[rpls]

Hi
I have this slam and in order to win I need to win two out of 3 finesses
Are the odds 50% or 25%?
One can reason like this: imagine declarer loses the first finesse; then he needs the second finesse to work AND the third too, which means 50*50% equals 25%. Do you agree?

If needed I can post the hands
Regards

[mikeh]

Assuming all else is equal, which it very rarely is, the odds of any one finesse working (or not working) is 50%. Therefore the odds of three finesses all working is 12.5% (50% of 50% of 50%)

The odds that all three fail is 12.5%

That leaves 75% for one working and two failing plus two working and one failing

Therefore the odds of two finesses working and one failing are 37.5%.

Of course that’s independent of the 12.5% for all three working, so the odds are: 50% that you make and 12.5% that you make an overtrick, assuming that winning all three gets you 13 tricks.

In real life it would be unusual to have a hand on which the best line was to take three different finesses. One is often able to find another line, especially if we’re finessing queens. kings are a little more difficult to avoid.

One of the hallmarks of a good declarer is that he takes fewer finesses than do most players and wins most of the ones he takes

[pilowsky]

This is a very old probability question.
If you toss a coin, the odds are 50:50 that it will come up heads.
The next time you toss it, the odds are unchanged.
In Tom Stoppard's famous play "Rosencrantz and Guildenstern are dead", the protagonists work out that they are dead because the coin always comes up heads.
The Rosencrantz and Guildenstern test is unavailable in a single Bridge deal because a sufficiently large number of finesses is unavailable.


In other words, the individual probability of a finesse working is 50:50 no matter the success or failure of a previous attempt.


On the other hand, if you are at the start of the hand and you are trying to devise a line of play that requires the success of three finesses working in a row the probability is 0.5*0.5*0.5 or P=0.125 (12.5%; or 13% if you're an optimist).
It may be better to find another line that depends on something else.

[rpls]

I dont mind looking dumb, so here I go:

Quote

The odds that all three fail is 12.5%

Yes, I agree

Quote

That leaves 75% for one working and two failing plus two working and one failing

Humm, dont get it
Could you please detail it a bit more?

[mikeh]

rpls, on 2021-July-20, 18:38, said:

I dont mind looking dumb, so here I go:
[/size]
Yes, I agree


Humm, dont get it
Could you please detail it a bit more?

There are four possible outcomes


All lose

2 lose

1 loses

All win

All lose is 12.5%

All win is 12.5%

Total of all lose plus all win is 25%

Therefore the total of two win plus two lose has to be 100-25 or 75%

There is a priori no reason two wins is more or less likely than two loses. Therefore the odds of either are one half of 75% or 37.5%

But of course that’s the odds of two winning and one losing. Since you make your contract not only when two win and one loses but also when all win, the odds of making your contract are 37.5% plus 12.5% or 50%

[rpls]

I tried a different approach and got a different result.



I will count the number of favorable cases and divide it by the total numbers of possible cases.



Aa means first finesse works

Af means first finesse fails

Same for B e C finesses



All possible cases are:

Aa Ba Ca

Aa Bf Ca

Aa Ba Cf

Aa Bf Cf

Af Ba Ca

Af Bf Ca

Af Ba Cf

Af Bf Cf

So I have 8 possible cases



All favorable cases are:

Aa Ba Ca

Aa Bf Ca

Aa Ba Cf

Af Ba Ca



So the quotient of favorable cases by the number of possible cases equals to one half –50%

[rpls]

Got it, thanks

mikeh, on 2021-July-20, 18:48, said:

There are four possible outcomes


All lose

2 lose

1 loses

All win

All lose is 12.5%

All win is 12.5%

Total of all lose plus all win is 25%

Therefore the total of two win plus two lose has to be 100-25 or 75%

There is a priori no reason two wins is more or less likely than two loses. Therefore the odds of either are one half of 75% or 37.5%

