?
one more question about this: assuming we now know that the % of players behind 4-betting is, 5%. I we now assume that villain folds to our 3-Bet 50% of the time(whenever none of the players behind 4bets). Our total fold equity in that spot is 50% minus the 5%, so 45% total. Is that right?
I mean assuming this, looks just like the first mistake I've made by thinking I could simply add possibitlies, which leads me to another question:
What if there are 4 players behind and they 4-bet 2%,3%,4%,5% ? How would I calculate this?
You can use the same method. All we're doing by raising a number (x) by a power (n) is multiplying (x), (n) times.
x^n when x= .98 and n =4 is just .98*.98*.98*.98
.98^4= .98 * .98 *.98 *.98
In the original problem we first needed to find the number of times each individual player does not 4 bet.
1-.02= 98%.
Then we multiplied them together.
.98^4
=98*98*98*98
=.92
We do this to find the combined probability that all the players do not 4 bet. Anytime events are independent of one another you can multiply those probabilities to get the combined probability.
Then it follows that if no one 4 bets ~92% of the time, at least one will 4bet the rest of the time which is ~8%
1-.92
=.08
The whole thing together is
1-(1-.02)^4
=1-.98^4
=1-.98 * .98 *.98 *.98
=.077
The new problem goes the same way. But instead of the (.98*.98*.98*.98) part
its now (.98*.97*.96*.95)
Now its 1-[(1-.02)*(1-.03)*(1-.04)*(1-.05)]
=1- (.98*.97*.96*.95)
=1- .866
=.134
EDIT: You might want to check out the sticky with poker prob primer. Whosnext and heehaw do a great job of breaking all this stuff down for the noobz.
Last edited by citamgine; 02-16-2018 at 09:42 AM.
Reason: credit where credit is due
You can use the same method. All we're doing by raising a number (x) by a power (n) is multiplying (x), (n) times.
x^n when x= .98 and n =4 is just .98*.98*.98*.98
.98^4= .98 * .98 *.98 *.98
In the original problem we first needed to find the number of times each individual player does not 4 bet.
1-.02= 98%.
Then we multiplied them together.
.98^4
=98*98*98*98
=.92
We do this to find the combined probability that all the players do not 4 bet. Anytime events are independent of one another you can multiply those probabilities to get the combined probability.
Then it follows that if no one 4 bets ~92% of the time, at least one will 4bet the rest of the time which is ~8%
1-.92
=.08
The whole thing together is
1-(1-.02)^4
=1-.98^4
=1-.98 * .98 *.98 *.98
=.077
The new problem goes the same way. But instead of the (.98*.98*.98*.98) part
its now (.98*.97*.96*.95)
Now its 1-[(1-.02)*(1-.03)*(1-.04)*(1-.05)]
=1- (.98*.97*.96*.95)
=1- .866
=.134
EDIT: You might want to check out the sticky with poker prob primer. Whosnext and heehaw do a great job of breaking all this stuff down for the noobz.
sorry, havent looked into here for a while. Thanks a lot!
