NewtoTown Joined: Aug 22, 2019
• Posts: 75
October 29th, 2019 at 7:38:21 PM permalink
If the house edge is 52% to 48% (less if basic strategy is perfect), assuming one is not counting cards for this example (if it would even matter), how would one calculate the odds of "how often" one will lose:

3 out of five hands
4 out of five hands
5 out of five hands?

Losing five hands in a row, I think is .52 to the fifth power.

But losing 3 in 5 hands (i.e., losing 60% of the time, when the house should only win 52% of the time), how can one calculate "how often" this deviation occurs? I am not referring to losing three in a row, just losing any 3 hands out of 5 hands played.

Same thing regarding losing 4 out of 5 hands (I guess whatever the calculation is for 3 in 5 will be similar).

Thank you. I didn't see anything like this in the calculator area.

(Thank you for the tip to post here versus the forum!)
ChesterDog Joined: Jul 26, 2010
• Posts: 922
Thanks for this post from: October 29th, 2019 at 8:12:02 PM permalink
Quote: NewtoTown

...Same thing regarding losing 4 out of 5 hands (I guess whatever the calculation is for 3 in 5 will be similar)...

You're right that the probability of losing five hands in a row is .52 to the fifth power. There is only one way to do that: lose-lose-lose-lose-lose.

Losing 4 out of 5 hands can be accomplished in any of five ways: win-lose-lose-lose-lose, lose-win-lose-lose-lose, lose-lose-win-lose-lose, lose-lose-lose-win-lose, and lose-lose-lose-lose-win.

The probability of each of these ways is (0.52)4(0.48). So, the probability of losing 4 out of 5 hands is 5(0.52)4(0.48).

You can check the above method by finding the probability of losing 0 out of 5, 1 out of 5, �, 5 out of 5, and add them up to verify that their sum is 1.
DogHand Joined: Sep 24, 2011
• Posts: 403
Thanks for this post from: October 29th, 2019 at 8:32:02 PM permalink
Quote: NewtoTown

If the house edge is 52% to 48% (less if basic strategy is perfect), assuming one is not counting cards for this example (if it would even matter), how would one calculate the odds of "how often" one will lose:

3 out of five hands
4 out of five hands
5 out of five hands?

Losing five hands in a row, I think is .52 to the fifth power.

But losing 3 in 5 hands (i.e., losing 60% of the time, when the house should only win 52% of the time), how can one calculate "how often" this deviation occurs? I am not referring to losing three in a row, just losing any 3 hands out of 5 hands played.

Same thing regarding losing 4 out of 5 hands (I guess whatever the calculation is for 3 in 5 will be similar).

Thank you. I didn't see anything like this in the calculator area.

(Thank you for the tip to post here versus the forum!)

NewtoTown,

First of all, welcome to the forum!

What are the odds of losing 3 of the next 5 hands, if we assume L% = 52%, W% = 48%, and we neglect ties? Well, we could have this arrangement: LWLWL... what are the odds of this?

P(LWLWL) = 0.52*0.48*0.52*0.48*0.52 = (0.52^3)*(0.48^2) = 0.0324.

But LWLWL is NOT the only way to lose 3 of the next 5. If we just look at the two W's, they can be on hands 12, 13, 14, 15, 23, 24 (this is the one we analyzed above), 25, 34, 35, or 45, so a total of 10 permutations. Naturally, each arrangement has the exact same probability, so

P(3L and 2W) = 10*(0.52^3)*(0.48^2) = 0.324, or 32.4%.

I'll leave it to you to calculate the probability of losing 4 of the next 5.

Hope this helps!

Dog Hand
NewtoTown Joined: Aug 22, 2019
• Posts: 75
October 29th, 2019 at 8:59:30 PM permalink
Thank you for the reply, but can I check if I have this correct?

(1) 5(0.52)^4(0.48) = .175 (17.5%) of the time one plays BJ for five hands could result in losing 4 in 5 of those hands? Is 17.5% the correct % for any 4 hands being lost out of any five hands played (excluding doubles/splits/pushes)?

ALSO
(2) If one plays BJ for five hands are the odds of 3 losses out of 5 of those hands? .337(33.7%)

5(0.52)^3(0.48) = .337(33.7%) (with the ^3 representing
win-win-lose-lose-lose,
lose-win-win-lose-lose,
lose-lose-win-win-lose,

or (3) does
lose-lose-lose-win-win also count and then the formula is 5(0.52)^4(0.48) = which is 17.5% (same as losing 4 in 5 hands) and I guess that can't be correct.

I couldn't figure out if there are other ways to arrange wins and losses to arrive at the odds of losing any 3 hands out of 5 hands?
NewtoTown Joined: Aug 22, 2019