But of course that's the odds of two winning and one losing. Since you make your contract not only when two win and one loses but also when all win, the odds of making your contract are 37.5% plus 12.5% or 50%

[Cyberyeti]

pilowsky, on 2021-July-20, 18:28, said:

This is a very old probability question.
If you toss a coin, the odds are 50:50 that it will come up heads.
The next time you toss it, the odds are unchanged.
In Tom Stoppard's famous play "Rosencrantz and Guildenstern are dead", the protagonists work out that they are dead because the coin always comes up heads.
The Rosencrantz and Guildenstern test is unavailable in a single Bridge deal because a sufficiently large number of finesses is unavailable.


In other words, the individual probability of a finesse working is 50:50 no matter the success or failure of a previous attempt.


On the other hand, if you are at the start of the hand and you are trying to devise a line of play that requires the success of three finesses working in a row the probability is 0.5*0.5*0.5 or P=0.125 (12.5%; or 13% if you're an optimist).
It may be better to find another line that depends on something else.

This is not always true in a bridge context but close enough. Imagine you are missing 4 of a suit including the K, you take the finesse in that suit which works, but the other hand shows out. Now vacant spaces suggest that since you're limited to 13 cards per hand, the other hand is 13:9 to hold any other specific cards so your odds may go up or down depending on who the other finesses are against.


This is most important when one opponent has preempted. If one opp opened 5♣ finesses against them in clubs and their partner for queens in other suits would be much better than 50:50. Also if you're missing 4 including the Q the odds of a 2:2 break would drop.

[AL78]

rpls, on 2021-July-20, 19:00, said:

I tried a different approach and got a different result.



I will count the number of favorable cases and divide it by the total numbers of possible cases.



Aa means first finesse works

Af means first finesse fails

Same for B e C finesses



All possible cases are:

Aa Ba Ca

Aa Bf Ca

Aa Ba Cf

Aa Bf Cf

Af Ba Ca

Af Bf Ca

Af Ba Cf

Af Bf Cf

So I have 8 possible cases



All favorable cases are:

Aa Ba Ca

Aa Bf Ca

Aa Ba Cf

Af Ba Ca



So the quotient of favorable cases by the number of possible cases equals to one half –50%

You get the same answer as MikeH. He says 37.5% for specifically two finesses out of three working. You then add the additional case of all three working which is 12.5%, 100*(0.5^3) to get 50% for the situation of at least two finesses working.

Alternatively go through the combinations and count the favourable ones:

LLL
WLL
LWL
LLW
WWL*
WLW*
LWW*
WWW*

* denotes a desirable combination, there are four out of eight possible combinations, so 50% probability. There are three out of eight cases where exactly two finesses work, 37.5%.

[pilowsky]

Cyberyeti, on 2021-July-20, 23:36, said:

This is not always true in a bridge context but close enough. Imagine you are missing 4 of a suit including the K, you take the finesse in that suit which works, but the other hand shows out. Now vacant spaces suggest that since you're limited to 13 cards per hand, the other hand is 13:9 to hold any other specific cards so your odds may go up or down depending on who the other finesses are against.


This is most important when one opponent has preempted. If one opp opened 5♣ finesses against them in clubs and their partner for queens in other suits would be much better than 50:50. Also if you're missing 4 including the Q the odds of a 2:2 break would drop.

Agree - any information from the bidding and play (quality of the information notwithstanding) will affect the odds.

[sfi]

rpls, on 2021-July-20, 17:49, said:

Hi
I have this slam and in order to win I need to win two out of 3 finesses
Are the odds 50% or 25%?

As other people have said, it's 50%. But it's easy to see without even calculating the odds.

Winning two finesses means losing one. Since each finesse is a 50% proposition, the odds must be the same as the chance to lose two finesses and win one. Similarly, winning all three finesses must be the same odds as losing all three finesses.

Therefore winning 2 or 3 finesses must be the same chance as losing 2 or 3 finesses. Since there are no other possibilities, it must be a 50% chance.