I’m doing some off table work and realized I have a hold in my math.

If I’m facing 5 opponents and the odds each of them fold independently is 75% then the odds of all of them folding is 23% (75%^5)... I think.

What are the odds that I end up facing 1 opponent, 2 opponents 3 opponents etc?

Thanks.

For one caller:

.75^4 * .25^1 *5 = ~ 39.5%

(Four folders, one caller, and *5 because there are five possible ways you can arrange one caller -- it can be person #1, person #2...)

For two callers:

.75^3 * .25^2 *10 (10 possible arrangements of 2 callers, exercise left to the reader)

...

Thanks but I think I’m doing something wrong.

I get a sum of 111%

Callers
0::::::::::23.7%
1::::::::::39.5%
2::::::::::26.3% .75^3*.25^2*10
3::::::::::13.2% .75^2*.25^3*15
4::::::::::05.8% .75^1*.25^4*20
5::::::::::02.4% .75^0*.25^5*25
Total:::::111.1%

Binomial Distribution
Callers
0 …. 0.75^5
1 …. 5*0.75^4*0 .25
2 …. 10*0.75^3 * 0.25^2
3 …. 10 *0.75^2 * 0.25^3
4 ….. OP you try it
5 …. 0.25^5

This has already been answered above, but it seems like OP may not be fully understanding the "multipliers" associated with each term.

Just echoing what others have said above, you need to multiply each term by the number of combinations of players that conform with the case in question.

So, to take the case of 3 callers, the correct multiplier is, in mathematical notation, C(5,3). This is "5 choose 3", which denotes the number of ways you can "choose" 3 different players of the 5 (without repeats).

The formula for C(x,y) is available on the internet and in many books and its value is given in many calculators, computer programs, etc.

Of course, in the 3 callers case, once you select the 3 callers, the other 2 players "automatically" are folders. So C(5,3) gives the total number of combinations of 3 callers and 2 folders among 5 players.

Perfect. I get it now. At least for functional sake.

I have done it long hand as descirbed in posts 2 and 4 and used the bingomdist function in google sheets alluded to by 5 (with a hint by 4.)

Thank you whosnext, statmanhal, Dwetzel.

Yes, although presumably the odds will change because the situation is different (the first guy has 5 players behind him, second 4 etc. making them all more likely to call than the last).